IPMAT Indore 2022 · Quantitative Ability · Question 39

Step 1 — reduce the two conditions to one. Divisible by both 2 and 3 means divisible by \text{lcm}\left( 2,3 \right)=6.

Step 2 — count multiples of 6 in range. The integers are four-digit and strictly greater than 1000, so they run from 1001 to 9999. The multiples of 6 in that range are 1002,1008,\ldots ,9996:

\frac{9996-1002}{6}+1=\frac{8994}{6}+1=1499+1=1500

Step 3 — remove those also divisible by 5. A multiple of 6 that is also a multiple of 5 is a multiple of \text{lcm}\left( 6,5 \right)=30. In range these are 1020,1050,\ldots ,9990:

\frac{9990-1020}{30}+1=\frac{8970}{30}+1=299+1=300

Step 4 — subtract.

1500-300=1200

Answer: Option 3 — 1200

Shortcut. Combine, count, then subtract the overlap. “Divisible by 2 and 3” is “divisible by 6”; “also by 5” is “divisible by 30”.

\underset{\text{multiples of}6}{\underset{⏟}{1500}}-\underset{\text{multiples of}30}{\underset{⏟}{300}}=1200

A faster route worth knowing: among multiples of 6, exactly one in five is also a multiple of 5, so the survivors are 4/5 of them — 4/5\times 1500=1200 directly.

Option (d) 1500 is the trap for anyone who forgets the “not by 5” clause, and it is also the count you get if you misread the range. Note that “greater than 1000” and “four-digit” happen to agree here, so the boundary costs nothing.