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

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\textbf{(E) }45\qquad</math>
 
\textbf{(E) }45\qquad</math>
  
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==Solution One==
+
==Solution 1==
  
 
For each digit <math>a=1,2,\ldots,9</math> there are <math>10\cdot10</math> (ways of choosing <math>b</math> and <math>c</math>) palindromes. So the <math>a</math>s contribute <math>(1+2+\cdots+9)(100)(10^4+1)</math> to the sum.
 
For each digit <math>a=1,2,\ldots,9</math> there are <math>10\cdot10</math> (ways of choosing <math>b</math> and <math>c</math>) palindromes. So the <math>a</math>s contribute <math>(1+2+\cdots+9)(100)(10^4+1)</math> to the sum.
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Similarly, for each <math>c=0,1,2,\ldots,9</math> there are <math>9\cdot10</math> palindromes, so the <math>c</math> contributes <math>(0+1+2+\cdots+9)(90)(10^2)</math> to the sum.
 
Similarly, for each <math>c=0,1,2,\ldots,9</math> there are <math>9\cdot10</math> palindromes, so the <math>c</math> contributes <math>(0+1+2+\cdots+9)(90)(10^2)</math> to the sum.
  
−
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>\boxed{\textbf{(B)}\; 18}</math>.
  
−
(Solution by AwesomeToad)
+
==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\cdot 10\cdot 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 <math> \boxed{\textbf{(B)}\; 18}</math>.
  
−
==Solution Two==
+
==Solution 3==
 +
As shown above, there are a total of <math>900</math> 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 <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>.
  
−
As there are only <math>9\cdot10\cdot10 = 900</math> five digit palindromes, it is sufficient to add up all of them.  
+
==Solution 4 (Variation of #2)==
−
<cmath>10001 + 10101 + 10201 + 10301 + 10401 + 10501 + 10601 + 10701 + 10801 + 10901 + 11011 + 11111 + 11211 + 11311 + 11411 + 11511 + 11611 + 11711 + 11811 + 11911 + 12021 + 12121 + 12221 + 12321 + 12421 + 12521 + 12621 + 12721 + 12821</cmath><cmath> + 12921 + 13031 + 13131 + 13231 + 13331 + 13431 + 13531 + 13631 + 13731 + 13831 + 13931 + 14041 + 14141 + 14241 + 14341 + 14441 + 14541 + 14641 + 14741 + 14841 + 14941 + 15051 + 15151 + 15251 + 15351 + 15451 + 15551 + 15651</cmath><cmath> + 15751 + 15851 + 15951 + 16061 + 16161 + 16261 + 16361 + 16461 + 16561 + 16661 + 16761 + 16861 + 16961 + 17071 + 17171 + 17271 + 17371 + 17471 + 17571 + 17671 + 17771 + 17871 + 17971 + 18081 + 18181 + 18281 + 18381 + 18481</cmath><cmath> + 18581 + 18681 + 18781 + 18881 + 18981 + 19091 + 19191 + 19291 + 19391 + 19491 + 19591 + 19691 + 19791 + 19891 + 19991 + 20002 + 20102 + 20202 + 20302 + 20402 + 20502 + 20602 + 20702 + 20802 + 20902 + 21012 + 21112 + 21212</cmath><cmath> + 21312 + 21412 + 21512 + 21612 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91919 + 92029 + 92129 + 92229 + 92329 + 92429 + 92529 + 92629 + 92729 + 92829 + 92929 + 93039 + 93139 + 93239 + 93339 + 93439 + 93539 + 93639 + 93739 + 93839 + 93939 + 94049</cmath><cmath> + 94149 + 94249 + 94349 + 94449 + 94549 + 94649 + 94749 + 94849 + 94949 + 95059 + 95159 + 95259 + 95359 + 95459 + 95559 + 95659 + 95759 + 95859 + 95959 + 96069 + 96169 + 96269 + 96369 + 96469 + 96569 + 96669 + 96769 + 96869</cmath><cmath> + 96969 + 97079 + 97179 + 97279 + 97379 + 97479 + 97579 + 97679 + 97779 + 97879 + 97979 + 98089 + 98189 + 98289 + 98389 + 98489 + 98589 + 98689 + 98789 + 98889 + 98989 + 99099 + 99199 + 99299 + 99399 + 99499 + 99599 + 99699</cmath><cmath> + 99799 + 99899 + 99999 = 49500000 = 4 + 9 + 5 = 18 \to \boxed{(B)}</cmath>.
+
First, allow <math>a</math> to be zero, and then subtract by how much we overcount. We'll also sum each palindrome with its <math>\textit{complement}</math>. If <math>\overline{abcba}</math> (the line means a, b, and c are digits and <math>abcba\ne a\cdot b\cdot c\cdot b\cdot a</math>) is a palindrome, then its complement is <math>\overline{defed}</math> where <math>d=9-a</math>, <math>e=9-b</math>, <math>f=9-c</math>. Notice how every palindrome has a unique compliment, and that the sum of a palindrome and its complement is <math>99999</math>. Therefore, the sum of our palindromes is <math>99999\times (10^3/2)</math>. (There are <math>10^3/2</math> pairs.)
−
Solution not by bestwillcui1
 
  
−
==Solution Three==
+
However, we have overcounted, as something like <math>05350</math> <math>\textit{isn't}</math> a palindrome by the problem's definition, but we've still included it. So we must subtract the sum of numbers in the form <math>\overline{0nmn0}</math>. By the same argument as before, these sum to <math>9990\times  (10^2/2)</math>. Therefore, the sum that the problem asks for is:
−
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>.
+
 
 +
<cmath>500\times99999-50\times 9990</cmath>
 +
<cmath>=500\times99999-500\times 999</cmath>
 +
<cmath>=500(99999-999)</cmath>
 +
<cmath>=500\times 99000</cmath>
 +
 
 +
Since all we care about is the sum of the digits, we can drop the <math>0</math>'s.
 +
 
 +
<cmath>5\times99</cmath>
 +
<cmath>=5\times(100-1)</cmath>
 +
<cmath>=495</cmath>
 +
 
 +
And finally, <math>4+9+5=\boxed{\textbf{(B)}18}</math>
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=14|num-a=16}}
 
{{AMC12 box|year=2014|ab=A|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 21:42, 2 August 2021

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 $\boxed{\textbf{(B)}\; 18}$.

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 $9\cdot 10\cdot 10$ palindromes, or $450$ pairs of palindromes summing to $110000.$ Performing the multiplication gives $49500000$, so the sum $\boxed{\textbf{(B)}\; 18}$.

Solution 3

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

Solution 4 (Variation of #2)

First, allow $a$ to be zero, and then subtract by how much we overcount. We'll also sum each palindrome with its $\textit{complement}$. If $\overline{abcba}$ (the line means a, b, and c are digits and $abcba\ne a\cdot b\cdot c\cdot b\cdot a$) is a palindrome, then its complement is $\overline{defed}$ where $d=9-a$, $e=9-b$, $f=9-c$. Notice how every palindrome has a unique compliment, and that the sum of a palindrome and its complement is $99999$. Therefore, the sum of our palindromes is $99999\times (10^3/2)$. (There are $10^3/2$ pairs.)

However, we have overcounted, as something like $05350$ $\textit{isn't}$ a palindrome by the problem's definition, but we've still included it. So we must subtract the sum of numbers in the form $\overline{0nmn0}$. By the same argument as before, these sum to $9990\times (10^2/2)$. Therefore, the sum that the problem asks for is:

\[500\times99999-50\times 9990\] \[=500\times99999-500\times 999\] \[=500(99999-999)\] \[=500\times 99000\]

Since all we care about is the sum of the digits, we can drop the $0$'s.

\[5\times99\] \[=5\times(100-1)\] \[=495\]

And finally, $4+9+5=\boxed{\textbf{(B)}18}$

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
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Problem 14 		Followed by
Problem 16
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All AMC 12 Problems and Solutions

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