CAT 2023 Slot 2 · Quant · Question 22

The first sequence starts at 13 and steps by 6, so its terms are 13, 19, 25, and so on.

Every such term leaves the same remainder as 13 on division by 6, and 13=2\times 6+1, so each term is of the form 6k+1.

The second sequence starts at 15 and steps by 7, so its terms are 15, 22, 29, and so on.

Here 15=2\times 7+1, so each term is of the form 7j+1.

A common term must be 1 more than a multiple of 6 and 1 more than a multiple of 7.

So it is 1 more than a common multiple of 6 and 7, that is of the form 42t+1.

Checking the smallest such value that actually appears in both lists: 43=13+6\times 5 and 43=15+7\times 4, so 43 is genuinely common.

So the common terms are 43, 85, 127, and so on, rising by 42 each time.

The largest three-digit value of 42t+1 needs 42t\le 998, so t\le \frac{998}{42}\approx 23.76.

Taking t=23, the value is 42\times 23+1=966+1=967.

The next term would be 967+42=1009, which has four digits, so 967 is the answer.

Shortcut. Both sequences land on 1 more than a multiple of their step, so common terms repeat every \text{LCM}\left( 6,7 \right)=42. Then just subtract 42 repeatedly from 1009, or divide 999 by 42 and adjust — 967 appears in a single step.