IPMAT Indore 2023 · Quantitative Ability · Question 2

Step 1 — find where the terms stop mattering. Note that

5!=120=15\times 8

so 5! is divisible by 15. And every factorial beyond it contains 5! as a factor:

6!=6\times 5!, 7!=7\times 6\times 5!, \ldots

So every term from 5! to 95! leaves remainder 0 and can be ignored.

Step 2 — add only the first four terms.

1!+2!+3!+4!=1+2+6+24=33

Step 3 — reduce.

33=2\times 15+3 \Rightarrow \text{remainder}3

Answer: 3

Shortcut. Find the cut-off, then ignore 91 of the 95 terms. Since 15=3\times 5, any factorial containing both a 3 and a 5 is divisible by 15 — that is every k! with k\ge 5.

So the sum collapses to 1+2+6+24=33\equiv 3 (\mod 15).

The general principle: in any “1!+2!+\cdots +n! modulo m” question, only the first few terms survive. Find the smallest k with m∣k! and stop there — the upper limit (95 here) is pure intimidation.