IPMAT Indore 2026 · Quantitative Ability · Question 14

Step 1 — count everything, then subtract. Total selections:

\left( \genfrac{}{}{0pt}{}{9}{3}\right)=84

Step 2 — classify by powers of 2. Among 1 to 9:

odd numbers: 1,3,5,7,9 — five of them, contributing no factor of 2

2 and 6 — contribute exactly one factor of 2

4 and 8 — contribute two or more factors of 2

The product is divisible by 4 when the total power of 2 is at least 2.

Step 3 — count the failures instead. The product is not divisible by 4 only if the total power of 2 is 0 or 1:

All three odd: \left( \genfrac{}{}{0pt}{}{5}{3}\right)=10

Exactly one even, and that even is 2 or 6 (two odds alongside): 2\times \left( \genfrac{}{}{0pt}{}{5}{2}\right)=2\times 10=20

(If the single even were 4 or 8 it would already supply {2}^{2}, and two evens always give at least {2}^{2}.)

\text{Not divisible by 4}=10+20=30

Step 4 — subtract.

84-30=54

Answer: 54

Shortcut. Always count the complement here — “divisible by 4” splits into several cases, but “not divisible by 4” has just two. The key insight is that 4 and 8 are worth two factors of 2 on their own, so only 2 and 6 can appear alone and still fail.