Difference between revisions of "2014 AMC 12A Problems/Problem 15"

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It just so happens that <cmath> (1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000 </cmath> so the sum of the digits of the sum is <math>18</math>, or <math>\boxed{\textbf{(B)}}</math>.
 
It just so happens that <cmath> (1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000 </cmath> so the sum of the digits of the sum is <math>18</math>, or <math>\boxed{\textbf{(B)}}</math>.
  
==Solution Two==
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==Solution 2==
 
Notice that <math>10001+ 99999 = 110000.</math> In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is <math>110000.</math> We have <math>9*10*10</math> palindromes, or <math>450</math> pairs of palindromes summing to <math>110000.</math> Performing the multiplication gives <math>49500000</math>, so the sum is <math>18</math> <math> \boxed{\textbf{(B)}} </math>.
 
Notice that <math>10001+ 99999 = 110000.</math> In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is <math>110000.</math> We have <math>9*10*10</math> palindromes, or <math>450</math> pairs of palindromes summing to <math>110000.</math> Performing the multiplication gives <math>49500000</math>, so the sum is <math>18</math> <math> \boxed{\textbf{(B)}} </math>.
  

Revision as of 15:13, 17 March 2015

Problem

A five-digit palindrome is a positive integer with respective digits $abcba$, where $a$ is non-zero. Let $S$ be the sum of all five-digit palindromes. What is the sum of the digits of $S$?

$\textbf{(A) }9\qquad \textbf{(B) }18\qquad \textbf{(C) }27\qquad \textbf{(D) }36\qquad \textbf{(E) }45\qquad$

Solution 1

For each digit $a=1,2,\ldots,9$ there are $10\cdot10$ (ways of choosing $b$ and $c$) palindromes. So the $a$s contribute $(1+2+\cdots+9)(100)(10^4+1)$ to the sum. For each digit $b=0,1,2,\ldots,9$ there are $9\cdot10$ (since $a \neq 0$) palindromes. So the $b$s contribute $(0+1+2+\cdots+9)(90)(10^3+10)$ to the sum. Similarly, for each $c=0,1,2,\ldots,9$ there are $9\cdot10$ palindromes, so the $c$ contributes $(0+1+2+\cdots+9)(90)(10^2)$ to the sum.

It just so happens that \[(1+2+\cdots+9)(100)(10^4+1)+(1+2+\cdots+9)(90)(10^3+10)+(1+2+\cdots+9)(90)(10^2)=49500000\] so the sum of the digits of the sum is $18$, or $\boxed{\textbf{(B)}}$.

Solution 2

Notice that $10001+ 99999 = 110000.$ In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is $110000.$ We have $9*10*10$ palindromes, or $450$ pairs of palindromes summing to $110000.$ Performing the multiplication gives $49500000$, so the sum is $18$ $\boxed{\textbf{(B)}}$.

Solution Three

As shown above, there are a total of $900$ five-digit palindromes. We can calculate their sum by finding the expected value of a randomly selected palindrome satisfying the conditions given, then multiplying it by $900$ to get our sum. The expected value for the ten-thousands and the units digit is $\frac{1+2+3+\cdots+9}{9}=5$, and the expected value for the thousands, hundreds, and tens digit is $\frac{0+1+2+\cdots+9}{10}=4.5$. Therefore our expected value is $5\times10^4+4.5\times10^3+4.5\times10^2+4.5\times10^1+5\times10^0=55,\!000$. Since the question asks for the sum of the digits of the resulting sum, we do not need to keep the trailing zeros of either $55,\!000$ or $900$. Thus we only need to calculate $55\times9=495$, and the desired sum is $\boxed{\textbf{(B) }18}$.

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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 12 Problems and Solutions

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