Test: Playing With Numbers- 2

For a number to be divisible by 9, the sum of its digits must be divisible by 9. 

Here, 2 + 4 = 6. Next multiple of 9 is 9. 

Therefore, we will add 9 -  6 = 3. So,  the value of x = 3.

The last digit is 5, the number is divisible by 5. 

Since 

311636 − 10 = 311626 ÷ 7 = 44568,

 the number is divisible by 7 also.   

A number divisible by 3 and 4 is divisible by 12.

ANSWER :- b

Solution :- If a number is divisible by 5 then it's unit digit must be 0 or 5. so if we need the remainder of 0 when divided by 5 then the numbers unit digit must be 0 or 5.

10 ×5 = 50
6 = 6
⇒ 50 + 6 = 56

B+1=8
B=7, A+7=-1 .So A=4 and B=7

Divisible by 2 and 3 both.

Since every multiple of 2 leaves no remainder when divided by 2.Therefore N can be ,0,2,4,6 or 8

We can write acb=100*a+10*c+b which say that a is at hundred’s place, c is at ten’s place and b is at one’s place.

Since every multiple of 5 ends with either 5 or 0. Therefore, the unit digit of N must be either 0+4=4 or 5+4=9. Hence the answer is either 4 or 9.