2009 AMC 10A Problems/Problem 5

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 $81\fbox{(E)}.$ (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 $81\fbox{(E)}.$

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+111111111\=

12,345,678,987,654,321$ (Error compiling LaTeX. Unknown error_msg) The digit sum is thus .

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 $81\longrightarrow \fbox{E}$

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

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