What is the remainder when (1^3+2^3+3^3 +…+95^3) Is divided by 97 ?

236

[victoramezcua]

Since you're reducing the expression (mod 97), notice that 95=-2 (97) and that (-2)^3=-2^3. This cancels 2^3. Similarly, the cubes from 3 to 47 is canceled by the cubes of k-97, k=48, ..., 94. So, only 1^3=1 is left (mod 97).

[Armcollector]

There is an identity (1+2+3+...+n)^2 = 1^3+...+n^3, using that and the sum of natural numbers we get that the sum = (95*96/2)^2 = 95*96*95*96/4, mod 27 this becomes -2*-1*-2*24 = -96 = 1 mod 97

[heltongama]

Excellent! This is thinking outside the box.

[meurdesoifphilippe]

Let B the sum. Then B = (B + 96^3) - 96^3. The sum between parentheses is equal to the square of (1+2+3+... +96) so a multiple of 97. Now 96 = -1 mod 97 and -(-1)^3=1, so B = 1 mod 97. The demander is 1.

[antoningarcic]

Well, that's pretty nonsense. Only 2492298. And 249298 divided by 97 = 1?. Perhaps 97x97x97 is missing from the assignment?

[oceanmath]

âm thanh toán học không có lời giải thích.

[김은영-dj]

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