Question:
The sum of all the numbers formed by the digits x, y and z of the number xyz is divisible by
(a) 11
(b) 33
(c) 37
(d) 74
Solution:
(c) We have, xyz + yzx + zxy
= (100x + 10y + z) + (100y + 10 z+ x) + (100z+ 10x + y) …(i)
= 100x + 10x + x + 10y + 100y + y + z+ 100z+ 10Z
= 111x + 111y + 111z = 111 (x + y + z)
= 3 x 37 x (x + y + z)
Hence, Eq. (i) is divisible by 37, but not divisible by 11,33 and 74.