Question

ABC is a 3-digit number such that its last digit Cis divisible by 2: the two digit part BC is divisible by 3: ABC tent in divisible by 4; The number of such 3 digit numbers is

Answer 1

(d) We know that, the sum of three-digit numbers taken in cyclic order can be written as 111 (a + b + c).

i.e. abc + pea + cab = 3 x 37 x (a + b + c)

Hence, the sum is divisible by 3, 37 and (a + b + c) but not divisible by 9.