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

(Solution Two)
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==Solution Three==
 
==Solution Three==
Notice that 10001+ 99999 = 111000. In fact, ordering the palindromes in ascending order, we find that the sum of the nth palindrome and the nth to last palindrome is 111000. We have 9*10*10 palindromes, or 450 pairs of palindromes summing to 111000. Performing the multiplication gives 49500000, so the sum is 18.
+
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.
  
 
==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}}

Revision as of 11:19, 2 September 2014

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

As there are only $9\cdot10\cdot10 = 900$ five digit palindromes, it is sufficient to add up all of them. \[10001 + 10101 + 10201 + 10301 + 10401 + 10501 + 10601 + 10701 + 10801 + 10901 + 11011 + 11111 + 11211 + 11311 + 11411 + 11511 + 11611 + 11711 + 11811 + 11911 + 12021 + 12121 + 12221 + 12321 + 12421 + 12521 + 12621 + 12721 + 12821\]\[+ 12921 + 13031 + 13131 + 13231 + 13331 + 13431 + 13531 + 13631 + 13731 + 13831 + 13931 + 14041 + 14141 + 14241 + 14341 + 14441 + 14541 + 14641 + 14741 + 14841 + 14941 + 15051 + 15151 + 15251 + 15351 + 15451 + 15551 + 15651\]\[+ 15751 + 15851 + 15951 + 16061 + 16161 + 16261 + 16361 + 16461 + 16561 + 16661 + 16761 + 16861 + 16961 + 17071 + 17171 + 17271 + 17371 + 17471 + 17571 + 17671 + 17771 + 17871 + 17971 + 18081 + 18181 + 18281 + 18381 + 18481\]\[+ 18581 + 18681 + 18781 + 18881 + 18981 + 19091 + 19191 + 19291 + 19391 + 19491 + 19591 + 19691 + 19791 + 19891 + 19991 + 20002 + 20102 + 20202 + 20302 + 20402 + 20502 + 20602 + 20702 + 20802 + 20902 + 21012 + 21112 + 21212\]\[+ 21312 + 21412 + 21512 + 21612 + 21712 + 21812 + 21912 + 22022 + 22122 + 22222 + 22322 + 22422 + 22522 + 22622 + 22722 + 22822 + 22922 + 23032 + 23132 + 23232 + 23332 + 23432 + 23532 + 23632 + 23732 + 23832 + 23932 + 24042\]\[+ 24142 + 24242 + 24342 + 24442 + 24542 + 24642 + 24742 + 24842 + 24942 + 25052 + 25152 + 25252 + 25352 + 25452 + 25552 + 25652 + 25752 + 25852 + 25952 + 26062 + 26162 + 26262 + 26362 + 26462 + 26562 + 26662 + 26762 + 26862\]\[+ 26962 + 27072 + 27172 + 27272 + 27372 + 27472 + 27572 + 27672 + 27772 + 27872 + 27972 + 28082 + 28182 + 28282 + 28382 + 28482 + 28582 + 28682 + 28782 + 28882 + 28982 + 29092 + 29192 + 29292 + 29392 + 2:)9492 + 29592 + 29692\]\[+ 29792 + 29892 + 29992 + 30003 + 30103 + 30203 + 30303 + 30403 + 30503 + 30603 + 30703 + 30803 + 30903 + 31013 + 31113 + 31213 + 31313 + 31413 + 31513 + 31613 + 31713 + 31813 + 31913 + 32023 + 32123 + 32223 + 32323 + 32423\]\[+ 32523 + 32623 + 32723 + 32823 + 32923 + 33033 + 33133 + 33233 + 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+ 94849 + 94949 + 95059 + 95159 + 95259 + 95359 + 95459 + 95559 + 95659 + 95759 + 95859 + 95959 + 96069 + 96169 + 96269 + 96369 + 96469 + 96569 + 96669 + 96769 + 96869\]\[+ 96969 + 97079 + 97179 + 97279 + 97379 + 97479 + 97579 + 97679 + 97779 + 97879 + 97979 + 98089 + 98189 + 98289 + 98389 + 98489 + 98589 + 98689 + 98789 + 98889 + 98989 + 99099 + 99199 + 99299 + 99399 + 99499 + 99599 + 99699\]\[+ 99799 + 99899 + 99999 = 49500000 = 4 + 9 + 5 = 18 \to \boxed{(B)}\].

Solution Three

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.

See Also

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