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

Revision as of 10:10, 3 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 One

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 by AwesomeToad)

Solution Two

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 $900$ $5$ 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 $\boxed{\textbf{(B) }18}$ .

@@ Line 21: / Line 21: @@
 ==Solution Two==
 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>.
+==Solution Three==
+As shown above, there are a total of <math>900</math> <math>5</math> 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 <math>900</math> to get our sum.  The expected value for the ten-thousands and the units digit is <math>\frac{1+2+3+\cdots+9}{9}=5</math>, and the expected value for the thousands, hundreds, and tens digit is <math>\frac{0+1+2+\cdots+9}{10}=4.5</math>.  Therefore our expected value is <math>5\times10^4+4.5\times10^3+4.5\times10^2+4.5\times10^1+5\times10^0=55,\!000</math>.  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 <math>55,\!000</math> or <math>900</math>.  Thus we only need to calculate <math>55\times9=495</math>, and the desired sum is <math>\boxed{\textbf{(B) }18}</math>.
 ==See Also==
 {{AMC12 box|year=2014|ab=A|num-b=14|num-a=16}}
 {{MAA Notice}}

2014 AMC 12A (Problems • Answer Key • Resources)
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