Difference between revisions of "2015 AMC 10A Problems/Problem 7"
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<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> | ||
− | ==Solution== | + | ==Solution 1== |
<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>. | <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 <math>3</math> going up to <math>60</math>, and there are <math>20</math> multiples of <math>3</math> in <math>60</math>. | 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, | + | However, the number 0 must also be included, adding another multiple. So, the answer is <math>\boxed{\textbf{(B)}\ 21}</math>. |
==Solution 2== | ==Solution 2== | ||
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==Solution 3== | ==Solution 3== | ||
− | Minus each of the terms by 12 to make the the sequence 1,4,7, | + | Minus each of the terms by <math>12</math> to make the the sequence <math>1 , 4 , 7,..., 61</math>. <math>\frac{61-1}{3}=20, 20 + 1 = 21</math> |
− | 61-1 | + | |
− | <math>\boxed{\textbf{(B)}}</math>. | + | <math>\boxed{\textbf{(B)}\ 21}</math>. |
==Solution 4== | ==Solution 4== | ||
− | Subtract each of the terms by 10 to make the sequence 3,6,9, | + | Subtract each of the terms by <math>10</math> to make the sequence <math>3 , 6 , 9,..., 60, 63</math>. Then divide the each term in the sequence by <math>3</math> to get <math>1, 2, 3,..., 20, 21</math>. Now it is clear to see that there are <math>21</math> terms in the sequence. |
+ | <math>\boxed{\textbf{(B)}\ 21}</math>. | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING)== | ||
+ | https://youtu.be/i7ItueJ6K8E | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/fcWPfgKeCmA | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}} | {{AMC10 box|year=2015|ab=A|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 23:05, 26 June 2023
Contents
Problem
How many terms are in the arithmetic sequence ,
,
,
,
,
?
Solution 1
, 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, the number 0 must also be included, adding another multiple. So, the answer is .
Solution 2
Using the formula for arithmetic sequence's nth term, we see that
.
Solution 3
Minus each of the terms by to make the the sequence
.
.
Solution 4
Subtract each of the terms by to make the sequence
. Then divide the each term in the sequence by
to get
. Now it is clear to see that there are
terms in the sequence.
.
Video Solution (CREATIVE THINKING)
~Education, the Study of Everything
Video Solution
~savannahsolver
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.