IPMAT Indore 2022 · Quantitative Ability · Question 9

Step 1 — relate position to value. If the new sequence’s {2022}^{\text{nd}} term is the integer N, then among 1,2,\ldots ,N exactly 2022 numbers survive. The ones removed are the perfect squares up to N, and there are ⌊\sqrt{N}⌋ of them. So

N-\left\lfloor \sqrt{N}\right\rfloor=2022

Step 2 — estimate. \sqrt{2022}\approx 44.97, so N is roughly 2022+45=2067.

Step 3 — verify N=2067.

{45}^{2}=2025\le 2067<2116={46}^{2}\Rightarrow \left\lfloor \sqrt{2067}\right\rfloor=45

2067-45=2022 ✓

Step 4 — check N itself is not a square. 2067 lies strictly between {45}^{2}=2025 and {46}^{2}=2116, so it is not a perfect square and therefore really is in the sequence.

Answer: 2067

Shortcut. The rule of thumb: the nth non-square is about n+\sqrt{n}. Compute \sqrt{2022}\approx 45, add, and you have 2067 immediately.

Then do the two checks that make it rigorous, both of which take seconds:

2067-⌊\sqrt{2067}⌋=2067-45=2022 ✔

2067 is not itself a square ✔

The trap is forgetting check 2. If the estimate lands exactly on a perfect square, that number has been deleted and you must move up by one.