IPMAT Indore 2026 · Quantitative Ability · Question 34

IPMAT Indore 2026 previous year question on Divisibility Rules, with the answer and a step-by-step solution.

Divisibility RulesNumber SystemMultiple choiceMedium
If m is a positive integer then the values of k for which 6m+k cannot be a perfect square are
  1. A
    2 and 5
  2. B
    3 and 4
  3. C
    1 and 5
  4. D
    1, 2 and 5
Pick an option, then check.
Answer & solution

Answer: Option A

Step 1 — ask what squares look like modulo 6. Square every residue:

n \mod 6

0

1

2

3

4

5

{n}^{2}\mod 6

0

1

4

3

4

1

So a perfect square is always congruent to 0,1,3 or 4 modulo 6 — never 2 and never 5.

Step 2 — apply it. 6m+k\equiv k (\mod 6), so 6m+k can only be a perfect square if k\in \{0,1,3,4\}.

Therefore it cannot be a square when

k=2\text{or}k=5

Answer: Option 1 — 2 and 5

Shortcut. Build the table of squares mod 6 once — it is six quick calculations and it answers the whole question. The same trick (squares mod 4 are 0 or 1; mod 8 are 0, 1 or 4) resolves a large family of IPMAT number-system questions.

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