Skip to main content
24Point

Browse all solutions

1 1 10 13 = 24: 14 solutions

Difficulty: HardOpen in Solver

1 × 1 + 10 + 13 = 24

  1. 1 × 1 = 1
  2. 10 + 13 = 23
  3. 1 + 23 = 24

14 solutions

  • 1 ÷ 1 + 10 + 13 = 24Steps
    1. 1 ÷ 1 = 1
    2. 10 + 13 = 23
    3. 1 + 23 = 24
  • 1 + 10 + 1 × 13 = 24Steps
    1. 1 + 10 = 11
    2. 1 × 13 = 13
    3. 11 + 13 = 24
  • 1 + 10 + 13 ÷ 1 = 24Steps
    1. 1 + 10 = 11
    2. 13 ÷ 1 = 13
    3. 11 + 13 = 24
  • 1 × 10 + 1 + 13 = 24Steps
    1. 1 × 10 = 10
    2. 1 + 13 = 14
    3. 10 + 14 = 24
  • 1 + 13 + 10 ÷ 1 = 24Steps
    1. 1 + 13 = 14
    2. 10 ÷ 1 = 10
    3. 14 + 10 = 24
  • 13 + 1 × (1 + 10) = 24Steps
    1. 1 + 10 = 11
    2. 1 × 11 = 11
    3. 13 + 11 = 24
  • 1 × (13 + 1 + 10) = 24Steps
    1. 1 + 10 = 11
    2. 13 + 11 = 24
    3. 1 × 24 = 24
  • (13 + 1 + 10) ÷ 1 = 24Steps
    1. 1 + 10 = 11
    2. 13 + 11 = 24
    3. 24 ÷ 1 = 24
  • 13 + (1 + 10) ÷ 1 = 24Steps
    1. 1 + 10 = 11
    2. 11 ÷ 1 = 11
    3. 13 + 11 = 24
  • 10 + 1 × (1 + 13) = 24Steps
    1. 1 + 13 = 14
    2. 1 × 14 = 14
    3. 10 + 14 = 24
  • 10 + (1 + 13) ÷ 1 = 24Steps
    1. 1 + 13 = 14
    2. 14 ÷ 1 = 14
    3. 10 + 14 = 24
  • 1 + 1 × (10 + 13) = 24Steps
    1. 10 + 13 = 23
    2. 1 × 23 = 23
    3. 1 + 23 = 24
  • 1 + (10 + 13) ÷ 1 = 24Steps
    1. 10 + 13 = 23
    2. 23 ÷ 1 = 23
    3. 1 + 23 = 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 1 10 13 make 24?

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

How many solutions does 1 1 10 13 have?

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

How difficult is 1 1 10 13?

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).