Problem 2 · 2026 AMC 8
Easy
Arithmetic & Operations
symmetrycareful-counting
In the array shown below, three 3s are surrounded by 2s, which are in turn surrounded by a border of 1s. What is the sum of the numbers in the array?
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 1 | 2 | 2 | 2 | 2 | 2 | 1 |
| 1 | 2 | 3 | 3 | 3 | 2 | 1 |
| 1 | 2 | 2 | 2 | 2 | 2 | 1 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
Show answer
Answer: C — The answer is 53.
Show hints
Hint 1 of 2
Before adding 35 numbers, look at the rows: some are exact copies of each other. Can you avoid adding the same row twice?
Still stuck? Show hint 2 →
Hint 2 of 2
Lean on the mirror symmetry — top matches bottom, 2nd matches 4th — so you only add three different rows and double two of them. (Or count how many 1s, 2s, and 3s there are.)
Show solution
Approach: use the mirror symmetry — only three rows are really different
- The grid is a mirror image top-to-bottom: row 1 = row 5, and row 2 = row 4. So you only have to add up three rows, then double two of them.
- Row 1 (all 1s): 7. Row 2: 1 + five 2s + 1 = 12. Middle row: 1+2+3+3+3+2+1 = 15.
- Total = 2×7 (top & bottom) + 2×12 (2nd & 4th) + 15 (middle) = 14 + 24 + 15 = 53.
- Why this transfers: whenever a figure repeats or mirrors, add one copy of each distinct piece and multiply — the symmetry does most of the counting for you.
Another way — count how many of each number, not row by row:
- The three 3s sit in the center: 3 × 3 = 9.
- The 2s form a ring — five in row 2, five in row 4, two in the middle row = 12 twos: 12 × 2 = 24.
- Everything else is a 1. The grid is 5×7 = 35 cells, so the 1s number 35 − 3 − 12 = 20, worth 20.
- Total: 9 + 24 + 20 = 53.
Mark:
· log in to save