(solved)Question 24 SSC-CGL 2020 March 3 Shift 1

Question 24
SSC-CGL 2020 Mar 3 Shift 1

200 leaves a remainder 8 when divided by x. How many such x are possible?

Solution

Divisor is always greater than the remainder, so we have to find all x greater than 8

If 200 ÷ x leaves 8 as remainder, then (200 - 8) should be fully divisible by 8.

so x should be a factor of 192

Hence, we have to count factors of 192

How many factors does 192 have?

To count the factors first write prime factors as:

$\displaystyle 192 = 2^6 \times 3^1$

*no. of factors = (6 + 1) x (1 + 1) = 14

we have to count only those that are greater than 8, so exclude these six: 1, 2, 3, 4, 6 and 8

$\displaystyle \therefore 14 - 6 = 8\:\underline{Ans}$

*Note on factors

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 them to be six: 1, 2, 3, 4, 6, 12