SSC CGL 2022 PYQ — Question 52

SSC CGL - 05 Dec 2022 - Shift 4 • Quantitative Aptitude • Q52

If the 9 - digit number 83P93678Q is divisible by 72, then what is the value of P2+Q2+12?\sqrt{P^2+Q^2+12}?

Show answer & explanation

Correct answer: C

Explanation

The correct answer is option 3 i.e. 8.

83P93678Q

The above number is divisible by 72 so it is also divisible by 9 and 8.

8 divisibility rule = The last three digits will divide by 8

9 divisibility rule = The sum of the digit will also divide by 9

So Q = 4

Because 784 is divisible by 8.

sum of the digits = 8 + 3 + P + 9 + 3 + 6 + 7 + 8 + 4 = P + 48

Put P = 6

6 + 48 = 54 

That is divisible by 9.

So,  P2+Q2+12=62+42+12\sqrt{P^2+Q^2+12} = \sqrt{6^2+4^2+12}

= 64=8\sqrt{64}=8