2 Marks

Question 12 Marks

Find A as required:
The number A08 is divisible by 4 and 9

Answer

08 is divisible by 4, so A08 is divisible by 4
If A = 1 then the sum of digits will be 9 which is divisible by 9
The number will be 108 and A = 1

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Question 22 Marks

Find A as required:
The greatest 2 digit number 9A is divisible by 2

Answer

A number is divisible by 2 if it is an even number.
Greatest 2 digit even number is 98.
∴ A = 8

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Question 32 Marks

Find A as required:
The least number 567A is divisible by 3

Answer

A number is divisible by 3 if the sum of its digits is divisible by 3 Sum of digits of $567 A=5+6+7+A$
$
=18+ A
$
$\therefore 18$ is divisible by 3
$\therefore$ A maybe 0
The number will be 5670

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Question 42 Marks

Find A as required:
The greatest 3 digit number 9A6 is divisible by 6

Answer

A number is divisible by 6 if it is divisible by both 2 and 3
9A6 is even and so divisible by 2
If A = 9 then the sum of digits will be 24 which is divisible by 3.
The number will be 996 and A = 9

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Question 52 Marks

Is 173 a prime? Why?

Answer

Yes, 173 is a prime.
Because it has only 1 and itself as factors.

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Question 62 Marks

Find whether the number 564872 is divisible by 88. (use of the test of divisibility rule for 8 and 11 will help!)

Answer

872 is divisible by 8 .
Hence $56+872$ is divisible by 8
$
(2+8+6)-(7+4+5)=16-16=0
$
$\therefore 564872$ is divisible by 11
$\therefore 564872$ is divisible by 88

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Question 72 Marks

The product of any three consecutive numbers is always divisible by 6. Justify this statement with an example

Answer

Yes.
Because one of every two consecutive integers is even and so the product of three consecutive integers is even and divisible by 2.
Also one of every 3 consecutive integers is divisible by 3.
Product of any three consecutive integers is divisible by 6.
Example: 5 × 6 × 7

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Question 82 Marks

Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single digit numbers.

Answer

Prime numbers between 10 and 20 are 11, 13, 17 and 19
Sum = 11 + 13 + 17 + 19 = 60
60 is divisible by 1, 2, 3, 4, 5 and 6.

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Question 92 Marks

The LCM of two numbers is $6$ times their HCF. If the HCF is $12$ and one of the numbers is $36,$ then find the other number

Answer

$H C F=12$
$\text { Product of two numbers }=L C M \times H C F$
$36 \times \text { other number }=72 \times 12$
$\text { Other number }=\frac{72 \times 12}{36}$
$\text { Other number }=24$

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Question 102 Marks

Find the LCM set of numbers using prime factorisation method.
6, 9

Answer

6,9
Prime factorisation of $6=2 \times 3$
Prime factorisation of $9=3 \times 3$
Product of common factors $=3$
Product of other factors $=2 \times 3=6$
$\operatorname{LCM}(6,9)=3 \times 6=18$

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Question 112 Marks

The sum of any three odd natural numbers is odd. Justify this statement with an example

Answer

True, as we know, that, "the sum of any three odd numbers is always an odd number".
Example: $3+7+9=19$ is odd.

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Question 122 Marks

Write the smallest and the biggest three digit composite number

Answer

Smallest three-digit number $=100$, divisible by $2,4,5,10,20,25,50$ and so it is composite biggest three-digit number is 999 , composite.
Smallest is 100
Biggest is 999

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Question 132 Marks

I am a two digit prime number and the sum of my digits is 10. I am also one of the factors of 57. Who am I?

Answer

Two-digit prime numbers with a sum of digits 10 are 19, 37, 73 of these factors of 57 is 19.
The number is 19.

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Question 142 Marks

Find A as required:
The number 225A85 is divisible by 11

Answer

5 + A + 2 – (8 + 5 + 2) = 7 + A – 15
= – 8 + A
∴ A = 8

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Maths 2 Marks

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