The unit digit of (3 × 1 + 4 × 2 × 1 + 5 × 3 × 2 × 1 + … + 20 × 18 × 17 × ⋯ × 2 × 1) is

Correct — D, 5. Read the pattern first. The terms are 3 x 1, then 4 x 2 x 1, then 5 x 3 x 2 x 1, and the last is 20 x 18 x 17 x … x 2 x 1 — so the general term is (k + 2) multiplied by k factorial, with k running from 1 up to 18. Now use the fact that 5! = 120 and every factorial after it contains both 2 and 5, so every term from k = 5 onwards ends in 0 and adds nothing to the unit digit. That leaves four terms: 3 x 1! = 3, 4 x 2! = 8, 5 x 3! = 30 and 6 x 4! = 144. Their last digits are 3, 8, 0 and 4, which add to 15. The unit digit of the whole sum is therefore 5.

• (a)2 — Does not arise from any reading of the series. Adding the last digits of the four contributing terms gives 15 whichever order they are taken in.
• (b)3 — The unit digit of the first term alone. Stopping after 3 x 1 ignores the three further terms that still end in a non-zero digit.
• (c)4 — The last digit of the fourth term, 6 x 4! = 144, taken by itself. The sum of all four contributing last digits is 3 + 8 + 0 + 4 = 15, so the answer is 5.

Last-digit questions are solved by throwing away everything that cannot affect the last digit. Factorials are the cleanest example: 5! = 120 already contains a 2 and a 5, so 5! and every larger factorial is a multiple of 10 and contributes a zero. Any long factorial series therefore collapses to its first four terms.

Two things can go wrong here and neither is arithmetic. The first is misreading the pattern: the leading multiplier runs 3, 4, 5 … 20 while the factorial runs 1!, 2!, 3! … 18!, and the last printed term confirms the offset of two. The second is stopping too early — the third term, 5 x 3! = 30, ends in zero, and a candidate who treats a zero as the signal that the series has gone quiet will drop the 144 that follows it and land on 5 by luck or on 3 by error. Write the four terms out; it takes ten seconds.

• The general term of the series is (k + 2) x k!, with k from 1 to 18; the last term is 20 x 18!.
• 5! = 120, so every factorial from 5! upwards is a multiple of 10 and ends in zero.
• The four contributing terms are 3, 8, 30 and 144.
• 3 + 8 + 0 + 4 = 15, so the unit digit of the sum is 5.
• The first four factorials are 1, 2, 6 and 24.

• Getting the offset wrong and reading the general term as (k + 1) x k!.
• Treating the first term ending in zero as the point where the series stops mattering.
• Trying to evaluate 20 x 18! rather than working modulo 10.

A unit-digit item on a long series, where all the work is in identifying which few terms are not multiples of ten.

• practice — not a real PYQ

  What is the unit digit of 1! + 2! + 3! + … + 100!?

  • (a)0
  • (b)1
  • (c)3
  • (d)5

  Answer(c) 3 — only 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 contributes, since every factorial from 5! on ends in zero.

• practice — not a real PYQ

  How many trailing zeros does 25! have?

  • (a)4
  • (b)5
  • (c)6
  • (d)7

  Answer(c) 6 — the count of factors of five is 25 divided by 5, which is 5, plus 25 divided by 25, which is 1.