# (solved)Question 12 SSC-CGL 2020 March 4 Shift 1

When 732 is divided by x, the remainder is 12. How many such values of x are possible?
(Rev. 18-Jun-2024)

## Categories | About Hoven's Blog

,

### Question 12SSC-CGL 2020 Mar 4 Shift 1

When 732 is divided by x, the remainder is 12. How many such values of x are possible?

### Solution in Detail

Given 732 leaves 12 as remainder

So 732 - 12 = 720 is fully divisible by x

So x is a factor of 720, but x > 12 because x is given as the divisor, and divisor is always greater than the remainder. Our task is to count the factors of 720, and exclude all those that are less than or equal to 12.

To count all the factors, first get the prime factorization of 720.

$\displaystyle 720 = 2^4 \times 3^2 \times 5^1$

Total factors are [see the discussion "Note on factors" after this solution for details]

$\displaystyle = (4 + 1) \cdot (2 + 1) \cdot (1 + 1)$

$\displaystyle = 5 \cdot 3 \cdot 2 = 30$

Verify that there are indeed 30 factors of 720: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 30, 36, 40, 45, 48, 60, 72, 80, 90, 120, 144, 180, 240, 360, 720

Exclude these ten: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12

Hence, possible values

$\displaystyle = 30 - 10 = 20\:\underline{Ans}$

### Note on factors

Remember this artifice for future exams

If prime factorization of N is $\displaystyle a^xb^yc^z\text{. . . }$ then the total factors of N are $\displaystyle (x + 1)(y + 1)(z + 1)\text{ . . . }$

Take for example $\displaystyle 12 = 2^2\cdot3^1$. The number of factors of 12 will be $\displaystyle (2 + 1)\times (1 + 1) = 6$. We can indeed verify their count to be these six: 1, 2, 3, 4, 6, 12

This Blog Post/Article "(solved)Question 12 SSC-CGL 2020 March 4 Shift 1" by Parveen is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.