Determine the minimum value of a positive integer N such that:
N ends with 56, and:
N is divisible by 56 and:
The sum of the digits of N is equal to 56.

Extra Challenge: Solving this puzzle without a computer program.

N ends with 56, so
N=100*m+56
N is divisible by 56, and lcm(56,100) = 1400, therefore
N=1400*k+56

The sum of N's digits is 56, so the sum of digits of 14*k is 45.

99999 ≠ 14*k , so we need at least six digits.  Let the six digits be abcdef.

We want
100000a + 10000b + 1000c + 100d + 10e + f = 0 (mod 14), with a+b+c+d+e+f = 45

Substituting a' = 9-a, b' = 9-b etc.:

a'+b'+c'+d'+e'+f' = 9
7 = -2a' + 4b' + 6c' + 2d' - 4e' + f' (mod 14)

Skipping a few steps, the best solution is abcdef = 298998

N = 29899856

Posted by Tristan on 2013-10-05 01:27:53