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Difference between revisions of "2009 AMC 10B Problems/Problem 11"

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== Solution ==
 
== Solution ==
  
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A seven-digit palindrome is a number of the form <math>\overline{abcdcba}</math>. Clearly, <math>d</math> must be <math>5</math>, as we have an odd number of fives. We are then left with <math>\{a,b,c\} = \{2,3,5\}</math>. Each of the <math>\boxed{(A)6}</math> permutations of the set <math>\{2,3,5\}</math> will give us one palindrome.
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A seven-digit palindrome is a number of the form <math>\overline{abcdcba}</math>. Clearly, <math>d</math> must be <math>5</math>, as we have an odd number of fives. We are then left with <math>\{a,b,c\} = \{2,3,5\}</math>. There are <math>3!</math> permutations of these three numbers, since each is reflected over the midpoint we only have to count the first there. Each of the <math>\boxed{(A)6}</math> permutations of the set <math>\{2,3,5\}</math> will give us one palindrome.
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==Solution 2==
 
==Solution 2==
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This means that our answer is <math>2+2+2=\boxed{\textbf{(A)}6}</math>
 
This means that our answer is <math>2+2+2=\boxed{\textbf{(A)}6}</math>
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~yofro
  
 
== See Also ==
 
== See Also ==

Latest revision as of 16:32, 28 June 2021

Problem

How many $7$-digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$, $2$, $3$, $3$, $5$, $5$, $5$?

$\text{(A) } 6 \qquad \text{(B) } 12 \qquad \text{(C) } 24 \qquad \text{(D) } 36 \qquad \text{(E) } 48$

Solution

A seven-digit palindrome is a number of the form $\overline{abcdcba}$. Clearly, $d$ must be $5$, as we have an odd number of fives. We are then left with $\{a,b,c\} = \{2,3,5\}$. There are $3!$ permutations of these three numbers, since each is reflected over the midpoint we only have to count the first there. Each of the $\boxed{(A)6}$ permutations of the set $\{2,3,5\}$ will give us one palindrome.

Solution 2

Say we have a 2 first. Then, we have a 2 pinned as the last digit, so we have to fill in the remaining digits with only 3's and 5's. We have 2 options for the second digit then, and the rest is fixed. This means that we have $2$ ways for this case.

Say we have a 3 first. By symmetry, this is the same as the 2 cases, so we have $2$ ways.

Say we have a 5 first. We then have a 5 in the middle. We can either have a 2 second or a 3 second. So we have $2$ ways.

This means that our answer is $2+2+2=\boxed{\textbf{(A)}6}$


~yofro

See Also

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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