IPMAT Indore 2025 · Quantitative Ability · Question 39

Step 1 — pin down the units digit. For divisibility by 5 the last digit must be 0 or 5. There is no 0 available, so

\text{units digit}=5 \text{(fixed)}

That leaves \{1,3,7,8,9\} for the other places.

Step 2 — which lengths are possible? With six available digits and no repetition, numbers can be 4, 5 or 6 digits long. (A 3-digit number is at most 987, below 5000.)

Case A — 4-digit numbers. To exceed 5000 the leading digit must be at least 5. The candidates are 5,7,8,9, but 5 is already used at the end, so the leading digit \in \{7,8,9\} → 3 ways. The two middle places are filled from the 4 remaining digits:

3\times 4\times 3=36

Case B — 5-digit numbers. Every 5-digit number already exceeds 5000, so there is no restriction on the leading digit. Choose the first four places from the remaining 5 digits:

5\times 4\times 3\times 2=120

Case C — 6-digit numbers. All five remaining digits are arranged in the first five places:

5!=120

Step 3 — total.

36+120+120=276

Answer: Option 1 — 276