Question 57

The sum of 3-digit numbers abc, cab and bca is not divisible by:

Solution

Sum of the numbers = $$\left(100a+10b+c\right)+\left(100c+10a+b\right)+\left(100b+10c+a\right)$$

= $$100\left(a+b+c\right)+10\left(a+b+c\right)+\left(a+b+c\right)$$

= $$111\left(a+b+c\right)$$

= $$37\times3\times\left(a+b+c\right)$$

Sum is not divisible by 31.

Hence, the correct answer is Option C

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