Question : If the seven-digit number 52A6B7C is divisible by 33, and A, B, and C are primes, then the maximum value of 2A + 3B + C is:
Option 1: 32
Option 2: 23
Option 3: 27
Option 4: 34
Correct Answer: 23
Solution :
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
Hence, the correct answer is 23.
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