CSAT Solved Papers/ 2023/Q54

2023 CSAT — Q54

Quant Number theory 2.5 marks Hard

$40$ children are standing in a circle and one of them (say child- $1$ ) has a ring. The ring is passed clockwise. Child- $1$ passes on to child- $2$ , child- $2$ passes on to child- $4$ , child- $4$ passes on to child- $7$ and so on. After how many such changes (including child- $1$ ) will the ring be in the hands of child- $1$ again?

A 14
B 15 Answer
C 16
D 17

Worked rationale

The step grows by $1$ each pass: $+1$ (to child- $2$ ), $+2$ (to child- $4$ ), $+3$ (to child- $7$ ), $\dots$ After $n$ passes the position is

1 + (1 + 2 + \cdots + n) = 1 + \frac{n(n+1)}{2} \pmod{40},

with positions read mod $40$ (position $41 \equiv 1$ ). The ring returns to child- $1$ when

\frac{n(n+1)}{2} \equiv 0 \pmod{40} \iff n(n+1) \equiv 0 \pmod{80}.

Search for the smallest $n$ : $80 = 2^4 \times 5$ , so the consecutive pair $n, n+1$ must together supply a factor $16$ and a factor $5$ .

$n = 15$ : $15 \times 16 = 240 = 80 \times 3$ ✓ (here $16$ gives the $2^4$ , $15$ gives the $5$ ).

No smaller $n$ works ( $n \le 14$ never makes $n(n+1)$ a multiple of $80$ ). Check: $\frac{15\cdot 16}{2} = 120 \equiv 0 \pmod{40}$ , position $= 1 + 120 \equiv 1$ . ✓

Answer: (b) 15.

Why the other options miss

A

off by one: stops one pass early; $\frac{14\cdot 15}{2} = 105 \equiv 25 \pmod{40}$ , not back at child- $1$ .
C

off by one: overshoots by one; $\frac{16\cdot 17}{2} = 136 \equiv 16 \pmod{40}$ .
D

an arithmetic slip: mis-solves $n(n+1)\equiv 0 \pmod{80}$ , taking the next coincidence instead of the first.

Specialist insight

The cumulative displacement after $n$ passes is the triangular number $\frac{n(n+1)}{2}$ , so “return to start” becomes $\frac{n(n+1)}{2} \equiv 0 \pmod{40}$ , i.e. $n(n+1)\equiv 0 \pmod{80}$ . Factor the modulus ( $80 = 2^4\cdot 5$ ) and find the first consecutive pair carrying both $16$ and $5$ — that is $15\cdot 16$ . Recognising the triangular-number structure turns a tedious step-by-step simulation into a one-line divisibility hunt.

The trap, in one line

Displacement $= \tfrac{n(n+1)}{2}$ ; return needs $n(n+1)\equiv 0 \pmod{80}$ , first at $n = 15$ ( $15\cdot 16 = 240$ ) $\Rightarrow$ (b).

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