CAT 2022 Slot 2 · Quant · Question 7

On day n, one particle in every n makes a copy, so the count grows by a factor of 1+\frac{1}{n}=\frac{n+1}{n}.

Starting from 100 on day 1, day 2 multiplies by \frac{3}{2}, giving 150.

Day 3 multiplies by \frac{4}{3}, giving 200.

Day 4 multiplies by \frac{5}{4}, giving 250.

The pattern is a telescoping product: after day m the count is 100\times \frac{3}{2}\times \frac{4}{3}\times \cdots \times \frac{m+1}{m}.

Every numerator cancels the next denominator, leaving 100\times \frac{m+1}{2}.

Setting this equal to 1000 gives \frac{m+1}{2}=10.

So m+1=20 and m=19 — the first option.

Shortcut. The counts run 100, 150, 200, 250, … — a plain arithmetic progression adding 50 a day from day 1. Reaching 1000 needs 18 steps of 50, so it happens on day 19.