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Difference between revisions of "1983 AHSME Problems/Problem 14"

(Created page with "First, we notice that <math>3^0</math> is congruent to <math>1</math> (mod 10), <math>3^1</math> is <math>3</math> (mod 10), <math>3^2</math> is <math>9</math> (mod 10), <math...")
 
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First, we notice that <math>3^0</math> is congruent to <math>1</math> (mod 10), <math>3^1</math> is <math>3</math> (mod 10), <math>3^2</math> is <math>9</math> (mod 10), <math>3^3</math> is <math>7</math> (mod 10), <math>3^4</math> is <math>1</math> (mod 10), and so on.... This turns out to be a cycle repeating every 4 powers.
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==Problem==
  
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Then, we have <math>3^1001</math> is congruent to <math>3</math> (mod 10).
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The units digit of <math>3^{1001} 7^{1002} 13^{1003}</math> is
  
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The number <math>7</math> has a similar cycle, going: <math>1, 7, 9, 3, 1, ...</math>. Following that, we have <math>7^1002</math> is congruent to <math>9</math> (mod 10).
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<math>\textbf{(A)}\ 1\qquad
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\textbf{(B)}\ 3\qquad
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\textbf{(C)}\ 5\qquad
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\textbf{(D)}\ 7\qquad
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\textbf{(E)}\ 9 </math>
  
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<math>13^1003</math> is congruent to <math>3^1003</math> (mod 10) = <math>7</math>.
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==Solution== 
  
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<math>3\cdot 9\cdot 7</math> is congruent to <math>\fbox9</math> (mod 10).
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First, we notice that <math>3^0</math> is congruent to <math>1 \ \text{(mod 10)}</math>, <math>3^1</math> is <math>3 \ \text{(mod 10)}</math>, <math>3^2</math> is <math>9 \ \text{(mod 10)}</math>, <math>3^3</math> is <math>7 \ \text{(mod 10)}</math>, <math>3^4</math> is <math>1 \ \text{(mod 10)}</math>, and so on. This turns out to be a cycle repeating every <math>4</math> terms, so <math>3^{1001}</math> is congruent to <math>3 \ \text{(mod 10)}</math>.
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The number <math>7</math> has a similar cycle, going <math>1, 7, 9, 3, 1, ...</math> Hence we have that <math>7^{1002}</math> is congruent to <math>9 \ \text{(mod 10)}</math>. Finally, <math>13^{1003}</math> is congruent to <math>3^{1003} \equiv 7 \ \text{(mod 10)}</math>. Thus the required units digit is <math>3\cdot 9\cdot 7 \equiv 9 \ \text{(mod 10)}</math>, so the answer is <math>\boxed{\textbf{(E)}\ 9}</math>.
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==See Also==
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{{AHSME box|year=1983|num-b=13|num-a=15}}
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{{MAA Notice}}

Latest revision as of 00:51, 20 February 2019

Problem

The units digit of $3^{1001} 7^{1002} 13^{1003}$ is

$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$

Solution

First, we notice that $3^0$ is congruent to $1 \ \text{(mod 10)}$, $3^1$ is $3 \ \text{(mod 10)}$, $3^2$ is $9 \ \text{(mod 10)}$, $3^3$ is $7 \ \text{(mod 10)}$, $3^4$ is $1 \ \text{(mod 10)}$, and so on. This turns out to be a cycle repeating every $4$ terms, so $3^{1001}$ is congruent to $3 \ \text{(mod 10)}$.

The number $7$ has a similar cycle, going $1, 7, 9, 3, 1, ...$ Hence we have that $7^{1002}$ is congruent to $9 \ \text{(mod 10)}$. Finally, $13^{1003}$ is congruent to $3^{1003} \equiv 7 \ \text{(mod 10)}$. Thus the required units digit is $3\cdot 9\cdot 7 \equiv 9 \ \text{(mod 10)}$, so the answer is $\boxed{\textbf{(E)}\ 9}$.

See Also

1983 AHSME (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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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