Problem 11 · 2002 AMC 8
Medium
Algebra & Patterns
perfect-squaredifference-of-squares
A sequence of squares is made of identical square tiles. Each square's edge is one tile longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?
Show answer
Answer: C — 13.
Show hints
Hint 1 of 2
Don't build the whole 7×7 square — think about what you *add* when growing a square by one. You glue tiles along two sides and one corner.
Still stuck? Show hint 2 →
Hint 2 of 2
Going from side 6 to side 7, you add a strip down one side (6), a strip across the top (6), plus 1 corner tile: 6 + 6 + 1.
Show solution
Approach: the nth square uses n² tiles
- A square n tiles on a side uses n² tiles, so the seventh needs 7² = 49 and the sixth needs 6² = 36. The extra tiles: 49 − 36 = 13.
- *The pattern to keep:* growing a square from side n to n+1 always adds an L-shaped border of (n) + (n) + 1 = 2n + 1 tiles — an odd number. Here that's 2·6 + 1 = 13, no squaring needed.
Another way — difference of squares:
- The gap between consecutive squares factors instantly: 7² − 6² = (7 + 6)(7 − 6) = 13 × 1 = 13.
- *Reusable:* any a² − b² = (a + b)(a − b), which turns a scary subtraction of big squares into one easy product.
Mark:
· log in to save