AMC 10 · 2009 · #11
Grade 7 arithmeticPick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question asks "how many ways," which points to Tool #2 (Make a Systematic List). But listing all 7-digit numbers blindly is hopeless, so first use Tool #5 (Look for a Pattern) to see that a palindrome is fixed by just its first half plus the middle digit. Then Tool #3 (Eliminate Possibilities) settles which digit is forced into the middle. What remains is a small, clean count of arrangements that Tool #2 finishes off.
See the palindrome's mirror structure
A 7-digit palindrome is d₁d₂d₃d₄d₃d₂d₁, so d₁, d₂, d₃ each appear twice while only d₄ stands alone in the middle.
A palindrome is just a mirror, so you only ever choose the front half and the middle — the back half copies itself.
A palindrome is a mirror, so you only ever choose the front half and the middle.
▸ Why?
Reversing the number lands it exactly on itself, so nothing about the digits is lost or gained.
▸ Why?
Each front digit pairs with exactly one back digit, so choosing the front settles the back.
Force the middle digit
The 2s and 3s come in pairs, but three 5s is odd, so the lone middle is forced to d₄ = 5, leaving {d₁,d₂,d₃} = {2,3,5}.
Only the digit with an odd count can stand alone in the center; the even-count digits must pair up.
2.OA.C.3Eliminate PossibilitiesCount the arrangements
Three different values in three front slots give 3! = 6 orders, and each builds exactly one palindrome — choice (A).
Once the middle is nailed down, the answer is simply how many ways three different digits can line up in three spots.
7.SP.C.8Make A Systematic ListA palindrome is a mirror, so pick only the front half and the middle — the odd-count digit takes the center, and the rest is just counting the ways to order what's left.
- See the palindrome's mirror structure
- Force the middle digit
- Count the arrangements