Difference between revisions of "2000 AMC 8 Problems/Problem 14"

(Created page with "Even powers of 19 have a units digit of 1, and odd powers of 19 have a units digit of 9. So, 19^19 has a units digit of 9. Powers of 99 have the exact same property, so 99^99 als...")
 
(Added problem, Latex'ed solution, added 'see also' box.)
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Even powers of 19 have a units digit of 1, and odd powers of 19 have a units digit of 9. So, 19^19 has a units digit of 9. Powers of 99 have the exact same property, so 99^99 also has a units digit of 9. 9+9=18 which has a units digit of 8, so the answer is D.
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==Problem==
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What is the units digit of <math>19^{19} + 99^{99}</math>?
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<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math>
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==Solution==
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Finding a pattern for each half of the sum, even powers of <math>19</math> have a units digit of <math>1</math>, and odd powers of <math>19</math> have a units digit of <math>9</math>. So, <math>19^{19}</math> has a units digit of <math>9</math>.  
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Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>.
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==See Also==
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{{AMC8 box|year=2000|num-b=13|num-a=15}}

Revision as of 20:15, 30 July 2011

Problem

What is the units digit of $19^{19} + 99^{99}$?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9$

Solution

Finding a pattern for each half of the sum, even powers of $19$ have a units digit of $1$, and odd powers of $19$ have a units digit of $9$. So, $19^{19}$ has a units digit of $9$.

Powers of $99$ have the exact same property, so $99^{99}$ also has a units digit of $9$. $9+9=18$ which has a units digit of $8$, so the answer is $\boxed{D}$.

See Also

2000 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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 AJHSME/AMC 8 Problems and Solutions