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

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== Solution ==
 
== Solution ==
  
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{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>. Each of the <math>\boxed{(A)6}</math> permutations of the set <math>\{2,3,5\}</math> will give us one palindrome.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 00:49, 16 January 2020

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 $\{a,b,c\} = \{2,3,5\}$. Each of the $\boxed{(A)6}$ permutations of the set $\{2,3,5\}$ will give us one palindrome.

See Also

2009 AMC 10B (ProblemsAnswer KeyResources)
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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