Summing cubes

Logic Level 3

Let $a_1, a_2,\ldots,a_9$ be positive integers such that the sum of their cubes is 239. Find the sum of all distinct values of $a_1 + a_2 + \cdots + a_9$ .

The answer is 23.

1 solution

Denton Young
Dec 6, 2015

There are two sets of numbers that provide a solution to the problem.

[5, 3, 3, 3, 2, 2, 2, 2, 1] and [4, 4, 3, 3, 3, 3, 1, 1, 1] both work. In both cases the sum of the numbers is 23.

How do you know these solution sets ?

Rohit Udaiwal - 5 years, 6 months ago

The largest cube used can't be 6, because 239-216 = 23, and 23 can't be expressed with 8 or fewer cubes (2,2,1,1,1,1,1,1,1 is only 9-cube solution for 23.) Since 216 =125 + 64 + 27, we can't have a 5, a 4 and a 3 simultaneously. The largest used has to be > 3, since we can't use 8 3's (= 216) and if we use 7 (=189) 239-189 = 50,which can't be expressed with 2 cubes.

so largest is either 5 or 4. With 5, 239-125 = 114, so if we use a 4 (-64 = 50) we have to use 2's and 1's, and can't express 50 with fewer than 8 additional cubes which is 1 too many. So: 5, 3 (-27 =87), 3 (-27 = 60), 3 (-27 = 33), and if we use another 3 we're left with 33-27 = 6. So 2,2,2,2,1.

with 4, 239 - 64 = 175. can't use 3's yet, so another 4 , -64 = 111. another 4 would yield 47, so we switch to 3's: 3, 3, 3, 3, 1, 1, 1.

Denton Young - 5 years, 6 months ago

There is other solution : ( 1 , 1 , 1 , 1 , 1 , 1 , 1 , 2 , 2 , 6 ) where the sum is : 17 .

mohamed aboalamayem - 5 years, 1 month ago

That uses 10 cubes, not 9.

Denton Young - 5 years, 1 month ago

Summing cubes

The answer is 23.

1 solution

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