346 likes cubes

Find the minimum number of perfect cubes such that their sum is equal to 34 6 346 346^{346} .

Bonus: Can you find the numbers?


The answer is 4.

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1 solution

Mark Hennings
Mar 12, 2019

Since a 3 1 , 0 , 1 ( m o d 9 ) a^3 \equiv 1,0,-1 \pmod{9} and 34 6 346 4 ( m o d 9 ) 346^{346} \equiv 4 \pmod{9} , it follows that 34 6 346 346^{346} cannot be written as a sum of one, two or three cubes. On the other hand 34 6 346 = ( 34 6 115 × 7 ) 3 + ( 34 6 115 ) 3 + ( 34 6 115 ) 3 + ( 34 6 115 ) 3 346^{346} \; = \; \big(346^{115} \times 7\big)^3 + \big(346^{115}\big)^3 + \big(346^{115}\big)^3 + \big(346^{115}\big)^3 using the fact that 346 = 7 3 + 1 3 + 1 3 + 1 3 346 = 7^3 + 1^3 + 1^3 + 1^3 , so the desired answer is 4 \boxed{4} .

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