Determine the total number of seven-digit base 10 positive integers divisible by 11 and having the sum of their digits equal to 59. (No computer programs)

The number is divisible by 11 if the sum of the odd-position digits minus the sum of the even-position digits is a multiple of 11.

o + e = 59
o - e = 11*k

k = 0: 2*o = 59: no solution
k = 1: 2*o = 70: o = 35: e = 24
k = -1: 2*o = 48: o = 24: e is odd: no solution
k = 2: 2*o = 81: no solution
k = -2: 2*o = 37: no solution
k = 3: 2*o = 92: o = 46: e is odd: no solutions
k = -3: 2*o = 26: o = 13: e = 46: bound exceeded
...  no solutions here
k = 5: 2*o = 114: o = 57: e = 2: bounds exceeded

That leaves k = 1 as the only possibility.

The sum of the odd-position digits is 35 and the sum of the even-position digits is 24.

There are four odd-position digits and three even-position digits.

a+c+e+g = 35
b+d+e = 24

There are four possibilities for aceg: 9998, 9989, 9899 and 8999.

The permutations within each of 996, 987 and 888, are the possibilities for bde. The first has 3 permutations; the second has 6 permutations and the last has only one. Altogether there are 10.

That makes 4*10 = 40 the answer to the puzzle.

Posted by Charlie on 2013-06-07 12:06:46