Question

Take a 3 digit number ${}^{\prime }ab{c}^{\prime }$. If we add $abc,cab$ and$bca$ together then the resulting sum is divisible by:

• A. 11
• B. 111
• C. 37
• D. 111 and 37

Answer 1

The correct option is D

111 and 37

The numbers can be written as:
$abc=100a+10b+c$
$cab=100c+10a+b$
$bca=100b+10c+a$
Adding all we get
$Sum=(100a+10a+a)+(100b+10b+b)+(100c+10c+c)$
$Sum=111a+111b+111c$
$Sum=111(a+b+c)$
$=37\times 3(a+b+c)$
$\frac{sum}{111}$ = (a + b + c)
$\frac{sum}{37}$= 3 (a + b + c)
The number will be divisible by $111$ and$37$.