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

(Added problem, Latex'ed solution, added 'see also' box.)
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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>
 
<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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==Video Solution by OmegaLearn==
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https://youtu.be/7an5wU9Q5hk?t=1552
  
 
==Solution==
 
==Solution==
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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>.
 
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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==Solution 2==
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Using modular arithmetic:
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<cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath>
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Similarly,
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<cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath>
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We have
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<cmath>(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath>
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 +
-ryjs
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 +
==Solution 3==
 +
Experimentation gives
 +
<cmath>\text{any number ending with }9^{\text{something even}} = \text{has units digit }1</cmath>
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 +
<cmath>\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9</cmath>
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 +
Using this we have
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<cmath>19^{19} + 99^{99}</cmath>
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<cmath>9^{19} + 9^{99}</cmath>
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Both <math>19</math> and <math>99</math> are odd, so we are left with
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<cmath>9+9=18,</cmath> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math>
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-ryjs
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==Solution 4==
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<math>19^{19}+99^{99}=</math>36972963764972677265718790562880544059566876428174110243025997242355257045527752342141065001012823272794097888
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9548326540119429996769494359451621570193644014418071060667659303363419435659472789623878 which ends in 8. :) If you are really stupid and have time (like about 3 hours), this is a real brainless bash.
 +
 +
--hefei417
  
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2000|num-b=13|num-a=15}}
 
{{AMC8 box|year=2000|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 19:24, 24 November 2024

Problem

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

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

Video Solution by OmegaLearn

https://youtu.be/7an5wU9Q5hk?t=1552

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 $\boxed{D}$.

Solution 2

Using modular arithmetic: \[99 \equiv 9 \equiv -1 \pmod{10}\]

Similarly, \[19 \equiv 9 \equiv -1 \pmod{10}\]

We have \[(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}\]

-ryjs

Solution 3

Experimentation gives \[\text{any number ending with }9^{\text{something even}} = \text{has units digit }1\]

\[\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9\]

Using this we have \[19^{19} + 99^{99}\] \[9^{19} + 9^{99}\]

Both $19$ and $99$ are odd, so we are left with \[9+9=18,\] which has units digit $\boxed{(\textbf{D}) \ 8}.$ -ryjs

Solution 4

$19^{19}+99^{99}=$36972963764972677265718790562880544059566876428174110243025997242355257045527752342141065001012823272794097888 9548326540119429996769494359451621570193644014418071060667659303363419435659472789623878 which ends in 8. :) If you are really stupid and have time (like about 3 hours), this is a real brainless bash.

--hefei417

See Also

2000 AMC 8 (Problems • Answer Key • Resources)
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

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