CSAT Solved Papers/ 2022/Q78

2022 CSAT — Q78

Quant Number theory 2.5 marks Medium

The sum of three consecutive integers is equal to their product. How many such possibilities are there?

  1. A Only one
  2. B Only two
  3. C Only three Answer
  4. D No such possibility is there

Worked rationale

Let the integers be n−1, n, n+1n-1,\ n,\ n+1. Their sum is 3n3n; their product is (n−1)n(n+1)=n3−n(n-1)n(n+1) = n^3 - n. Set equal:

3n=n3−n  ⇒  n3−4n=0  ⇒  n(n2−4)=0  ⇒  n∈{0, 2, −2}.3n = n^3 - n \;\Rightarrow\; n^3 - 4n = 0 \;\Rightarrow\; n(n^2 - 4) = 0 \;\Rightarrow\; n \in \{0,\ 2,\ -2\}.

Each gives a valid integer triple:

  • n=0n = 0: (−1,0,1)(-1, 0, 1) — sum 00, product 00 ✓
  • n=2n = 2: (1,2,3)(1, 2, 3) — sum 66, product 66 ✓
  • n=−2n = -2: (−3,−2,−1)(-3, -2, -1) — sum −6-6, product −6-6 ✓

Three possibilities.

Answer: (c) Only three.

Why the other options miss

  • A
    missed a case: finds (1,2,3)(1,2,3) and stops, missing the zero and negative triples.
  • B
    missed a case: catches the positive and zero cases but forgets that negatives are integers too.
  • D
    solved the wrong question: assumes “integers” means positive only and rejects all.

Specialist insight

“Integers,” not “natural numbers” — so 00 and negatives are in play. Factoring n3−4n=n(n−2)(n+2)n^3 - 4n = n(n-2)(n+2) exposes all three roots at once. The examiner banks on candidates restricting to positives and reporting “one.” Always solve the full polynomial and keep every integer root.

The trap, in one line

3n=n3−n⇒n(n2−4)=0⇒n=0,±23n = n^3 - n \Rightarrow n(n^2-4)=0 \Rightarrow n = 0, \pm 2 — three triples ⇒\Rightarrow (c).

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