The sum of 3-digit numbers abc, bca, and cab is always divisible by:
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 37.
abc = 100a + 10b + c
bca = 100b + 10c + a
cab = 100c + 10a + b
Sum of the numbers = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)
= 111(a + b + c)
= 3 × 37(a + b + c)
Hence, the required number is 37.