Difference between revisions of "2015 AMC 12A Problems/Problem 10"
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<math> \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26</math> | <math> \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Use [[SFFT]] to get <math>(x+1)(y+1)=81</math>. The terms <math>(x+1)</math> and <math>(y+1)</math> must be factors of <math>81</math>, which include <math>1, 3, 9, 27, 81</math>. Because <math>x > y</math>, <math>x+1</math> is equal to <math>27</math> or <math>81</math>. But if <math>x+1=81</math>, then <math>y=0</math> and so <math>x=\boxed{\textbf{(E)}\ 26}</math>. | Use [[SFFT]] to get <math>(x+1)(y+1)=81</math>. The terms <math>(x+1)</math> and <math>(y+1)</math> must be factors of <math>81</math>, which include <math>1, 3, 9, 27, 81</math>. Because <math>x > y</math>, <math>x+1</math> is equal to <math>27</math> or <math>81</math>. But if <math>x+1=81</math>, then <math>y=0</math> and so <math>x=\boxed{\textbf{(E)}\ 26}</math>. | ||
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| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Plug in values of <math>x</math> and solve for <math>y</math>, noting <math>x > y</math> and that <math>y</math> is an integer. | ||
| + | |||
| + | <math>x = 8</math>: | ||
| + | |||
| + | <math> | ||
| + | 8 + y + 8y = 80 | ||
| + | </math> | ||
| + | |||
| + | <math>y = 8</math> (Does not work because <math>x > y</math>) | ||
| + | |||
| + | |||
| + | <math>x = 10</math>: | ||
| + | |||
| + | <math> | ||
| + | 10 + y + 10y = 80 | ||
| + | </math> | ||
| + | |||
| + | <math>y = 70/11</math> (Does not work because <math>y</math> is not an integer) | ||
| + | |||
| + | |||
| + | <math>x = 15</math>: | ||
| + | |||
| + | <math> | ||
| + | 15 + y + 15y = 80 | ||
| + | </math> | ||
| + | |||
| + | <math>y = 65/16</math> (Does not work because <math>y</math> is not an integer) | ||
| + | |||
| + | |||
| + | <math>x = 18</math>: | ||
| + | |||
| + | <math> | ||
| + | 18 + y + 18y = 80 | ||
| + | </math> | ||
| + | |||
| + | <math>y = 62/19</math> (Does not work because <math>y</math> is not an integer) | ||
| + | |||
| + | |||
| + | Thus <math>x = 26</math>, <math>\boxed{\textbf{(E)}\ 26}</math>. | ||
| + | |||
| + | |||
| + | ~ Solution by CYB3RFLARE7408 | ||
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/ba6w1OhXqOQ?t=4512 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2015|ab=A|num-b=9|num-a=11}} | {{AMC12 box|year=2015|ab=A|num-b=9|num-a=11}} | ||
Latest revision as of 15:47, 28 September 2025
Problem
Integers 
and 
with 
satisfy 
. What is ?
Solution 1
Use SFFT to get 
. The terms 
and 
must be factors of 
, which include 
. Because 
, 
is equal to 
or . But if 
, then 
and so 
.
Solution 2
Plug in values of and solve for , noting and that is an integer.
:
(Does not work because )
:
(Does not work because is not an integer)
:
(Does not work because is not an integer)
:
(Does not work because is not an integer)
Thus 
, 
.
~ Solution by CYB3RFLARE7408
Video Solution by OmegaLearn
https://youtu.be/ba6w1OhXqOQ?t=4512
~ pi_is_3.14
See Also
| 2015 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 9 |
Followed by Problem 11 |
| 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 12 Problems and Solutions | |
