Difference between revisions of "2015 AMC 10A Problems/Problem 7"
(Problem and Solution 7) |
BeastX-Men (talk | contribs) (Added See Also and LaTeX in Problem and Solution.) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | How many terms are in the arithmetic sequence 13, 16, 19, | + | 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> | ||
Line 9: | Line 9: | ||
==Solution== | ==Solution== | ||
− | 73-13 | + | <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>. |
− | In this sequence, the terms are the multiples of 3 going up to 60, and there are 20 multiples of 3 in 60. | + | 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>. |
− | However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B) }21}</math>. | + | However, one more must be added to include the first term. So, the answer is <math>\boxed{\textbf{(B)}\ 21}</math>. |
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Revision as of 17:51, 4 February 2015
Problem
How many terms are in the arithmetic sequence , , , , , ?
Solution
, so the amount of terms in the sequence , , , , , is the same as in the sequence , , , , , .
In this sequence, the terms are the multiples of going up to , and there are multiples of in .
However, one more must be added to include the first term. So, the answer is .
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.