The remainder, when 1 + (1 × 2) + (1 × 2 × 3) + … + (1 × 2 × 3 × … × 500) is divided by 8, is

Correct — A, 1. The series is 1! + 2! + 3! + … + 500!. From the fourth term on, every factorial contains both 2 and 4 among its factors, so each is divisible by 8 and contributes nothing to the remainder: 4! = 24 = 8 × 3, 5! = 120 = 8 × 15, and so on for all 497 remaining terms. That leaves only 1! + 2! + 3! = 1 + 2 + 6 = 9. Dividing 9 by 8 gives quotient 1 and remainder 1. The size of the series is decoration; the answer is fixed by its first three terms.

• (b)2 — 2 is the value of the second term alone. The three surviving terms add to 9, and no reading of 9 modulo 8 gives 2.
• (c)3 — 3 would follow from taking 3! as 8 instead of 6, which makes the surviving sum 11 and the remainder 3. The factorial of 3 is 1 × 2 × 3 = 6.
• (d)4 — 4 matches no step of the working. It is the remainder of 9 on division by 5, not by 8.

A remainder question over a long series is a divisibility question about its terms. Once a term is a multiple of the divisor it can be dropped, because adding a multiple of 8 never changes a remainder modulo 8. For factorials the cut-off is easy to locate: n! for n ≥ 4 contains the factors 2 and 4, whose product is 8, so every factorial from 4! upward vanishes modulo 8.

The 500 terms are there to frighten. A candidate who starts adding factorials loses the question to arithmetic, while one who asks 'from where do the terms stop mattering?' finishes in two lines. The habit worth carrying is to test small n first: 4! = 24 is already a multiple of 8, and every later factorial has 4! as a factor, so the tail is dead. Note the wording of the series — it is written out as 1 + (1 × 2) + (1 × 2 × 3) rather than in factorial notation, which is the paper's way of stating the same sum without assuming the symbol.

• If a ≡ r (mod m) and b ≡ 0 (mod m), then a + b ≡ r (mod m) — multiples of the divisor can be discarded.
• n! is divisible by 8 for every n ≥ 4, because 4! = 24 already carries the factor 8.
• 1! + 2! + 3! = 9, and 9 = 8 × 1 + 1, so the remainder is 1.
• The same technique with divisor 5 would keep 1! + 2! + 3! + 4! = 33, since 5! is the first factorial divisible by 5.
• The last digit of 1! + 2! + … + n! is 3 for every n ≥ 4, by the same argument applied to 10.

• Beginning to add the factorials term by term instead of looking for the cut-off.
• Reading 3! as 8 or 3 × 3, which changes the surviving sum.
• Assuming the answer depends on 500 being even, or on the final term at all.

A remainder item disguised as a long summation — the examiner is checking whether you know that multiples of the divisor can be thrown away before any arithmetic starts.

• practice — not a real PYQ

  What is the remainder when 1! + 2! + 3! + … + 100! is divided by 5?

  • (a)0
  • (b)1
  • (c)3
  • (d)4

  Answer(c) 3 — factorials from 5! on are multiples of 5, leaving 1 + 2 + 6 + 24 = 33, and 33 = 5 × 6 + 3.

• practice — not a real PYQ

  What is the unit digit of 1! + 2! + 3! + … + 50!?

  • (a)0
  • (b)1
  • (c)3
  • (d)5

  Answer(c) 3 — every factorial from 5! on ends in 0, so the unit digit comes from 1 + 2 + 6 + 24 = 33.