Problem 2 · 2025 AMC 8
Easy
Arithmetic & Operations
place-valuenumber-systems
Show answer
Answer: B — 10,423.
Show hints
Hint 1 of 2
Unlike our digits, here position doesn't matter — a symbol is worth the same wherever it appears. So you only need to count how many of each kind.
Still stuck? Show hint 2 →
Hint 2 of 2
Tally each symbol type and add their table values, just like the example (three ∩ arches and two | strokes made 32). Notice there's no thousands symbol — what does that force the thousands digit to be?
Show solution
Approach: count each symbol type, then add the values
- The key idea: in this Egyptian system a symbol's value doesn't depend on where it sits — you just count how many of each kind there are. So tally: one 10,000, four 100s, two 10s, three 1s.
- Add the values: 10,000 + 4×100 + 2×10 + 3×1 = 10,000 + 400 + 20 + 3 = 10,423.
- Sanity check: there's no thousands symbol, so the thousands digit must be 0 — that alone rules out everything except 10,423.
Mark:
· log in to save