The largest number that exactly divides each number of the sequence $1^5 - 1, 2^5 - 2, 3^5 - 3, \dots, n^5 - n, \dots$ is

## Concept & Formula To find the largest number that divides every term in a sequence generated by a formula like $n^5 - n$, you must find the Greatest Common Divisor (GCD) of the first few valid non-zero terms. ## Step-by-Step Solution * The sequence is defined by the term $T_n = n^5 - n$. Let us calculate the first few terms by substituting $n = 1, 2, 3, \dots$ * For $n = 1$: $$1^5 - 1 = 0$$ *(0 is divisible by everything, so move to the next term).* * For $n = 2$: $$2^5 - 2 = 32 - 2 = 30$$ * For $n = 3$: $$3^5 - 3 = 243 - 3 = 240$$ * For $n = 4$: $$4^5 - 4 = 1024 - 4 = 1020$$ * We now need the largest number that divides 30, 240, and 1020. * Notice that the first non-zero term is 30. The greatest divisor of a sequence cannot be larger than its smallest non-zero term. * Let us verify if 30 divides the next terms: $240 \div 30 = 8$ and $1020 \div 30 = 34$. It works perfectly. ## Exam Strategy & Shortcut The largest common divisor for a sequence of this type is universally dictated by its first non-zero result. Simply calculate $n = 2 \Rightarrow 2^5 - 2 = 30$. The answer cannot exceed 30. Look at the options: 1, 15, 30, 120. The largest option that is $\le 30$ and divides 30 is 30 itself. ## Common Pitfall A frequent mistake is calculating the GCD algebraically by factoring $n^5 - n$ into $n(n-1)(n+1)(n^2+1)$, which proves divisibility by 6 (since $n(n-1)(n+1)$ is divisible by $3! = 6$), but missing the hidden factor of 5 that makes the full divisor 30. Value substitution avoids this blind spot entirely. ## Final Answer **Therefore, the correct answer is 30.**