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Class 8 Maths Chapter 5 Number Play Solutions
Ganita Prakash Class 8 Chapter 5 Solutions
Class 8 Maths Ganita Prakash Chapter 5 Solutions Number Play
Intext (NCERT Page 112-114)
Question 1.
(i) “Can I write every natural number as a sum of consecutive numbers?”
Solution:
No.
e.g. Powers of 2 (like 2, 4, 8, 16, ) cannot be expressed as the sum of two or more consecutive numbers.
So, every natural number cannot be written as a sum of consecutive numbers.
(ii) “Which numbers can I write as the sum of consecutive numbers in more than one way?”
Solution:
Numbers that have multiple odd factors (other than 1) can be written as the sum of consecutive numbers in more than one way.
e.g. 15 can be written as 1 + 2 + 3 + 4 + 5 or 7 + 8.
(iii) “Ohh, I know all odd numbers can be written as a sum of two consecutive numbers. Can we write all even numbers as a sum of consecutive numbers?”
Solution:
No.
e.g. Even numbers that are powers of 2 (like, 2,4, 8, 16,……….) cannot be written as a sum of consecutive numbers.
(iv) “Can I write 0 as a sum of consecutive numbers? Maybe I should use negative numbers.”
Explore these questions and any others that may occur to you. Discuss them with the class.
Solution:
Yes.
e.g. If we use negative numbers then 0 can be written as a sum of consecutive numbers.
e.g. -1 + 0 + 1 = 0 or -2 + (-1) + 0 + 1 + 2 = 0
Question 2.
Take any 4 consecutive numbers. For example, 3, 4, 5, and 6. Place ‘+’ and signs in between the numbers. How many different possibilities exist? Write all of them.
3 + 4 – 5 + 6
3 – 4 – 5 – 6
Eight such expressions are possible. You can use the diagram below to systematically list all the possibilities.
Evaluate each expression and write the result next to it. Do you notice anything interesting?
Solution:
Now, we have the following expressions :
3 + 4 + 5 + 6 = 18
3 + 4 + 5-6 = 12 – 6 = 6
3+4-5 + 6 = 13 – 5 = 8
3 + 4 – 5 – 6 = 7 – 11 = -4
3 – 4 + 5 + 6 = 14 – 4 = 10
3 – 4 + 5 – 6 = 8 – 10 = -2
3 – 4 – 5 + 6 = 9 – 9 = 0
3 – 4 – 5 – 6 = 3 – 15 = -12
The interesting observation is that by systematically placing’+’ and signs between four consecutive numbers, we can achieve a variety of results, including, positive, negative, even and zero.
Also, we notice that each expression gives even numbers.
Question 3.
Now, take four other consecutive numbers. Place the ‘+’ and signs as you have done before. Find out the results of each expression. What do you observe?
Solution:
Let us take four consecutive numbers 1, 2, 3 and 4.
We have the following expressions :
1 + 2 + 3 + 4 = 10
1 + 2 + 3 – 4 = 6 – 4 = 2
1 + 2 – 3 + 4 = 7 – 3 = 4
1 + 2 – 3 – 4 = 3 – 7 = – 4
1 – 2 + 3 + 4 = 8 – 2 = 6
1 – 2 + 3 – 4 = 4 – 6 = -2
1 – 2 – 3 + 4 = 5 – 5 = 0
1 – 2 – 3 – 4 = 1 – 9 = -8
The result of each each expression is an even number.
Question 4.
Replace any negative sign in the expression a + to – c- d with a positive sign and find the difference between the two numbers.
Solution:
Given, expression is a + b – c – d
Replacing a negative sign with a positive sign, we get a + b + c – d
Required difference = a + b – c – d – (a + b + c – d) = a + b – c – d – a – b – c + d = – 2c,
which is an even number.
Breaking Even (NCERT Page 115-116)
Question 1.
We know how to identify even numbers. Without computing them, find out which of the following arithmetic expressions are even.
(i) 43 + 37
(ii) 672 – 348
(iii) 4 × 347 × 3
(iv) 708 – 477
(v) 809 + 214
(vi) 119 × 303
(vii) 543 – 479
(viii) 5133
Solution:
(i) We have, 43 + 37
Since, both 43 and 37 are odd numbers.
∴ 43 + 37 is even [v sum of two odd numbers is even]
(ii) We have, 672 – 348
Since, both 672 and 348 are even numbers.
∴ 672 – 348 is even.
[v difference of two even numbers in even]
(iii) We have, 4 × 347 × 3
Since, 4 is an even number and 347 is an odd number.
∴ 4 × 347 is an even number.
[v product of an even number and an odd number is even]
Since, 3 is an odd number.
∴ 4 × 347 × 3 is even.
(iv) We have, 708 – 477
Since, 708 is even number and 477 is an odd number.
∴ 708 – 477 is odd.
[∵ difference of an even number and an odd number is odd]
(v) We have, 809+214
Since, 809 is an odd number and 214 is an even number.
∴ 809 + 214 is odd.
[∵ sum of an odd number and an even number is odd]
(vi) We have, 119 × 303
Since, both 119 and 303 are odd numbers.
∴ 119 × 303 is odd.
[∵ product of two odd numbers is odd]
(vii) We have, 543 – 479
Since, both 543 and 479 are odd numbers.
∴ 543 – 479 is even.
[∵ difference of two odd numbers is even]
(viii) We have, 5133 =513 × 513 × 513
∵ Product of three odd numbers is odd.
∴ 513 × 513 × 513 is an odd number.
Question 2.
Using our understanding of how parity behaves under different operations, identify which of the following algebraic expressions give an even number for any integer values for the letter-numbers.
Solution:
(i) We have, 2a + 2b
Since, both 2a and 2b are even.
[∵ 2 × any number = even]
So, 2a + 2b is even. [∵ even + even = even]
(ii) We have, 3g + 5h
Since, 3g and 5h can be either even or odd depending on g and h.
So, result is not always even.
(iii) We have, 4m + 2n
Since, both 4m and 2n are even.
So, 4m + 2n is even. [∵ even + even = even]
(iv) We have, 2u – 4v
Since, both 2u and 4v are even.
So, 2u – 4v is even. [∵ even – even = even]
(v) We have, 13k – 5k = 8k
So, 8k is even. [∵ 8 x any number = even]
(vi) We have, 6m – 3n
Since, 6m is even, but 3n may be odd or even,
∵ Even – Odd = Odd
and Even – Even = Even
So, 6m – 3n is not always even.
(vii) We have, x2 +2
x2 can be either even or odd depending on x.
∵ Even + 2 = Even and Odd + 2 = Odd
So, x2 + 2 is not always even.
(viii) We have, b2 + 1
b2 can be even or odd.
∵ Even + 1 = Odd and Odd + 1 = Even
So, b2 + 1 is not always even.
(ix) We have, 4k × 3j = 12kj
∵ 12 is even.
Since, 12 × any number = even
So, 4k × 3j is even.
Question 3.
Similarly, determine and explain which of the other expressions always give even numbers. Write a couple of examples and non-examples, as appropriate, for each expression.
Solution:
Let us take a few algebraic expressions and check if they always give even numbers or not.
Expression 1
2n
This expression will always give even numbers because any number multiplied by 2 is even.
Examples
If n = 1 then 2 × 1 = 2 (even)
If n = 2, then 2 × 2 = 4 (even)
Non-examples
There are no non-examples, because this expression will always be even for all values of n.
Expression 2
n + n
This is equal to 2n, so it will also always give even numbers.
Examples
If n = 4, then 4 + 4 = 8 (even)
If n = 5, then 5 + 5 = 40 (even)
Non-examples
No non-examples because this always gives an even number.
Question 4.
Write a few algebraic expressions which always give an even number.
Solution:
Here are a few algebraic expressions that always result in an even number.
(i) 2n
(ii) 2n + 2m
(iii) 2n – 2m
(iv) (2n)2
(v) (n + m)2 – (n – m)2
Intext (NCERT Page 117-118)
Question 1.
Look at the following expressions and the visualisation. Write the corresponding explanation and examples.
Explanation with Algebra and Visualisation Examples
Solution:
Adding two even numbers that one is multiple of 4 and other is not multiple of 4 will not give a multiple of 4.
e.g. 4, 6, 8, 10, 12
4 + 6 = 10,
6 + 8 = 14
and 10 + 12 = 22
What Remains? (NCERT Page 121)
Question 1.
Find a number that has a remainder of 3 when divided by 5. Write more such numbers.
Solution:
Let N be the number.
When N is divided by 5, the remainder is 3.
∴ N = 5q + 3, where q is non negative integer.
Let q = 0
Then, N = 5 × 0 + 3 = 3
Let q = 1
Then, N = 5 × 1 + 3 = 5 + 3 = 8
Let q = 2
Then, N = 5 × 2 + 3 = 10 + 3 = 13
Let q = 3
Then N = 5 × 3 + 3 = 15 + 3 = 18
Let q = 4
Then, N = 5 × 4 + 3 = 20 + 3 = 23
∴ Numbers are 3, 8, 13, 18, 23, ………………
Figure it Out (NCERT Page 122-123)
Question 1.
The sum of four consecutive numbers is 34. What are these numbers?
Solution:
Let the four consecutive numbers be x, x + 1, x + 2 and x + 3.
According to the question,
x + x + 1 + x + 2 + x + 3 = 34 ,
⇒ 4x + 6 = 34
⇒ 4x = 34 – 6
⇒ 4x = 28
⇒ x = 7
So, the four consecutive numbers are 7, 7 + 1, 7 + 2, 7 + 3
i. e. 7, 8, 9, 10.
Question 2.
Suppose pis the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Solution:
Since, p is the greatest number. So, the numbers beforep will be p – 1, p – 2, p – 3 and p – 4.
Therefore, the other four numbers are p – 1, p – 2, p – 3 and p – 4.
Question 3.
For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.
(i) The sum of two even numbers is a multiple of 3.
Solution:
Let the two even numbers be 2m and 2n.
∴ Sum =2m + 2n = 2(m + n), which is a multiple of 2.
A multiple of 3 can be even (6,12, …………….) or odd (3, 9, ………….)
For the sum of two even numbers to be a multiple of 3, it must be an even multiples of 3.
Example 2 + 4 = 6, which is a multiple of 3.
2 + 8 = 10 which is not a multiple of 3.
So, the given statement is sometimes true.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
Solution:
If a number is not divisible by 18, then it may or may not be divisible by 9.
Example 27 is not divisible by 18 but it is divisible by 9. 10 is not divisible by 18 and it is also not divisible by 9.
So, the given statement is sometimes true.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
Solution:
It’s possible that two numbers individually are not divisible by 6, but their sum is divisible by 6.
Example 5 and 7 are not divisible by 6, but 5+7 = 12 is divisible by 6.
1 and 2 are not divisible by 6 and their sum 1 + 2 = 3 is also not divisible by 6.
So, the given statement is sometimes true.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Solution:
Let one number be 6a and other number be 9 b.
Sum = 6a + 9b = 3 (2a + 3b), which is a multiple of 3.
Example 6 + 9 = 15, which multiple of 3.
12 + 27 = 39, which is multiple of 3.
So, the given statement is always true.
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Solution:
Let one number be 6a and other number be 3b.
Sum = 6a + 3b = 3(2a + b), which is a multiple of 3.
Example 6 + 3 = 9, which is a multiple of 9.
12 + 3 = 15, which is not a multiple of 9.
So, the given statement is sometimes true.
Question 4.
Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Solution:
Let x be the number.
When x is divided by 3, the remainder is 2.
∴ x = 3k + 2, where k is any whole number.
If k = 0, x = 3 × 0 + 2 = 2
If k = 1, x = 3 × 1 + 2 = 5
If k = 2, x = 3 × 2 + 2 = 8
If k = 3, x = 3 × 3 + 2 = 11
If k = 4, x = 3 × 4 + 2 = 14
And so on.
When x is divided by 4, the remainder is 2.
x = 4m + 2, where m is any whole number.
If m = 0, then x = 4 × 0 + 2 = 2
If m = 1, then x = 4 × 1 + 2 = 6
If m = 2, then x = 4 × 2 + 2 = 10
If m = 3, then x = 4 × 3 + 2 = 14
If m = 4, then x = 4 × 4 + 2 = 18 And so on.
Now, the numbers that appear in both lists are 2 and 14. It we continue the lists, the next one would be 26.
Now, the difference between these numbers are 14 – 2 = 12 and 26 – 14 = 12 The difference between consecutive numbers is 12 and 12 is the LCM of the divisors 3 and 4.
We have, the common numbers are 2, 14, 26
Here, each number is 12 more than the previous one.
All these numbers leave a remainder of 2 when divided by both 3 and 4.
Required algebraic expression is x = LCM (3 and 4) n + 2
= 12n + 2, where n is any whole number.
Question 5.
“I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up-it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around, But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”
Solution:
Here, “When I group them in 3’s, one stays with me.” This means the number of pebbles leaves a remainder of 1 when divided by 3.
Next, “Try pairing them up it simple won’t do.
A stubborn odd pebble remains in my view.”
This means the number of pebbles is odd and this leaves a remainder of 1 when divided by 2.
Next, “Group them by 5, yet one’s still around.”
This means the number of pebbles leaves a remainder of 1 when divided by 5.
Next, “But grouping by 7, perfection is found.”
This means the number of pebbles is perfectly divisible by 7 i.e. leaves a remainder of 0 when divided by 7.
At last, “More than one hundred would be far too bold.” This means the number of pebbles is less than or equal to 100.
Since, the number leaves a remainder of 1 when it is divided by 3, 2 and 5, it means that if we subtract 1 from the number, the result is divisible by 3, 2 and 5.
∴ LCM of 3, 2 and 5 = 3 × 2 × 5 = 30
Therefore, the number of pebbles can be expressed as
30n + 1 for some integer n.
If n = 1, then 30 × 1 + 1 = 31, which is not divisible by 7.
If n = 2, then 30 × 2 + 1 = 61, which is not divisible by 7.
If n = 3, then 30 × 3 + 1 = 91 = 7 × 13, which is divisible by 7.
∴ Number of pebbles = 91
Question 6.
Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?
Solution:
Let x be the number.
When x is divided by 6, then remainder is 2.
∴ x = 6k + 2, where k is any whole number.
Let us take three numbers.
First number (x1) = 6a + 2
Second number (x2) = 6b + 2
Third number (x3) = 6c + 2
Here, a, b and c are any whole numbers.
.’. Sum of x1, x2 and x3 = x1 + x2 + x3
= 6a + 2 + 6b + 2 + 6c + 2
= 6 (a + b + c) + 6
= 6(a + b + c + 1)
Since, a, b and c are whole numbers.
∴ a + b + c is a whole number.
Also, a + h + c + 1 is a natural number.
So, sum will always be a multiple of 6.
Hence, Tathagat’s claim is true.
Question 7.
When divided by 7, the number 661 leaves a remainder of 3 and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661
(ii) 4779 – 661
Solution:
Given, when 661 is divided by 7, the remainder is 3
∴ 661= (7 × q1) + 3 …(i)
Here, q1 is the quotient when 661 is divided by 7.
When 4779 is divided by 7, the remainder is 5.
∴ 4779 – (7 × q2) + 5 …(ii)
Here, q2 is the quotient when 4779 is divided by 7.
(i) On adding Eqs (i) and (ii), we get
4779 + 661 = 7q2 + 5 + 7q1 + 3
= 7(q1 + q2) +8
= 7(q1 + q2) + 7 + 1
= 7(q1 + q2 + 1) + 1
∴ Required remainder is 1 when divided by 7.
(ii) On subtracting Eq. (i) from Eq. (ii), we get
4779 – 661 = 7q2 + 5 – (7q1 + 3)
= 7q2 + 5 – 7q1 – 3
= 7(q2 – q1) + 2
Required number is 2 when divided by 7.
Question 8.
Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number?
Can you give a simple explanation of why it is the smallest?
Solution:
When divided by 3, the remainder is 2.
∴ The number is the form 3k + 2.
When divided by 4, the remainder is 3.
∴ The number is the form 4m + 3.
When divided by 5, the remainder is 4.
∴ The number is the form 5n + 4.
Here, in each case, the remainder is one less than the divisor,
i.e. 3 – 2 = 1, 4 – 3 = 1, 5 – 4 = 1
If we add 1 to the number, it will be perfectly divisible by 3, 4 and 5.
Since, adding 1 to the number makes it divisible by 3, 4 and 5, the remainder plus 1 must be a multiple of the LCM of 3, 4 and 5.
LCM (3, 4, 5) = 3 × 4 × 5 = 60
∴ Required number = 60 – 1 = 59
Explanation of why it is the smallest.
The property of the remainders being less than the respective divisors means that if we add 1 to the number, it becomes a multiple of all three divisors.
To find the smallest such number, we need the smallest multiple of all three divisors, which is their LCM.
Since, 60 is the smallest positive common multiples of 3, 4, and 5, subtracting 1 from it gives the smallest number satisfying the condition.
Intext (NCERT Page 125-126)
Question 1.
Look at each of the following statements. Which are correct and why?
(i) If a number is divisible by 9, then the sum of its digits is divisible by 9.
(ii) If the sum of the digits of a number is divisible by 9, then the number is divisible by 9.
(iii) If a number is not divisible by 9, then the sum of its digits is not divisible by 9.
(iv) If the sum of the digits of a number is not divisible by 9, then the number is not divisible by 9.
Solution:
(i) Correct [by divisibility rule for 9]
(ii) Correct [by divisibility rule for 9]
(iii) Correct [by divisibility rule for 9]
(iv) Correct [by divisibility rule for 9]
Figure it Out (NCERT Page 126)
Question 1.
Find, without dividing, whether the following numbers are divisible by 9.
(i) 123
Solution:
We know that a number is divisible by 9, if the sum of its digits is divisible by 9:
Sum of the digits of 123 = 1 + 2 + 3 = 6, which is not divisible by 9.
So, 123 is not divisible by 9.
(ii) 405
Solution:
Sum of the digits of 405 = 4 + 0 + 5 = 9, which is divisible by 9.
So, 405 is divisible by 9.
(iii) 8888
Solution:
Sum of the digits of8888 = 8 + 8 + 8 + 8 = 32, which is not divisible by 9.
So, 8888 is not divisible by 9.
(iv) 93547
Solution:
Sum of the digits of 93547 = 9 + 3 + 5 + 4 + 7 = 28, which is not divisible by 9.
So, 93547 is not divisible by 9.
(v) 358095
Solution:
Sum of the digits of 358095 = 3 + 5 + 8 + 0 + 9 + 5 = 30, which is not divisible by 9.
So, 358095 is not divisible by 9.
Question 2.
Find the smallest multiple of 9 with no odd digits.
Solution:
To find the smallest multiple of 9 with no odd digits,
we can list multiples of 9 and check their digits.
We need a number that is a multiple of 9 and all its digits must be even i.e. 0, 2, 4, 6, 8.
So, multiples of 9 are 9 contains odd digit 9,
18 contains odd digit 1,
27 contains odd digit 7,
36 contains odd digit 3,
.
.
.
81 contains odd digit 1,
90 contains odd digit 9,
.
.
.
108 contains odd digit 1.
288 has all digits even.
Sum of digits = 2 + 8 + 8 = 18, which is a multiple of 9.
So, the smallest multiple of 9 with no odd digits is 288.
Question 3.
Find the multiple of 9 that is closest to the number 6000.
Solution:
To find the multiple of 9 closest to 6000, follow these steps:
1. Divide 6000 by 9.
6000 ÷ 9 = 666 with a remainder of 6
2. Identify the two closest multiples.
One multiple = 9 × 666 = 5994
Next multiple = 9 × 667 = 6003
3. Calculate the difference from 6000 for each multiple.
Difference for 5994 = 6000 – 5994 = 6
Difference for 60003 = 6003 – 6000 = 3
4. Determine the closest multiple.
Since, 3 < 6
So, the multiple 6003 is closer to 6000 than 5994. Therefore, the multiple of 9 that is closest to the number 6000 is 6003.
Question 4.
How many multiples of 9 are there between the numbers 4300 and 4400?
Solution:
To find the number of multiples of 9 between 4300 and 4400, we need to find the first multiple of 9 greater than 4300.
On dividing 4300 by 9, we get 4300 – 9 = 477.78
∴ Next integer = 478
Multiplying 478 by 9, we get 478 × 9 = 4302
So, the first multiples of 9 greater than 4300 = 4302
Now, the multiples of 9 after 4302 are
4302 + 9, 4302 + 9 × 2, 4302 + 9 × 3, 4302 + 9 × 4, 4302 + 9 × 5, 4302 +9 × 6, 4302 + 9 × 7, 4302 + 9 × 8, 4302 +9 × 9, 4302 +9 × 10
i.e. 4311, 4320, 4329, 4338, 4347, 4356, 5365, 4374, 4383, 4392.
So, there are total 11 multiples of 9 between 4300 and 4400.
Intext (NCERT Page 127-128)
Question 1.
Using these observations, can you tell whether the number 462 is divisible by 11?
Solution:
Units place (2)
The unit place is 1 more than a multiple of 11.
[1 = 11 × 0 + 1]
So, the contribution of 2 units = 2 × (+1) = + 2
Tens place (6)
The tens place is 1 less then a multiple of 11.
[10 = 11 × 1 – 1]
So, the contribution of 6 tens = 6 × (-1) = – 6
Hundreds place (4)
The Hundred place is 1 more than a multiple of 11.
[100 = 11 × 9 + 1]
So, the contribution of 4 hundreds = 4 × (+1) = + 4
Now, sum of these contributions = 2 – 6 + 4 = 0
Since, the sum is 0 and 0 is divisible by 11.
So, the number 462 is divisible by 11.
Question 2.
What could be a general-method or shortcut to check divisibility by 11?
Solution:
General rule If the difference between the sum of the digits at the odd places (from right to left) and the sum of the digits at even places (from right to left) of the number is either 0 or divisible by 11, then the number is divisible by 11.
Question 3.
If this difference is 11 or a multiple of 11, what does that say about the remainder obtained when the number is divisible is by 11?
Solution:
If difference is 11 or a multiple of 11, then the number is divisible by 11.
Therefore, the remainder obtained when the number is divisible by 11, is 0.
Question 4.
Using this shortcut, find out whether the following numbers are divisible by 11. Further, find the remainder if the number is not divisible by 11.
(i) 158
(ii) 841
(iii) 481
(iv) 5529
(v) 90904
(vi) 857076
Solution:
A number is divisible by 11, if the difference between the sum of the digits at odd places (from the right to left) and the sum of the digits at even places (from the right to left) of the number is either 0 or divisible by 11.
(i) We have, 158
Sum of digits at odd places = 8 + 1 = 9
Sum of digits at even places = 5
∴ Difference = 9 – 5 = 4, which is not divisible by 11.
So, 158 is not divisible by 11.
Remainder = 4
(ii) We have, 841
Sum of digits at odd places = 1 + 8 = 9
Sum of digits at even places = 4
∴ Difference = 9 – 4 = 5, which is not divisible by 11.
So, 841 is not divisible by 11.
Remainder = 5
(iii) We have, 481
Sum of digits at odd places = 1 + 4 = 5
Sum of digits at even places = 8
Difference = 8 – 5 = 3, which is not divisible by 11.
So, 481 is not divisible by 11.
Remainder = 3
(iv) We have, 5529
Sum of digits at odd places = 9 + 5 = 14
Sum of digits at even places = 5 + 2 = 7
Difference = 14 – 7 = 7, which is not divisible by 11.
So, 5529 is not divisible by 11.
Remainder = 7
(v) We have, 90904
Sum of digits at odd places = 4 + 9 + 9 = 22
Sum of digits at even places =0 + 0 = 0
Difference = 22 – 0 = 22, which is divisible by 11.
So, 90904 is divisible by 11.
Remainder = 7
(vi) We have, 857076
Sum of digits at odd places = 6 + 0 + 5 = 11
Sum of digits at even places = 7 + 7+ 8 = 22
Difference = 22 – 11 = 11, which is divisible by 11.
So, 857076 is divisible by 11.
Intext (NCERT Page 129)
Question 1.
Look at the following procedure.
Is this method similar to or different from the method we saw just before?
Solution:
Yes, this method is similar to from the method we saw just before.
Question 2.
Find the following table. Find a quick way to do this?
Solution:
We know that a number is divisible by
- 2, if it has digits 0, 2, 4, 6 or 8 at ones place.
- 3, if the sum of the digits is a multiple of 3 or it is divisible by 3.
- 4, if last two digits of the number is completely divisible by 4.
- 5, if a number has 0 or 5 in its ones place.
- 6, if it is divisible by 2 and 3 both.
- 8, if last three digits of the number is completely divisible by 8.
- 9, if the sum of the digits of the number is divisible by 9.
- 10, if a number has 0 in its ones place.
- 11, if the difference of sum of the digits at even places and sum of the digits at odd places is either 0 or multiple of 11.
Now, complete table is shown below :
Question 3.
How can we find out if a number is divisible by 6?
Solution:
If a number is divisible by 2 and 3 both, then the number is divisible by 6.
Question 4.
Will checking its divisibility by its factors 2 and 3 work? Use the shortcuts for 2 and 3 on these numbers and divide each number by 6 to verify 38, 225, 186, 64.
Solution:
A number is divisible by 6, if it is divisible by both of its factors 2 and 3.
We have, 38
Since, its last digit is 8 (even).
So, it is divisible by 2.
Now, sum of digits of 38 = 3 + 8 = 11, which is not divisible by 3.
So, 38 is not divisible by 6.
Verification
On dividing 38 by 6, we get remainder is 2.
Next, we have, 225 Since, its last digit is 5 (odd).
So, it is not divisible by 2.
Hence, 225 is not divisible by 6.
Verification
On dividing 225 by 6, we get remainder is 3.
Next, we have, 186
Since, its last digit is 6 (even).
So, it is divisible by 2.
Now, sum of digits of 186 = 1 + 8 + 6 = 15, which is divisible by 3.
So, 186 is divisible by 6.
Verification
On dividing 186 by 6, we get remainder is 0.
So, it is divisible by 6.
Next, we have, 64
Since, its last digit is 4 (even).
So, it is divisible by 2.
Now, sum of digits of 64 = 6 + 4 = 10, which is not divisible by 3.
Hence, 64 is not divisible by 6.
Verification
On dividing 64 by 6, we get remainder is 4.
Digital Roots (NCERT Page 130)
Question 1..
What property do you think this digital root will have? Recall that we did this while finding the divisibility shortcut for 9.
Solution:
Property The digital root of a number is equal to the remainder when that number is divided by 9.
e.g. 125
Sum of digits = 1 + 2+ 5 = 8
∴ Digital root of 125 = 8
On dividing 125 by 9, we get remainder = 8
So, the digital root is the same as the remainder when the number is divided by 9.
Question 2.
Between the numbers 600 and 700, which numbers have the digital root
(i) 5
(ii) 7
(iii) 3
Solution:
Let the number between 600 and 700 be N.
Again, let N = 6xy, where x and y are digits.
Sum of digits = 6 + x + y
So, digital root of N = 6 + x + y
(i) Digital root = 5
A number has a digital root of 5 if the sum of its digits is 5, 14, 23, 32 etc.
Possible sums for a digital root of 5,
6 + x + y = 14
⇒ x + y = 8
Possible numbers are 608, 617,626, 635,644, 653, 662, 671, 680.
Now, 6 + x + y = 23 ⇒ x + y = 17
Possible numbers are 689, 698.
(ii) Digital root = 7
A number has a digital root of 7 if the sum of its digits is 7,16,25 etc.
Possible sums for a digital root of 7,
6 + x + y = 7 ⇒ x + y = 1
Possible numbers are 601,610.
Now, 6 + x + y = 16 ⇒ x + y = 10
Possible numbers are 619, 628, 637, 646, 655, 664, 673, 682, 691.
Also, 6 + x + y = 25 ⇒ x + y = 19,
which is not possible as maximum of x + y is 9 + 9 = 18.
(iii) Digital root = 3
A number has a digital root of 3 if the sum of its digits is 3,12, 21 etc.
Possible sums for a digital root of 3,
6 + x + y = 3, not possible Next,
6 + x + y = 12 ⇒ x + y = 6
Possible numbers are 606, 615, 624, 633, 642, 651, 660.
Now, 6 + x + y = 21 ⇒ x + y = 15
Possible numbers are 669, 678, 687, 696.
Question 3.
Write the digital roots of any 12 consecutive numbers. What do you observe?
Solution:
Let us consider the 12 consecutive numbers starting from 1 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.
∴ Digital root of 1 = 1
Digital root of 2 = 2
.
.
.
.
Digital root of 9 = 9
Now, digital root of 10 = 1 + 0 = 1
Digtial root of 11 = 1 + 1 = 2
Digital root of 12 = 1 + 2 = 3
The digital roots of consecutive numbers follow a
repeating pattern of 1, 2, 3, 4, 5, 6, 7, 8, 9.
This pattern repeats after 9 numbers.
When the number is a multiple of 9, its digital root is 9.
For numbers greater than 9, the digital root essentially resets back to 1 after reaching 9.
Question 4.
Find the digital roots of some consecutive multiples of
(i) 3
Solution:
We know, multiples of 3 are 3, 6, 9, 12, 15, 18
Digital root of 3 = 3
Digital root of 6 = 6
Digital root of 9 = 9
Digital root of 12 = 1 + 2 = 3
Digital root of 15 = 1 + 5 = 6
Digital root of 18 = 1 + 8 = 9, and so on.
The pattern of digital roots for multiples of 3 is 3, 6, 9, 3, 6, 9
(ii) 4
Solution:
We know, multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40,
∴ Digital root of 4 = 4
Digital root of 8 = 8
Digital root of 12 = 1 + 2 = 3
Digital root of 16 = 1 + 6 = 7
Digital root of 20 = 2 + 0 = 2
Digital root of 24 = 2 + 4 = 6
Digital root of 28 = 2 + 8 = 10
and sum of digits of 10 = 1 + 0 = 1
So, digital root of 28 = 1
Digital root of 32 = 3 + 2 = 5
Digital root of 36 = 3 + 6 = 9
Digital root of 40 = 4 + 0 = 4, and so on.
The pattern of digital roots for consecutive multiples of 4 is 4, 8, 3, 7, 2, 6, 1, 5, 9, 4, 8, 3, 7,………..
(iii) 6
Solution:
Do same as above.
The pattern of digital roots for consecutive multiples of 6 is 6, 3, 9,6, 3, 9, ……….
Question 5.
What are the digital roots of numbers that are 1 more than a multiple of 6? What do you notice?
Try to explain the patterns noticed.
Solution:
The numbers that are 1 more than a mutiple of 6 are
6 × 1 + 1 = 7
6 × 2 + 1 = 13
6 × 3 + 1 = 19
6 × 4 + 1 = 25
6 × 5 + 1 = 31
6 × 6 + 1 = 37
6× 7 + 1 = 43
6 × 8 + 1 =49
6 × 9 + 1 = 55
6 × 10 + 1 = 61
6 × 11 + 1 = 67,
and so on.
Now, digital root of 7 = 7
Digital root of 13 = 1 + 3 = 4
Digital root of 19 = 1 + 9 = 10,1 + 0 = 1
Digital root of 25 = 2 + 5 = 7
Digital root of 31 = 3 + 1 = 4
Digital root of 37 = 3 + 7 = 10, 1 + 0 = 1
Digital root of 43 = 4 + 3 = 7
Digital root of 49 = 4 + 9 = 13, 1 + 3 = 4
igital root of 55 = 5 + 5 = 10, 1 + 0 = 1
Digital root of 61 = 6 + 1 = 7
Digital root of 67 = 6 + 7 = 13, 1 + 3 = 4, and so on.
∴ The digital roots follow a repeating patten 7, 4, 1, 7, 4, 1, ……….
Question 6.
I’m made of digits, each tiniest and odd,
No shared ground with root #l-how odd!
My digits count, their sum, my root- All point to one bold number’s pursuit- The largest odd single-digit I proudly claim.
What’s my number? What’s my name?
Solution:
Here, “I’m made of digits, each tineist and odd.”
This means the number is composed of single odd digits (1, 3, 5, 7, 9).
Next, “No shared ground with root # l-how odd!”
This mean the digital root of the number is not 1.
Next, “My digits count, their sum, my root-
All point to one bold number’s pursuit-.”
This means that the count of digits, the sum of digits and the digital root are all related to a specific number.
Next, “The largest odd single-digit I proudly claim.”
This means the number itself is 9 or its digital root is 9.
∴ Required number is 9 and its number is Nine.
Figure it Out (NCERT Page 131)
Question 1.
The digital root of an 8-digit number is 5. What will be the digital root of 10 more than that number?
Solution:
Let the original 8-digit number be n.
New number = n + 10
Digital root of (n + 10)
= Digital root of n + Digital root of 10
= 5 + (1 + 0)
[∵ digital root of n = 5, given] = 5 + 1 = 6
Question 2.
Write any number. Generate a sequence of numbers by repeatedly adding 11. What would be the digital roots of this sequence of numbers? Share your observations.
Solution:
Let 5 be any number.
On adding 11 repeatedly, we get
5 + 11 = 16,
16 + 11 = 27,
27 + 11 = 38,
38 + 11 = 49,
49 + 11 = 60,
60 + 11 = 71,
71+ 11 = 82,
82 + 11 = 93,
93 + 11 = 104,
104 + 11 = 115
and so on.
Now, digital root of 5 = 5
Digital root of 16 = 1 + 6 = 7
Digital root of 27 = 2 + 7 = 9
Digital root of 38 = 3 + 8 = 11, 1 + 1 = 2
Digital root of 49 = 4 + 9 = 13, 1 + 3 = 4
Digital root of 60 = 6 + 0 = 6
Digital root of 71 = 7 + 1 = 8
Digital root of 82 = 8 + 2 = 10, 1 + 0 = 1
Digital root of 93 = 9 + 3 = 12, 1 + 2 = 3
Digital root of 104 = 1 + 0+ 4 = 5 digital root of 115 = 1 + 1 + 5 = 7
∴ The sequence of digital roots is
5, 7, 9, 2, 4, 6, 8, 1, 3, 5, 7, …………..
Hence, the digital roots repeat in a cycle of 9,
i. e. 5, 7, 9, 4, 6, 8, 1, 3 …………….
Question 3.
What will be the digital root of the number 9a + 36b+ 13?
Solution:
We have, 9a + 36b + 13
Since, 9a is a multiple of 9.
∴ Digital root of 9a = 9
Also, 36b is a multiple of 9.
∴ Digital root of 36b = 9
Now, digital root of 13 = 1 + 3 = 4
∴ Digital root of 9 a + 36
b + 13 = 9 + 9 + 4 = 22
and digital root of 22 = 2 + 2 = 4
∴ Digital root of 9a + 36b + 13 = 4
Question 4.
Make conjectures by examining if there are any patterns or relations between
(i) the parity of a number and its digital root.
(ii) the digital root of a number and the remainder obtained when the number is divided by 3 or 9.
Solution:
(i) Parity means whether the number is even or odd.
Let us take some examples.
1.36(even)
Digital root of 36 = 3 + 6 = 9 (odd)
2. 47 (odd)
Digital root of 47 = 4 + 7 = 11 and digital root of 11 = 1 + 1 = 2
So, digital root of 47 = 2 (even)
3. 44 (even)
Digital root of 44 = 4 + 4 = 8 (even)
4. 19 (odd)
Digital root of 19 = 1 + 9 = 10
and digital root of 10 = 1 + 0 = 1 (odd)
∴ There is no fixed relation between parity of a number and its digital root.
(ii) Let us take some examples.
1.86
Digital root of 86 = 8 + 6 = 14
and digital root of 14 = 1 + 4 = 5
∴ Digital root of 86 = 14
On dividing 86 by 9 and 3, we get remainders
5 and 2, respectively.
2.123
Digital root of 123 = 1 + 2 + 3 = 6
On dividing 123 by 9 and 3, we get remainders
6 and 0, respestively.
3. 59
Digital root of 59 = 5 + 9 = 14
and digital root of 14 = 1+ 4 = 5
∴ Digital root of 59 = 5
On dividing 59 by 9 and 3, we get remainders
5 and 2, respectively.
∴ The digital root is same as the remainder when the number is divided by 9.
Digits in Disguise (NCERT Page 131 – 132)
Question 1.
Solve the cryptarithms given below.
Solution:
(i) We have,
In first column, 1 + B = O
In second column, A + 1 = B
Let us assume 11 = 2
∴ 1 + 2 = O
⇒ O = 3
Now, A + 1 = 2
⇒ A = 2 – 1
⇒ A = 1
Therefore, the puzzle is solved as shown below.
(ii) We have,
Here, we have two letters A and 5 whose values are to be found.
Studying the addition in one’s column, we have 5 + 7 and we get A from this i.e. a number whose one’s digit is A
Also, from addition in ten’s column, we have A + 3 and we get 6 from this.
Therefore, A must be 0,1,2 or 3.
If A = 0, then puzzle becomes
It is not possible because first number AB becomes B
i. e. one digit number.
If A = 1, then puzzle becomes
Then, B gives 1, so B must be 4 and sum in ten’s column = 1 + 3 + 1 = 55 ≠ 6
So, it is not possible.
If A = 2, then puzzle becomes
Then, B + 7 gives 2, so B must be 5 and sum in ten’s column = 2 + 3 + 1 = 6
So, it is correct.
Therefore, the puzzle is solved as shown below.
Hence, A = 2 and B = 5
(iii) We have,
∴ 3 × ON = PO
So, PO must be a two-digit number divisible by 3.
Let us assume ON =17
3 × 17 = PO
⇒ 51 = PO
∴ P = 5 and 0 = 1
Therefore, the puzzle is solved as shown below.
Hence, O = 1, N = 7 and P = 5
(iv) We have
In first column, R + R + R = R
⇒ 3R = R i.e. a number whose one’s digits is R.
So, R must be 5.
In second column, 1 + Q + Q + Q = 5
i. e. a number whose one’s digit is R.
So, Q must be 8.
Therefore, the puzzle is solved as shown below. 85 85
So, 5 = 5, Q = 8 and P = 2.
Question 2.
Solve the following.
(i) UT × 3 = PUT
Solution:
We have, UT × 3 = PUT
Here, we have three letters P, U and T whose values are to be found.
Since, units digit of 3 × T is T.
So, it must be T = 0 or T =5.
When T = 0, then I/O × 3 = PU0
and when T = 5, then 1/5 × 3 = PU5
Now, units digit of 3 × H is U.
So, U must be 0 or 5 but U cannot be 0 because if U = 0
then UT becomes one-digit number.
So, U must be 5 and multiplication is either 50 × 3 or 55 × 3.
Here, the second possibility fails.
50 × 3 = 150
Therefore, the puzzle is solved as shown below
(ii) AB × 5 = BC
Solution:
We have, AB × 5 = BC
The units digit of B × 5 is C.
Since, it is a multiple of 5.
So, C must be 0 or 5.
If C = 0, then 5 × 5 ends in 0.
So, B can be 0, 2, 4, 6, 8.
If C = 5, then 5 × 5 ends in 5.
So, 5 can be 1, 3, 5, 7, 9.
Also, AB × 5 = BC
⇒ 5(10A + B) = 10B + C
⇒ 50A + 5B = 10B + C
⇒ 50A = 5B + C
Consider, C = 0
Then, 50A = 5 B
⇒ 10A = B
Since, A and B are single digits.
If A = 1 ⇒ B = 10, not possible.
Similarly, A = 2,3 are also not possible for this case.
Consider, C = 5
Then, 50A = 5B + 5
⇒ 10A = B + 1
IfA = 1, then 10 = B + 1
⇒ B = 9
So, A = 1, B = 9, C = 5
The equation becomes 19 × 5 = 95
(iii) L2N × 2 = 2NP
Solution:
We have, L2N × 2 = 2NP
From the multiplication, N × 2 must result in a number ending in P.
Also, 2 × 2 = 4 or 5 if there is a carry from N × 2.
And L × 2 must result in a number starting with 2, sol must be 1.
If L = 1, then 12N × 2 = 2NP.
If JV = 4, then 4 × 2 = 8, so P = 8.
Then, 124 × 2 = 248.
This fits the pattern.
Therefore, L = 1, N = 4, P = 8.
The equation becomes 124 × 2 = 248.
(iv) XV × 4 = ZX
Solution:
We have, XT × 4 = ZX
Since, XT is a two-digit number and ZX is also a two-digit number and multiplying XT by 4 results in ZX, then Z must be a digit such that Z is the result of multiplyingX by 4 (or a multiple of 4 plus a carry).
Also, 4 × T must result in a number ending in X
Let’s consider the possible values for T.
If T = 0, 4 × T = 0, so x = 0.
Then, XT becomes 00, which is meaningless.
If T = 1, 4 × T = 4, so X = 4.
Then, 4 × X = 16, so Z = 16, which is not possible as Z is a single digit number.
If T = 2, 4 × T = 8, so X = 8.
Then, 4 × X = 32, so Z = 32, which is not possible as Z is a single digit number.
If T = 3, 4 × T = 12, so X= 2 (carry 1).
Then, (4 × X) + 1 = 4 (2) + 1 = 9, so Z = 9
If X = 2,T = 3, then 23 × 4 = 92
Here, X = 2,T = 3, Z = 9
This satisfies the condition XT × 4 = ZX.
(v) PP × QQ = PRP
Solution:
We have, PP × QQ = PRP
Here, PP = 10P + P = 1 IP
and QQ = 10Q + Q= 11Q
So, 1 IP × 1 IQ = PRP
⇒ 121 × P × Q = PPP.
Here, PRP is a three-digit number where the hundreds digit and units digit are the same (P).
Let’s test values for P and Q.
If P = 1, 121 × Q = 1P1.
Also, if Q = 2, 121 × 2 = 242.
HereP = 2, which contradicts P = 1.
IfP = 2,121 × 2 × Q = 2B2, so 242 × Q = 2R2.
This means Q must be 1.
If Q = 1,242 × 1 = 242.
Here P = 2, Q = 1 and P = 4.
This satisfies the condition PP × QQ = PRP.
So, P = 2, Q = 1 and P = 4
(vi) JK × 6 = KKK
Solution:
We have, JK × 6 = KKK
Since, KKKis a multiple of 6, it must be an even number.
Also, KKK can be written as 111 × K.
Therefore, 111 × K must be divisible by 6.
Since, 111 is not divisible by 2, K must be an even digit
i. e. 2,4, 6, 8 (0 is not possible).
Since, 111 is divisible by as 3 (as 1 + 1 + 1 = 3), KKK is always divisible by 3.
So, for KKK to be divisible by 6, K must be an even digit.
If K = 2, KKK = 222.
222 + 6 = 37.
So, JK = 37.
If X = 4, KKK = 444.
444 + 6 = 74.
So, JK = 74.
If X = 6, KKK = 666.
666 + 6 = 111.
This is hot a two-digit number, so Xcannot be 6.
IfX = 8,XXX = 888.
888 + 6 = 148
This is not a two-digit number, so X cannot be 8.
For X = 2, JK = 37
Here, J = 3 and X = 7.
This contradicts X = 2.
For X =4, JX = 74
Here, J = 7 and X = 4
This is consistent with X = 4.
J = 7, X = 4.
Figure It Out (NCERT Page 132-134)
Question 1.
If 31z5 is a multiple of 9, where z is a digit, what is the value of z? Explain why there are two answers to this problem.
Solution:
We know that, if a number is a multiple of 9, then the sum of its digits is divisible by 9.
Sum of digits of 31z5 = 3 + 1 + z + 5 = 9 + z
So, 9 + z should be a multiple of 9.
This is possible when 9 + z is any one of these numbers 0, 9, 18, 27
Since, z is a single digit number.
So, this sum can be either 9 or 18.
∴ 9 + z = 9
⇒ z = 0 and 9 + z = 18
⇒ z = 9
So, values of z can be either 0 or 9.
That’s why there are two answers to this question.
Question 2.
“I take a number that leaves a remainder of 8 when divided by 12.1 take another number which is 4 short of a multiple of 12. Their sum will always be a multiple of 8”, claims Snehal. Examine his claim and justify your conclusion.
Solution:
Let the first number be N1 and the second number be N2.
According to the question,
N1 = 12m + 8, for some integer m
N2 = 12M – 4, for some integer n
∴ Sum of N1 and N2 = 12m + 8 + 12n – 4
⇒ N + N2 = 12 (m + n) + 4
= 12 k + 4, where k = m + n
= 8 k + (4k + 4)
For the sum to be multiple of 8, 4k + 4 must be a multiple of 8.
Let us consider different values of k.
If k = 0, then N1 + N2 = 8 × 0 + (4 × 0 + 4) = 4, which is not multiple of 8.
If k = 1, then N1+ N2 = 8 × 1 + (4 × 1 + 4) = 8 + 8 = 16, which is a multiple of 8.
If k = 2, then N1 + N2 = 8 × 2 +(4 × 2 + 4) = 16 + 12 = 28, which is not multiple of 8.
If k = 3, then N1 + N2 = 8 × 3 + (4 × 3 + 4)
= 24 + 16 = 40,
which is a multiple of 8.
∴ The sum of two numbers is not always a multiple of 8.
Hence, Snehal’s claim is not always true.
Question 3.
When is the sum of multiples of 3, a multiple of 6 and when is it not? Explain the different possible cases, and generalise the pattern.
Solution:
Let us represent the two multiples of 3 as 3n and 3m, where n and m are integers.
∴ Sum = 3n + 3m = 3(n + m)
v A number is a multiple of 6, if it is a multiple of both 2 and 3.
Since, sum = 3(n + m) is already a multiple of 3.
So, we only need to check whether it is also a multiple of 2 (i.e. even number or not).
The term 3(n + m) will be an even number if and only if the term (n + m) is an even number.
The sum of two integers (n + m) is even in two scenarios.
1. When both n and m are even.
2. When both n and m are odd.
Question 4.
Sreelatha says, “I have a number that is divisible by 9. If I reverse its digits, it will still be divisible by 9”.
(i) Examine if her conjecture is true for any multiple of 9.
Solution:
A number is divisible by 9 if the sum of its digits is divisible by 9.
When we reverse the digits of number then sum of its digits remains the same.
Since, the sum of the digits remains the same and the original number is divisible by 9, the reversed number will also have a sum of digits divisible by 9.
Thus, it will also be divisible by 9.
So, Sreelatha’s conjecture is true for any multiple of 9.
(ii) Are any other digit shuffles possible such that the number formed is still a multiple of 9?
Solution:
Yes, any shuffle of the digits of a number will result in a new number that has the same set of digits and therefore, the same sum of digits.
∴ If the original number is a multiple of 9, any number formed by shuffling its digits will also be a multiple of 9 because the sum of its digits will remain divisible by 9.
Question 5.
If 48a23b is a multiple of 18, list all possible pairs of values for a and b.
Solution:
We know that, if a number is a multiple of 18 then it is multiple of both 2 and 9.
For 48a23h to be a multiple of 2, the last digit b must be an even number.
So, b must be 0, 2, 4, 6, 8.
For 48a23h to be a multiple of 9, the sum of its digits must be a multiple of 9.
Sum of digits of 48a23b = 4 + 8 + a+ 2+ 3 + b
= 17+a + b …(i)
So, 17 + a + b must be a multiple of 9.
If b = 0, then 17+a + 0=17+o, [from Eq. (i)]
which is a multiple of 9.
So, a must be 1.
∴ (a,b) = (1, 0)
If b = 2, then 17 + a + 2 = 19 + a, which is a multiple of 9. So, a must be 8.
∴ (a, b) = (8, 1)
If b = 4, then 17 + a + 4 = 21 + a, which is a multiple of 9. So, a must be 6.
(a, b) = (6, 4)
If b = 6, then 17 + a + 6 = 23 + a, which is a multiple of 9. So, a must be 4.
∴(a, b) = (4, 6)
If = 8, then 17 + a + 8 = 25 + a, which is a multiple of 9. So, a must be 2.
∴(a, b) = (2, 8)
Question 6.
If 3p7q8 is divisible by 44, list all possible pairs of values for pand q.
Solution:
We know that, if a number is divisible by both 4 and 11 then it is divisible by 44.
Divisiblity by 4
A number is divisible by 4 if its last two digits are divisible by 4.
So, q8 must be divisible by 4.
If q = 0, then 08 is divisible by 4.
If q = 1, then 81 is not divisible by 4.
If q = 2, then 28 is divisible by 4.
If q = 3, then 38 is not divisible by 4.
If q = 4, then 48 is divisible by 4.
If q = 5, then 58 is not divisible by 4.
If q = 6, then 68 is divisible by 4.
If q = 7, then 78 is not divisible by 4.
If q = 8, then 88 is divisible by 4.
If q = 9, then 98 is not divisible by 4.
∴ The possible values for q are 0, 2, 4, 6, 8.
Divisibility by 11
A number is divisible by 11, if the difference between the sum of the digits at odd places and the sum of the digits at even places of the number is either 0 or divisible by 11.
Here, sum of its digits at odd places = 8 + 7 + 3 = 18
Sum of its digits at even places = q + p
Difference in sums = 18 – p – q
So, 18 – p – q must be a multiple of 11.
Possible multiples of 11 are 0, 11, 22, …………
Since, p and q are single digits.
So, maximum value of p + q is 9 + 9 = 18
and minimum value of p + q is 0 + 0 = 0
Therefore, 18 – p – q can only be 0 or 11.
Case 1 If 18 – p – q = 0
⇒ p + q = 18
Also, q = 0, 2, 4, 6, 8
If q = 8 then p + 8 = 18
⇒ p = 10 (not possible)
In the same way, we can find that there is no solution for other values of q in this case.
Case 2 If 18 – p – q = 11
⇒ p + q = 7
Also, q = 0, 2, 4, 6, 8
If q = 0, then p + 0 = 7 ⇒ p = 7
If q = 2, then p + 2= 7 ⇒ p = 5
If q = 4, then p + 4 = 7 ⇒ p = 3
If q = 6, then p + 6 = 7 ⇒ p = 1
If q = 8, then p + 8 = 7 ⇒ p = -1 (not possible)
Possible pairs of (p, q) are (7, 0), (5, 2), (3, 4), (1, 6).
Question 7.
Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3 and the third number is a multiple of 4.
Are there more such numbers? How often do they occur?
Solution:
Let the three consecutive numbers be n, n + 1 and n + 2. Since, first number is a multiple of 2.
So, n = 2m1, for some whole number m,.
Also given, second number is a multiple of 3.
So, n + 1 = 2m1 + 1 = 3m2, for some whole number m2
⇒ 2m1 = 3m2 – 1
⇒ Numbers of the form 2ml are one less than a multiple of 3
= m1 is one more than the multiples of 3.
∴ m1 = 3p + 1 for some whole number p Also, given, third number is a multiple of 4.
So, n + 2 = 2m1 +2 = 2(m1 + 1) is a multiple of 4, which is possible only when m1 + 1 is even i.e. m1 is odd.
Now, on combining m1 = 3p + 1 and m1 is odd, we get 3p is even.
⇒ p is even
Let us say p = 2k
Then, m1 = 3(2 k) + 1 = 6k + 1
Hence, n = 2m1 = 2 (6k + 1) = 12k + 2
Therefore, the required consecutive numbers are
12k + 2, 12k + 3, 12k + 4, where k = 0, 1, 2, ……
Such triples exist infinitely many times and appear after every 12 numbers:
(2, 3, 4), (14, 15, 16), (26, 27, 28), …………
Question 8.
Write five multiples of 36 between 45,000 and 47,000. Share your approach with the class.
Solution:
To find multiple of 36 between 45000 and 47000, first divide the lower bound of the range by 36 to find the starting multiple. Then, add 36 repeatedly to find subsequent multiples until we reach the upper bound of the range.
Now, on dividing 45000 by 36, we get
\(\frac{1}{2}\) = 1250
∴ First multiple = 36 × (1250 + 1) = 36 × 1251 = 45036
Second multiple = 36 × (1250 + 2) = 36 × 1252 = 45072
Third multiple = 36 × (1250 + 3) = 36 × 1253 = 45108
Fourth multiple = 36 × (1250 + 4) = 36 × 1254 = 45144
Fifth multiple = 36 × (1250 + 5) = 36 × 1255 = 45180
Therefore, any five multiples of 36 between 45000 and 47000 are 45036, 45072, 45108, 45144 and 45180.
Question 9.
The middle number in the sequence of 5 consecutive even numbers is 5p. Express the other four numbers in sequence in terms of p.
Solution:
Given, the middle number in the sequence of 5 consecutive even number = 5p
We know that consecutive even numbers differ by 2.
Even number immediately before 5p = 5p -2
Even number immediately before 5p – 2 =5p – 4
Even number immediately after 5p = 5p + 2
Even number immediately after 5p + 2 = 5p + 4
Therefore, other four numbers in the sequence are 5p – 4, 5p – 2, 5p + 2, 5p + 4.
Question 10.
Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.
Solution:
We know that, a number is divisible by 15 if it is divisible by both 3 and 5 and a number is divisible by 6 if it is divisible both 2 and 3.
Let 6-digit number be abcdef.
For abcdef to be divisible by 15, f must be 0 or 5 and (a + b + c + d + e + f) must be divisible by 3.
The reversed number is fedcba.
For fedcba to be divisible by 6, a must be even and (f + e + d + c + b + a) must be divisible by 3.
Since, a must be even (for the reversed number to be divisible by 2) and f must be 0 or 5 (for the original number to be divisible by 5), we can choose values that satisfy these.
Ths sum of digits (a + b + c + d + e + f) being divisible by 3 is required for both the original and reversed number when is consistent.
Let’s choose f = 0 (to satisfy divisibility by 5 for the original number).
Let’s choose a = 2 (to satisfy divisibility by 2 for the reversed number).
Now we need to place b, c, d, e such that the sum of digits (2 + b + c + d + e + 0) is divisible by 3.
Let’s pick simple digits to make the sum divisible by 3.
For example, Let b = 1, c = 1, d = 1, e = 1
The-sum of digits would be 2 + 1 + 1 + 1 + 1 + 0 = 6, which is divisible by 6.
So, the original number is 211110.
Reversed number is 011112 i.e. 11112
∴ Required 6-digit number = 211110
Question 11.
Deepak claims, “There are some multiples of 11 which, when doubled, are still multiples of 11.
But other multiples of 11 don’t remain multiples of 11 when doubled”. Examine if his conjecture is true, explain your conclusion.
Solution:
Part 1 Let’s take multiples of 11 be 11k.
Double of 11k = 2 × 11k = 22k = 11 × (2k)
Since, 22k is expressed as 11 × (an integer).
So, its is always a multiple of 11.
Part 2 Based on the analysis of part 1, this part of the claim is false. Any multiple of 11, when doubled, will always be a multiple of 11.
Question 12.
Determine whether the statements below are ‘Always True’, ‘Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.
(ii) The sum of three consecutive even numbers will be divisible by 6.
(iii) If abcdef is a multiple of 6, then badcef will be a multiple of 6.
(iv) 8 (7b – 3) – 4 (lib + 1) is a multiple of 12.
Solution:
(i) Let the multiple of 6 be 6 k1, and multiple of 3 be 3k2, where k1 and k2 are integers.
∴ Product = 6k1 × 3k2 = 18k, k2
= 9 × (2k1k2), which is a multiple of 9.
So, the given statement is always true.
(ii) Let the three consecutive even numbers 2n, 2n + 2 and 2 n + 4.
∴ Sum = 2n + 2n + 2 + 2n + 4
= 6n + 6
= 6(n + 1), which is divisible by 6.
So, the given statement is always true.
(iii) A number is a multiple of 6, if it is divisible by both 2 and 3.
For abcdef to be a multiple of 6, f must be an even digit and the sum of its digits must be divisible by 3.
For badcef to be a multiple of 6, f must be an even digit. The sum of its digits (b + a + d + c + e + f) is the same as (a + b + c + d + e + f).
So, if the original sum was multiple of 3, the new sum will also be a multiple of 3.
So, if abcdef is a multiple of 6, then badcef will also be a multiple of 6.
So, the given statement is always true.
(iv) We have, 8 (7h – 3) – 4 (11 b + 1)
= 56b – 24 – 44b – 4 = 12b – 28
Here, 12b is always a multiple of 12, but 28 is not a multiple of 12.
12b – 28 is not always a multiple of 12.
So, the given statement is never true.
Question 13.
Choose any 3 numbers. When is their sum divisible by 3? Explore all possible cases and generalise.
Solution:
When we divide any number by 3, the possible remainders are 0,1, 2.
So, any number is either of the form 3k, 3k + 1 or 3k + 2, where k is any whole number.
Case 1: All three numbers give remainder 0.
Example Let three numbers are 3, 6, 9.
∴ Sum = 3+ 6 + 9 = 18, which is divisible by 3.
Case 2: All three numbers give remainder 1
Example Let three numbers are 4, 7, 10.
∴ Sum = 4 + 7 + 10=21, which is divisible by 3.
∴ Remainder = 0
Case 3: All three humbers give remainder 2.
Example Let three numbers are 5, 8, 11.
Sum = 5 + 8 + 11 = 24, which is divisible by 3.
∴ Remainder = 0
Case 4:The numbers give remainder 0, 1 and 2, respectively.
Example Let three numbers are 3, 4, 5.
Sum = 3 + 4 + 5 = 12, which is divisible by 3.
Case 5:Two give remainder 1 and one gives 2.
Example Let three numbers are 4, 7, 5.
∴ Sum = 4+7 + 5 = 16, which is not divisible by 3.
Case 6 : Two give remainder 2 and one gives 1.
Example Let three numbers are 5, 8, 4.
∴ Sum = 5+ 8 + 4 = 17, which is not divisible by 3.
Case 7: One gives remainder 0 and two give 1.
Example Let three numbers are 3, 4, 7.
∴ Sum = 3 + 4 + 7 = 14, which is not divisible by 3.
Case 8: One gives remainder 0 and two give 2.
Example Let three’riumbers are 3, 5, 8.
∴ Sum = 3 + 5+8 = 16, which is not divisible by 3. Therefore, the sum of three numbers is divisible by 3 if the sum of their remainders is divisible by 3.
That means either all remainders are same.
(0 + 0 + 0, 1 + 1 + 1 + 2 + 2 + 2 = divisible by 3)
or the remainder are 0, 1, 2 .(0 + 1 + 2 = 3, divisible by 3).
So, either all remainders are same or the remainders are 0, 1,2.
Question 14.
Is the product of two consecutive integers always multiple of 2? Why? What about the product of three consecutive integers? Is it always a multiple of 6? Why or why not? What can you say about the product of 4 consecutive integers? What about the product of five consecutive integers?
Solution:
Yes, the product of two consecutive integers is always a multiple of 2. In any pair of consecutive integers, one integer must be even and other must be odd.
When an even number is multiplied by any other integer, the result is always even and it is divisible by 2.
Yes, the product of three consecutive integers is always a multiple of 6. Among three consecutive integers, at least one is a multiple of 2 and exactly one is a multiple of 3. Since, the product contains a factor of 2 and a factor of 3. So, it must be a multiple of 6.
The product of 4 consecutive integers is always a multiple of 24. Among any four consecutive integers, at least two are even numbers, one of which is a multiple of 4.
So, the product is a multiple of 8.
And at least one is multiples of 3.
Since, the product contains factors of 8 and 3.
So, it must be multiple of their LCM, which is 24.
The product of five consecutive integers is always a multiple of 120, respectively.
Among any five consecutive numbers, at least two are even numbers, one of which is a multiple of 4, atleast one is a multiple of 3 and exactly one is a multiple of 5.
So, the product is a multiple of 2 × 3 × 4 × 5 = 120
Question 15.
Solve the cryptarithms
(i) EF × E = GGG
(ii) WOW × 5 = MEOW
Solution:
(i) We have,EF × E = GGG
For EF × E, to result in a three-digit number GGG.
So, GGG must be a multiple of 111.
(i.e. 111, 222, 333,…)
If GGG = 111, then we find 111 = 37 × 3.
This fits the criteria of unique digits and satisfies all the conditions.
Here, E= 3, E = 7 and G = L
(ii) We have, WOW × 5 = MEOW
When any digit is multiplied by 5, then we find resultant number ends in either 0 or 5.
Therefore, W = 0 or 5.
If W = 0, then MEOW would end in zero.
However, WO W starts with W, so W can not be 0 because WOW is a three-digit number.
Therefore, W = 0 is not possible.
Thus, W = 5
Now, 505 × 5 = ME05
The possible values of O are 0, 1, 2, 3, 4, 6, 7, 8, 9. Since, W is 5, so O can not be 5.
If O = 0,505 × 5 = 2525
If O = 1,515 × 5 = 2575
If O = 2,525 × 5 = 2625
If O = 3,535 × 5 = 2675
If O = 4,545 × 5 =2725
If O = 6,565 × 5 = 2825
If O = 7,575 × 5 =2875
If O = 8,585 × 5 = 2925
If O = 9,595 × 5 =2975
Here, only 2 conditions are satisfied.
i. e.525 × 5 = 2625 and 575 × 5 = 2875
Therefore, O can be 2 or 7.
Case I: 525 × 5 = 2625
On comparing with WOW × 5 = MEOW,
we get M = O = 2, but distinct letters must represent distinct digits.
So, O ≠ 2 and this case is not possible.
Case II: 575 × 5 = 2875
On comparing with WOW × 5 = MEOW,
we get M = 2, E = 8,0 = 7 and W = 5.
Hence, W = 5,0 = 7,M = 2 and E = 8 give the solution of cryptarithms
WOW × 5 = MEOW.
Question 16.
Which of the following Venn diagrams captures the relationship between the multiplies of 4, 8, and 32?
Solution:
(iv) Since, 32 = 8 x 4
∴ Any multiple of 32 is also a multiple of 8 and 4. Also, 8 = 4×2
∴ Any multiple of 8 is also a multiple of 4 and 4. So, the Venn diagram (iv) correctly represents the relationship.