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

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 05 Dec 2022 Shift 4)

Answer 1

Given:

9-digit number = 83P93678Q

Divisor = 72

Concept Used:

Divisibility of 8 = Last three digits should be divisible by 8.

Divisibility of 9 = Sum of digits is divisible by 9.

Calculation:

As the divisor 72, is divisible by 8 and 9, so the divisibility will be checked.

For divisible by 8,

78Q should be divisible by 8, so, Q should be 4 as 784 is divisible by 8.

For divisible by 9,

⇒ 8 + 3 + P + 9 + 3 + 6 + 7 + 8 + 4 = 48 + P

For being divisible by 9, the nearest number to be added is 6 which gives 54.

Now, \(\sqrt{P^2+Q^2+12}=\sqrt{6^2+4^2+12}\)

⇒ \(\sqrt{36+16+12}=\sqrt{64}=8\)

Therefore, the required value is 8.

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