Difference between revisions of "2009 AMC 10A Problems/Problem 5"
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We can apply the same fact into 111,111,111, receiving | We can apply the same fact into 111,111,111, receiving | ||
<math>111111111+1111111110+11111111100... = 12,345,678,987,654,321</math> whose digits sum up to <math>81\longrightarrow \fbox{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== | ||
| + | We can also do something a little bit more clever rather than just adding up the digits though. Realize too that | ||
| + | <math>1^2 = 1</math> | ||
| + | <math>11^2 = 121</math> | ||
| + | <math>111^2 = 12321</math> | ||
| + | <math>1,111^2 = 1234321</math> | ||
| + | |||
| + | We clearly see the pattern, the number of digits determines how high the number goes, as with <math>111^2</math>, it has <math>3</math> digits so it goes up to <math>1,2,3</math> then decreases back down. If we start adding up the digits, we see that the first one is <math>1</math>, the second is <math>2 + 1 + 1 = 4</math>, the third one is <math>1 + 2 + 3 + 2 + 1 = 9</math>, and the fourth one is <math>16</math>. We instantly see a pattern and find that these are all square numbers. If the number you square has <math>4</math> digits, you do <math>4^2</math> to see what the added digits of that particular square will be. In this case, we are dealing with <math>111,111,111</math> which has <math>9</math> digits so <math>9^2</math> equals <math>81\longrightarrow \fbox{E}</math> | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}} | {{AMC10 box|year=2009|ab=A|num-b=4|num-a=6}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:23, 20 July 2018
Problem
What is the sum of the digits of the square of 
?
Solution 1
Using the standard multiplication algorithm, 
whose digit sum is 
(I hope you didn't seriously multiply it outright... ;) )
Solution 2(Pattern)
Note that:
We see a pattern and find that 
whose digit sum is
Solution 3
You can see that
can be written as
, which is 
.
We can apply the same fact into 111,111,111, receiving
whose digits sum up to
Solution 4
We can also do something a little bit more clever rather than just adding up the digits though. Realize too that
We clearly see the pattern, the number of digits determines how high the number goes, as with 
, it has 
digits so it goes up to 
then decreases back down. If we start adding up the digits, we see that the first one is 
, the second is 
, the third one is 
, and the fourth one is 
. We instantly see a pattern and find that these are all square numbers. If the number you square has 
digits, you do 
to see what the added digits of that particular square will be. In this case, we are dealing with 
which has 
digits so 
equals 
See also
| 2009 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
