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Supreme Seventy Sum (Posted on 2009-10-30) Difficulty: 3 of 5
Determine the maximum value that is obtained by multiplying together a set of positive integers which are all different and whose sum is 70.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Some Thoughts General method | Comment 3 of 4 |
Basically you want the numbers to be as small as possible.

2+3+4+5+6+7+8+9+10+11 = 65 (note consecutive)
These numbers give the maximum product for a sum of 65.
Increase the 11 to a 12 and you have the numbers that give the maximum product for a sum of 66.
Increase the 10 to an 11 and the sum is 67.
Increase the 9 to a 10 and the sum is 68.
Increase the 8 to a 9 and the sum is 69.
Increase the 7 to an 8 and the sum is 70.

2+3+4+5+6+8+9+10+11+12 = 70 and the product is maximal.

This continues to
3+4+5+6+7+8+9+10+11+12 = 75
The trick is getting to the next level
3+4+5+6+7+8+9+10+11+13 = 76
2+3+4+5+6+7+8+9+10+11+12 = 77
then shift the numbers up one by one as before.

  Posted by Jer on 2009-10-30 15:23:27
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