NTT Data Divisibility Quiz -1

A number which is divisible by 2 and 3 both is also divisible by 6

235686 unit digit is 6 so it is divisible by 2

2+3+5+6+8+6 = 30 which is divisible by 3.

The factors of 45 are 3, 5, and 9.

In 752310,

From the above-given options, 7+5+2+3+1+0=18

As 18 is completely divisible by 3 and 9.

last digit is 0 so it is divisible by 5

The given condition can be written as = n= 5x + 3 where x is an integer.

We add 2 to both the sides in the equation

n + 2 = 5x + 5

= n +2 = 5(x +1)

The above equation states that n + 2 divided by 5 gives a remainder zero. Therefore, the term is n + 2.

All the numbers that end with a 0 are completely divisible by 10. Thus, only 85280 is divisible by 10.

All the numbers that end with 5 or 0 are completely divisible by 5. Thus, only 452195 is divisible by 5.

Let the missing digit be x

Sum of digits =5+6+2+x+1+4+9+7=34+x

For 34+x to be divisible by 9, x must be replaced by 2.

There are 11 numbers divisible by 9. (100 mod 9 = 11)

The numbers are : 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99.

But in number 99,

9+9 is not equal to 9.

So, the answer is 10 numbers.

We need to calculate the LCM of 3, 8, and 16 to find out the smallest integer.

LCM= 48

Therefore, the smallest common integer is 48.

Let the number be 65k + 32

= (13 x 5k + 26) + 6 = 13 x (5k + 2) + 6

Therefore the required remainder is 6

n = 8 k + 3

multiply all terms by 6

6 n = 6(8 k + 3) = 8(6k) + 18

Write 18 as 16 + 2 since 16 = 8 * 2.

= 8(6k) + 16 + 2

Factor 8 out.

= 8(6k + 2) + 2

The above indicates that if 6n is divided by 8, the remainder is 2.