Difference between revisions of "2009 AMC 10A Problems/Problem 5"

Revision as of 12:18, 30 December 2018

Problem

What is the sum of the digits of the square of $\text 111,111,111$ ?

$\mathrm{(A)}\ 18\qquad\mathrm{(B)}\ 27\qquad\mathrm{(C)}\ 45\qquad\mathrm{(D)}\ 63\qquad\mathrm{(E)}\ 81$

Solution 1

Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$ whose digit sum is $81\longrightarrow \fbox{(E)}.$ (I hope you didn't seriously multiply it outright...)

Solution 2

Note that

$11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321.$

We observe a pattern with the squares of number comprised of only $1$ 's and use it to find that $111,111,111^2=12,345,678,987,654,321$ whose digit sum is $81\longrightarrow \fbox{(E)}.$

Solution 3

We see that $111^2$ can be written as $111(100+10+1)=11100+1110+111=12321$ .

We can apply this strategy to find $111,111,111^2$ , as seen below. $111111111^2=111111111(100000000+10000000\cdots+10+1)=11111111100000000+1111111110000000+\cdots+1111111110+111111111 = 12,345,678,987,654,321$ The digit sum is thus $1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}$ .

Solution 4

We can also do something a little bit more clever rather than just adding up the digits though. Realize too that

$1^2 = 1$

$11^2 = 121$

$111^2 = 12321$

$1,111^2 = 1234321$

We clearly see the pattern, the number of digits determines how high the number goes, as with $111^2$ , it has $3$ digits so it goes up to $1,2,3$ then decreases back down. If we start adding up the digits, we see that the first one is $1$ , the second is $2 + 1 + 1 = 4$ , the third one is $1 + 2 + 3 + 2 + 1 = 9$ , and the fourth one is $16$ . We instantly see a pattern and find that these are all square numbers. If the number you square has $4$ digits, you do $4^2$ to see what the added digits of that particular square will be. In this case, we are dealing with $111,111,111$ which has $9$ digits so $9^2$ equals $81\longrightarrow \fbox{E}$

@@ Line 16: / Line 16: @@
 ==Solution 3==
-You can see that
+We see that <math>111^2</math> can be written as <math>111(100+10+1)=11100+1110+111=12321</math>.
-<math>111*111</math> can be written as
-<math>111+1110+11100</math>, which is <math>12321</math>.
+We can apply this strategy to find <math>111,111,111^2</math>, as seen below.
-We can apply the same fact into 111,111,111, receiving
+<cmath>111111111^2=111111111(100000000+10000000\cdots+10+1)=11111111100000000+1111111110000000+\cdots+1111111110+111111111 = 12,345,678,987,654,321</cmath> The digit sum is thus <math>1+2+3+4+5+6+7+8+9+8+7+6+5+4+3+2+1=81 \boxed{(E)}</math>.
-<math>111111111+1111111110+11111111100... = 12,345,678,987,654,321</math> whose digits sum up to <math>81\longrightarrow \fbox{E}.</math>
 ==Solution 4==

2009 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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