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Divisibility and Digit Sum (Posted on 2013-06-07) Difficulty: 3 of 5
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)

No Solution Yet Submitted by K Sengupta    
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Solution solution | Comment 4 of 8 |

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
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