Difference between revisions of "2003 AMC 10A Problems/Problem 16"
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<math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9 </math> | <math> \mathrm{(A) \ } 1\qquad \mathrm{(B) \ } 3\qquad \mathrm{(C) \ } 7\qquad \mathrm{(D) \ } 8\qquad \mathrm{(E) \ } 9 </math> | ||
| − | == Solution == | + | == Solution 1== |
<math>13^{2003}\equiv 3^{2003}\pmod{10}</math> | <math>13^{2003}\equiv 3^{2003}\pmod{10}</math> | ||
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<math>3^{2003}=(3^{4})^{500}\cdot3^{3}\equiv1^{500}\cdot27\equiv7\pmod{10}</math> | <math>3^{2003}=(3^{4})^{500}\cdot3^{3}\equiv1^{500}\cdot27\equiv7\pmod{10}</math> | ||
| − | Therefore, the units digit is <math>7 \Rightarrow C</math> | + | Therefore, the units digit is <math>7 \Rightarrow\boxed{\mathrm{(C)}\ 7}</math> |
| + | |||
| + | == Solution 2 (Patterns) == | ||
| + | Since we are looking for the units digit of <math>13^{2003}</math>, we only have to focus on the units digit of the base (13) as none of the other digits of the base affect the units digit of the resulting value. | ||
| + | |||
| + | By testing the first few values or through previous knowledge, you might see that the units digit of exponents with base 3 follow this pattern: | ||
| + | <cmath>3^1=3</cmath> | ||
| + | <cmath>3^2=9</cmath> | ||
| + | <cmath>3^3=27</cmath> | ||
| + | <cmath>3^4=81,</cmath> | ||
| + | giving us the rotation <math>3-9-7-1.</math> | ||
| + | |||
| + | As this cycle resets every time the index increases by 4, we know that this cycle ends on 2000, and starts once again on 2001. As our expression is raised to the power of 2003, we know that the units digit of our expression must end with the third term of our pattern: <math>7</math>. | ||
| + | |||
| + | Therefore, the units digit of our expression is <math>7 \Rightarrow\boxed{\mathrm{(C)}\ 7}</math> | ||
| + | |||
| + | ~ JinhoK | ||
== See Also == | == See Also == | ||
| − | + | {{AMC10 box|year=2003|ab=A|num-b=15|num-a=17}} | |
| − | |||
| − | |||
[[Category:Introductory Number Theory Problems]] | [[Category:Introductory Number Theory Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 15:52, 19 August 2023
Problem
What is the units digit of 
?
Solution 1
Since 
:
Therefore, the units digit is 
Solution 2 (Patterns)
Since we are looking for the units digit of , we only have to focus on the units digit of the base (13) as none of the other digits of the base affect the units digit of the resulting value.
By testing the first few values or through previous knowledge, you might see that the units digit of exponents with base 3 follow this pattern:
![\[3^1=3\]](http://latex.artofproblemsolving.com/f/d/0/fd06b7f0a326c11154c8bae770fff73158d8592c.png)
![\[3^2=9\]](http://latex.artofproblemsolving.com/b/7/1/b7157f3301a75c2ea1c2c8c5e70ad15334a3b284.png)
![\[3^3=27\]](http://latex.artofproblemsolving.com/3/5/e/35ed4d2f9b1f54dff1a9eded0e7777ef89290f77.png)
![\[3^4=81,\]](http://latex.artofproblemsolving.com/2/f/a/2fa1ea0b060f67290542a76d1131bda3295649b3.png)
giving us the rotation 
As this cycle resets every time the index increases by 4, we know that this cycle ends on 2000, and starts once again on 2001. As our expression is raised to the power of 2003, we know that the units digit of our expression must end with the third term of our pattern: 
.
Therefore, the units digit of our expression is
~ JinhoK
See Also
| 2003 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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. 
