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1 5 8 11 = 24: 15 solutions

Difficulty: HardOpen in Solver

1 × 5 + 8 + 11 = 24

  1. 1 × 5 = 5
  2. 8 + 11 = 19
  3. 5 + 19 = 24

15 solutions

  • 1 × 8 + 5 + 11 = 24Steps
    1. 1 × 8 = 8
    2. 5 + 11 = 16
    3. 8 + 16 = 24
  • 1 × 11 + 5 + 8 = 24Steps
    1. 1 × 11 = 11
    2. 5 + 8 = 13
    3. 11 + 13 = 24
  • 5 ÷ 1 + 8 + 11 = 24Steps
    1. 5 ÷ 1 = 5
    2. 8 + 11 = 19
    3. 5 + 19 = 24
  • 5 + 8 + 11 ÷ 1 = 24Steps
    1. 5 + 8 = 13
    2. 11 ÷ 1 = 11
    3. 13 + 11 = 24
  • 5 + 11 + 8 ÷ 1 = 24Steps
    1. 5 + 11 = 16
    2. 8 ÷ 1 = 8
    3. 16 + 8 = 24
  • 11 + 1 × (5 + 8) = 24Steps
    1. 5 + 8 = 13
    2. 1 × 13 = 13
    3. 11 + 13 = 24
  • 1 × (11 + 5 + 8) = 24Steps
    1. 5 + 8 = 13
    2. 11 + 13 = 24
    3. 1 × 24 = 24
  • (11 + 5 + 8) ÷ 1 = 24Steps
    1. 5 + 8 = 13
    2. 11 + 13 = 24
    3. 24 ÷ 1 = 24
  • 11 + (5 + 8) ÷ 1 = 24Steps
    1. 5 + 8 = 13
    2. 13 ÷ 1 = 13
    3. 11 + 13 = 24
  • 8 + 1 × (5 + 11) = 24Steps
    1. 5 + 11 = 16
    2. 1 × 16 = 16
    3. 8 + 16 = 24
  • 8 + (5 + 11) ÷ 1 = 24Steps
    1. 5 + 11 = 16
    2. 16 ÷ 1 = 16
    3. 8 + 16 = 24
  • 5 × (8 − 1) − 11 = 24Steps
    1. 8 − 1 = 7
    2. 5 × 7 = 35
    3. 35 − 11 = 24
  • 5 + 1 × (8 + 11) = 24Steps
    1. 8 + 11 = 19
    2. 1 × 19 = 19
    3. 5 + 19 = 24
  • 5 + (8 + 11) ÷ 1 = 24Steps
    1. 8 + 11 = 19
    2. 19 ÷ 1 = 19
    3. 5 + 19 = 24

Solving hints

  • Build a factor pair: split the four numbers into two groups and try to make a factor pair of 24 — 3 × 8, 4 × 6, or 12 × 2 — then multiply the two groups.

  • No factor pair divides cleanly? Switch to sums: make two numbers add up to 24, or try shapes like (a ± b) × (c ± d) and (a + b + c) × d.

Related hands

FAQ

Can 1 5 8 11 make 24?

Yes. Every one of them is listed above with step-by-step working.

How many solutions does 1 5 8 11 have?

15 — the list above is exhaustive, so every mathematically distinct way to combine the numbers is shown.

How difficult is 1 5 8 11?

It is rated “Hard” on our four-tier scale, based on whether whole-number intermediate steps suffice, how many distinct solutions exist, and whether the hand includes a two-digit card (10–13).