CSAT Solved Papers/ 2021/Q67

2021 CSAT — Q67

Quant Number theory 2.5 marks Medium

Consider the following statements:

The sum of $5$ consecutive integers can be $100$ .
The product of three consecutive natural numbers can be equal to their sum.

Which of the above statements is/are correct?

A 1 only
B 2 only
C Both 1 and 2 Answer
D Neither 1 nor 2

Worked rationale

Statement 1: five consecutive integers $n, n+1, n+2, n+3, n+4$ sum to $5n + 10 = 5(n+2)$ . Set equal to $100$ : $n + 2 = 20 \Rightarrow n = 18$ , giving $18+19+20+21+22 = 100$ . Possible — correct.

Statement 2: three consecutive naturals $a, a+1, a+2$ have product $a(a+1)(a+2)$ and sum $3(a+1)$ . Equate:

a(a+1)(a+2) = 3(a+1) \;\Rightarrow\; a(a+2) = 3 \;\Rightarrow\; a^2 + 2a - 3 = 0 \;\Rightarrow\; (a-1)(a+3) = 0.

The natural-number root is $a = 1$ : product $1\cdot 2\cdot 3 = 6$ , sum $1+2+3 = 6$ . Possible — correct.

Both statements hold.

Answer: (c) Both 1 and 2.

Why the other options miss

A

stopped one step short: verifies the sum- $100$ case but doesn’t solve $a(a+2) = 3$ , missing the $a = 1$ solution to Statement 2.
B

reached for the wrong rule: mis-sums the five consecutive integers (e.g. $5n + 5$ ) and wrongly rejects Statement 1.
D

an arithmetic slip: errs in both setups, rejecting two true existence claims.

Specialist insight

Both parts are existence claims (“can be”), so a single witness settles each. Statement 1 reduces to $5(n+2) = 100$ — instantly solvable; Statement 2 cancels the shared factor $(a+1)$ to leave the quadratic $a(a+2) = 3$ , whose natural root $a = 1$ gives the famous $1\cdot2\cdot3 = 6 = 1+2+3$ . The trap on part 2 is not cancelling $(a+1)$ and getting bogged in a cubic; factor it out and the answer is one line.

The trap, in one line

$5(n+2)=100\Rightarrow n=18$ (yes); $a(a+2)=3\Rightarrow a=1$ , $1\cdot2\cdot3=6=1+2+3$ (yes) $\Rightarrow$ both $\Rightarrow$ (c).

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