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

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Bonus:
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Can you find the numbers?

The answer is 4.

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