If the seven-digit number 52A6B7C is divisible by 33, and A, B, C are primes, then the maximum value of 2A + 3B + C is:

Question

Question

This question was previously asked in

SSC CGL 2022 Tier-I Official Paper (Held On : 12 Dec 2022 Shift 3)

Answer 1

Given:

The seven-digit number 52A6B7C is divisible by 33

Concept used:

If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.

If the sum of digits of a number is a multiple of 3, the number will be completely divisible by 3.

Calculation:

5 + 2 + A + 6 + B + 7 + C = 20 + A + B + C

Possible values of (A + B + C) = 1, 4, 7, 10, 13, 16, 19, 22, 25

Again,

5 + A + B + C - 2 - 6 - 7 = 0 or Multiple of 11

⇒ 5 + A + B + C - 15 = 0 or Multiple of 11

From the above possible values, for (A + B + C) = 10 we get 0

So, A + B + C = 10

Now,

If A = 5, B = 3, C = 2

2A + 3B + C = 10 + 9 + 2

⇒ 21

If A = 3, B = 5, C = 2

2A + 3B + C = 6 + 15 + 2

⇒ 23

If A = 3, B = 2, C = 5

2A + 3B + C = 6 + 6 + 2

⇒ 17

If A = 2, B = 5, C = 3

2A + 3B + C = 4 + 15 + 3

⇒ 22

So, The maximum value of 2A + 3B + C is 23.

∴ The required answer is 23.

Download Solution PDF