3-Digit Numbers Divisible by 2 and 3, Not 4 and 9

JEE Main 2025 — 24 January, Morning Shift. Previous Year Question.

Problem

The number of 3-digit numbers which are divisible by 2 and 3 but not divisible by 4 and 9 is _.

Key insight. “Divisible by 2 and 3” really means “divisible by their LCM, 6,” and “not divisible by 4 and 9” means “not divisible by their LCM, 36.” Once the problem is rephrased this way, it becomes two arithmetic-progression counting problems and a subtraction.

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Approach

A number divisible by both 2 and 3 is exactly a number divisible by $\mathrm{lcm}(2,3) = 6$, and a number divisible by both 4 and 9 is exactly a number divisible by $\mathrm{lcm}(4,9) = 36$. So the count needed is:

$$(\text{3-digit multiples of }6) – (\text{3-digit multiples of }36)$$

since every multiple of 36 is automatically a multiple of 6, subtracting removes exactly the numbers that shouldn’t be counted. Both counts are found the same way: identify the first and last 3-digit multiple, treat them as an arithmetic progression, and use the AP term-count formula.

Solution

Step 1 — Count the 3-digit multiples of 6

The smallest 3-digit multiple of 6 is $102$, and the largest is found by checking downward from $999$: since $999 = 6 \times 166 + 3$, the remainder is 3, so subtracting it gives $996 = 6 \times 166$ as the largest multiple of 6 under 1000.

These multiples form an AP with first term $a = 102$, common difference $d = 6$, and last term $996$. Using $a_n = a + (n-1)d$:

$$996 = 102 + (n-1)\times 6 \implies n – 1 = \frac{894}{6} = 149 \implies n = 150$$

So there are 150 three-digit multiples of 6.

Step 2 — Count the 3-digit multiples of 36

The smallest 3-digit multiple of 36 is $108$ ($36 \times 3$), and the largest is $972$ ($36 \times 27$).

This is an AP with $a = 108$, $d = 36$, last term $972$:

$$972 = 108 + (n-1)\times 36 \implies n – 1 = \frac{864}{36} = 24 \implies n = 25$$

So there are 25 three-digit multiples of 36 — and since every multiple of 36 is also a multiple of 6 and of 4 and 9, every one of these 25 numbers is exactly a number that needs to be excluded.

Step 3 — Subtract

$$150 – 25 = 125$$

Answer

Common mistakes

• Checking divisibility by 4 and 9 separately with inclusion–exclusion, rather than noticing “divisible by both 4 and 9” is just “divisible by 36.” Working with 4 and 9 separately turns a one-line subtraction into a much longer inclusion–exclusion calculation.
• Getting the largest multiple wrong by rounding instead of finding the exact remainder. For 999 divided by 6, the remainder is 3, not 0, so 999 itself is not a multiple of 6 — the true largest multiple is $999 – 3 = 996$.

Practise next

• How many 4-digit numbers are divisible by both $2$ and $5$, but not by $100$? The same count-the-multiples-and-subtract method applies.