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Difficulty: MediumProperties of Integers and Divisibility

For all positive integers nn, what is the greatest integer that MUST divide n5−nn^5 - n?

  1. A
    6
  2. B
    10
  3. C
    15
  4. 30Answer
  5. E
    60

Answer

The greatest integer that must divide n5−nn^5 - n for all positive integers nn is 30.
Factoring the expression yields n5−n=(n−1)n(n+1)(n2+1)n^5 - n = (n-1)n(n+1)(n^2+1). The factor (n−1)n(n+1)(n-1)n(n+1) consists of three consecutive integers, ensuring divisibility by both 2 and 3 (and thus by 6). Furthermore, the units digit of n5n^5 is always identical to the units digit of nn for any positive integer nn, which guarantees that n5−nn^5 - n is divisible by 5. Since 6 and 5 share no common factors, n5−nn^5 - n must be divisible by 6×5=306 \times 5 = 30. Testing n=2n = 2 gives 25−2=302^5 - 2 = 30, proving that 30 is the largest integer that divides the expression for all positive integers nn.

Step-by-Step Solution

1
Factor the algebraic expression n5−nn^5 - n.
n5−n=n(n4−1)=n(n2−1)(n2+1)=(n−1)n(n+1)(n2+1)n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = (n - 1)n(n + 1)(n^2 + 1).
Factoring helps isolate product terms with known divisibility rules.
2
Analyze divisibility by 2 and 3.
The sub-expression (n−1)n(n+1)(n - 1)n(n + 1) is the product of three consecutive integers.
Among any three consecutive integers, at least one is divisible by 2 and exactly one is divisible by 3. Therefore, (n−1)n(n+1)(n - 1)n(n + 1) is always divisible by 2×3=62 \times 3 = 6.
3
Analyze divisibility by 5.
By Fermat's Little Theorem (or analyzing last-digit repeating cycles of powers), for any integer nn, n5≡n(mod5)n^5 \equiv n \pmod 5, which implies 55 divides n5−nn^5 - n.
The units digit of n5n^5 is always equal to the units digit of nn, so n5−nn^5 - n always ends in 0 or 5, making it a multiple of 5.
4
Combine the common prime factors and evaluate the maximum lower bound.
Since 2, 3, and 5 are pairwise coprime, any number divisible by 2, 3, and 5 must be divisible by 2×3×5=302 \times 3 \times 5 = 30. Evaluating at n=2n = 2 yields 25−2=302^5 - 2 = 30, showing no integer greater than 30 can divide n5−nn^5 - n for all nn.
The greatest common divisor across all generated values of n5−nn^5 - n is 30.

Key Concept

Divisibility properties of consecutive integers and power mod rules
Estimated Time:1m 30s
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Topics

Properties of Integers and Divisibility

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