AMC 10 · 2009 · #11
Grade 7 arithmeticHow many 7-digit palindromes (numbers that read the same backward as forward) can be formed using the digits 2, 2, 3, 3, 5, 5, 5?
Pick an answer.
AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Using exactly the seven digits $2$, $2$, $3$, $3$, $5$, $5$, $5$, count how many $7$-digit palindromes (numbers that read the same left-to-right and right-to-left) can be built.
Givens: The number has $7$ digits.; It must read the same backward as forward (a palindrome).; The digits used are exactly $2$, $2$, $3$, $3$, $5$, $5$, $5$ — two $2$s, two $3$s, and three $5$s.; Answer choices: (A) $6$, (B) $12$, (C) $24$, (D) $36$, (E) $48$.
Unknowns: The number of different $7$-digit palindromes that use this exact set of digits.
Understand
Restated: Using exactly the seven digits $2$, $2$, $3$, $3$, $5$, $5$, $5$, count how many $7$-digit palindromes (numbers that read the same left-to-right and right-to-left) can be built.
Givens: The number has $7$ digits.; It must read the same backward as forward (a palindrome).; The digits used are exactly $2$, $2$, $3$, $3$, $5$, $5$, $5$ — two $2$s, two $3$s, and three $5$s.; Answer choices: (A) $6$, (B) $12$, (C) $24$, (D) $36$, (E) $48$.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
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.
Execute — Answer: A
4.OA.C.5 Step 1 See the palindrome's mirror structure
- A $7$-digit palindrome reads the same both ways, so it has the form $d_1 d_2 d_3 d_4 d_3 d_2 d_1$.
- Position $1$ copies position $7$, position $2$ copies position $6$, and position $3$ copies position $5$.
- That means three of the digits ($d_1$, $d_2$, $d_3$) each appear twice — once in each half — while the middle digit $d_4$ appears just once.
- So the whole number is decided by the first four slots.
💡 A palindrome is just a mirror, so you only ever choose the front half and the middle — the back half copies itself.
2.OA.C.3 Step 2 Force the middle digit
- Because $d_1$, $d_2$, $d_3$ each get used twice, they account for an even count of every digit outside the middle.
- But our supply is two $2$s, two $3$s, and three $5$s — and three is odd.
- The only way to use up an odd number of $5$s is to place one $5$ alone in the middle.
- The $2$s and $3$s come in even amounts, so neither can sit in the middle.
- So $d_4 = 5$, which leaves one $2$, one $3$, and one $5$ to fill the pair-slots $d_1$, $d_2$, $d_3$.
💡 Only the digit with an odd count can stand alone in the center; the even-count digits must pair up.
7.SP.C.8 Step 3 Count the arrangements
- Now just arrange the three leftover values $2$, $3$, $5$ into the front slots $d_1$, $d_2$, $d_3$; the back half and the middle are already set, so each front arrangement makes exactly one palindrome.
- Three distinct values in three slots give $3 \times 2 \times 1 = 6$ orders: $235$, $253$, $325$, $352$, $523$, $532$.
- Each one builds a valid palindrome, for example $235 \to 2355532$.
- That is $6$ palindromes, choice (A).
💡 Once the middle is nailed down, the answer is simply how many ways three different digits can line up in three spots.
4.OA.C.5 A $7$-digit palindrome reads the same both ways, so it has the form $d_1 d_2 d_3 2.OA.C.3 Because $d_1$, $d_2$, $d_3$ each get used twice, they account for an even count 7.SP.C.8 Now just arrange the three leftover values $2$, $3$, $5$ into the front slots $d Review
Reasonableness: List all six and check each uses two $2$s, two $3$s, three $5$s: $2355532$, $2535352$, $3255523$, $3525253$, $5235325$, $5325235$. Every one is a palindrome and uses the exact digit supply, and there are no others because the front trio $(d_1,d_2,d_3)$ has only $6$ orderings. So $6$ is right, matching (A).
Alternative: Count with the permutations formula instead. The front three slots hold the multiset $\{2,3,5\}$, all distinct, so the number of orderings is $\tfrac{3!}{1!\,1!\,1!} = 6$. Since the middle digit is forced to be $5$ and the back half is a forced mirror, this single factor is the whole count — again $6$.
CCSS standards used (min grade 7)
4.OA.C.5Generate a number or shape pattern following a given rule (Reading the palindrome rule as the mirror pattern $d_1 d_2 d_3 d_4 d_3 d_2 d_1$ so only the front half plus middle need choosing.)2.OA.C.3Determine whether a group of objects has an odd or even number (Using the odd count of $5$s (three) to force a single $5$ into the middle while the even-count $2$s and $3$s pair up.)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Counting the $3! = 6$ ways to arrange the three distinct leftover digits in the front slots via an organized list.)
⭐ A 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.
⭐ A 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.
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