# Recreational Mathematics 1

A mate of mine, Jim Grime, posted this on twitter.

Been asked if there is a six digit number abcdef such that a+b+c+d+e+f = y, ab+cd+ef=10y and abc+def=100y. Any solutions?

Well, I’ve been missing having my recreational mathematics head on, and I thought that there couldn’t be many solutions, so I gave it a whack.

a+b+c+d+e+f = y (A)

ab+cd+ef = 10y means 10(a+c+e) + b+d+f = 10y (B)

abc+def = 100y means 100(a+d) + 10(b+e) + c+f = 100y (C)

Substituting A into B gives

10(a+b+c+d+e+f) = 10(a+c+e) + b+d+f

Cancelling the a, c, e gives

10(b+d+f) = b+d+f

Because we’re working with digits, which are whole and non-negative, this means that b, d, f = 0

Now substitute A into C

100(a+b+c+d+e+f) = 100(a+d) + 10(b+e) + c+f

Remove the digits that we know = 0

100(a+c+e) = 100a + 10e + c

Cancel the a

100c + 100e = c + 10e

The only non-negative solutions to this are c, e = 0

The a has cancelled out of everything, so can be any digit, so all solutions are of the form a00000.