Numbers divisible by several are common multiples

3.OA.B.63.OA.C.7 · take · grade 3

Archetype: Divisibility and Remainder Reasoning · step in a 8-type progression

From the numbers below, find the one that can be divided evenly by $3$ and also by $4$ .

$18 \quad 16 \quad 24 \quad 21$

Show solution

Understand

Out of the numbers 18, 16, 24, and 21, we must find the single number that can be divided evenly (no remainder) by both 3 and 4.

Givens

The candidate numbers are 18, 16, 24, and 21.
The answer must divide evenly by 3.
The answer must also divide evenly by 4.

Unknowns

Which of the four numbers is divisible by both 3 and 4.

Constraints

Even division means a remainder of 0.
Exactly one of the four numbers should satisfy both conditions.

Plan

#3 Eliminate Possibilities · also uses: #2 Make a Systematic List

There is a small, genuinely finite set of four numbers and two clear logic clues (divisible by 3, divisible by 4). Testing each clue lets us cross off numbers until one survives both.

Execute

#3 Eliminate Possibilities 3.OA.C.7

Check which numbers divide evenly by 4: 18 divided by 4 leaves a remainder, 16 is 4 times 4, 24 is 4 times 6, 21 leaves a remainder. So 16 and 24 pass the 'divisible by 4' clue; 18 and 21 are eliminated.

16 = 4 \times 4, \quad 24 = 4 \times 6

Grade 3 times-tables: a number is divisible by 4 only if it appears in the 4 times table.

#3 Eliminate Possibilities 3.OA.B.6

Now test 16 and 24 against the 'divisible by 3' clue: 16 is not in the 3 times table, but 24 is 3 times 8. So 16 is eliminated and 24 survives.

24 = 3 \times 8, \quad 16 \ne 3 \times (\text{whole number})

Grade 3 unknown-factor sense: 24 divided by 3 is 8 with nothing left over, so 24 is a multiple of 3.

Answer: 24

Review

Check 24 against both clues directly: 24 divided by 3 is 8 (exact) and 24 divided by 4 is 6 (exact). Both are whole numbers, so 24 genuinely meets both conditions, and it is the only survivor.

A number divisible by both 3 and 4 is a multiple of 12. List the multiples of 12 (12, 24, 36, ...) and notice only 24 appears among the candidates.

Standards · min grade 3

3.OA.C.7 Fluently multiply and divide within 100 — Recalling the 4 times table to test divisibility by 4.
3.OA.B.6 Understand division as an unknown-factor problem — Checking divisibility by 3 by finding the missing factor.

💡 Divisible by both 3 and 4 just means it shows up in both times tables, and 24 is the one that does!