Patterns & Reasoning

Problem

Numbers divisible by several are common multiples

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.
Operations
Your answer
How to solve
Strategy Eliminate Possibilities — 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.
1STEP 1

Keep only the multiples of 4

Among 18, 16, 24, 21, only 16 and 24 are multiples of 4 — 18 and 21 leave a remainder.

16 = 4 × 4, 24 = 4 × 6
2STEP 2

Of those, keep the multiples of 3

Of 16 and 24, only 24 is a multiple of 3 (24 = 3 × 8) — 16 is eliminated.

24 = 3 × 8, 16 ≠ 3 × (whole number)
Answer
24
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.
Takeaway

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

  • Keep only the multiples of 4
  • Of those, keep the multiples of 3
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