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Difference between revisions of "2015 AMC 10A Problems/Problem 7"

(Problem and Solution 7)
(Added See Also and LaTeX in Problem and Solution.)
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==Problem==
 
==Problem==
  
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How many terms are in the arithmetic sequence 13, 16, 19,..., 70, 73?
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How many terms are in the arithmetic sequence <math>13</math>, <math>16</math>, <math>19</math>, <math>\dotsc</math>, <math>70</math>, <math>73</math>?
  
 
<math> \textbf{(a)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 </math>
 
<math> \textbf{(a)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61 </math>
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==Solution==
 
==Solution==
  
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73-13 is 60, so the amount of terms in the sequence 13, 16, 19,..., 70, 73 is the same as in the sequence 0, 3, 6,..., 57, 60.  
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<math>73-13 = 60</math>, so the amount of terms in the sequence <math>13</math>, <math>16</math>, <math>19</math>, <math>\dotsc</math>, <math>70</math>, <math>73</math> is the same as in the sequence <math>0</math>, <math>3</math>, <math>6</math>, <math>\dotsc</math>, <math>57</math>, <math>60</math>.  
  
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In this sequence, the terms are the multiples of 3 going up to 60, and there are 20 multiples of 3 in 60.  
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In this sequence, the terms are the multiples of <math>3</math> going up to <math>60</math>, and there are <math>20</math> multiples of <math>3</math> in <math>60</math>.  
  
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However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B) }21}</math>.
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However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B)}\ 21}</math>.
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==See Also==
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{{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}}
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{{MAA Notice}}

Revision as of 17:51, 4 February 2015

Problem

How many terms are in the arithmetic sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$?

$\textbf{(a)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$


Solution

$73-13 = 60$, so the amount of terms in the sequence $13$, $16$, $19$, $\dotsc$, $70$, $73$ is the same as in the sequence $0$, $3$, $6$, $\dotsc$, $57$, $60$.

In this sequence, the terms are the multiples of $3$ going up to $60$, and there are $20$ multiples of $3$ in $60$.

However, one more must be added to include the first term. So, the answer is $\boxed{\textbf{(B)}\ 21}$.

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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. AMC logo.png

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