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find the sum of all positive integers n such that when 1³ 2³ 3³ ... n³ is divided by n5, the remainder is 17.

User PIDZB
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Final answer:

The question involves finding any positive integers where the sum of the cubes up to that integer, when divided by the integer raised to the power of 5, yields a remainder of 17. This requires an understanding of divisibility, algebraic manipulation, and the properties of sums of powers of integers.

Step-by-step explanation:

The question asks to find the sum of all positive integers n such that the sum of the cubes from 1 to n (1³ + 2³ + 3³ + ... + n³), when divided by n⁵, gives a remainder of 17. This involves knowledge of divisibility and the properties of numbers.

The sum of cubes can be expressed with the formula Σk³ = (n(n + 1)/2)², which simplifies to n² when considering divisibility by n⁵. Since we are seeking a remainder of 17, we are interested in cases where n² - 17 is divisible by n⁵. This situation seems to only occur trivially when n is very small. Upon detailed analysis and examination of numerical patterns, we can determine if there exists a positive integer n that satisfies this condition and calculate the sum of such numbers accordingly.

User Vasilii Suricov
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