Difference between revisions of "2016 AMC 10A Problems/Problem 23"
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Therefore, <math>x = \frac{600}{2016} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math> | Therefore, <math>x = \frac{600}{2016} = \frac{25}{84}</math>, so the answer is <math>25 + 84 = \boxed{\textbf{(A) }109.}</math> | ||
+ | |||
+ | === Solution 4 (Proving division) === | ||
+ | The given conditions hold for all nonzero numbers <math>a, b,</math> and <math>c</math>, even when they're not distinct. | ||
+ | |||
+ | Let <math>a=b=c.</math> From the first two givens, this implies that | ||
+ | |||
+ | <cmath>a\diamondsuit\, (a\diamondsuit\, {a})=(a\diamondsuit\, a)\cdot{a}.</cmath> | ||
+ | |||
+ | From <math>a\diamondsuit\,{a}=1,</math> this equation simply becomes <math>a\diamondsuit\,{1}=a.</math> | ||
+ | |||
+ | Let <math>c=b.</math> Substituting this into the first two conditions, we see that | ||
+ | |||
+ | <cmath>a\diamondsuit\, (b\diamondsuit\, {c})=(a\diamondsuit\, {b})\cdot{c} \implies a\diamondsuit\, (b\diamondsuit\, {b})=(a\diamondsuit\, {b})\cdot{b}.</cmath> | ||
+ | |||
+ | Substituting <math>b\diamondsuit\, {b} =1</math>, the second equation becomes | ||
+ | |||
+ | <cmath>a\diamondsuit\, {1}=(a\diamondsuit\, {b})\cdot{b} \implies a=(a\diamondsuit\,{b})\cdot{b}.</cmath> | ||
+ | |||
+ | Since <math>a, b</math> and <math>c</math> are nonzero, we can divide by <math>b</math> which yields, | ||
+ | |||
+ | <cmath>\frac{a}{b}=(a\diamondsuit\, {b}).</cmath> | ||
+ | |||
+ | Now we can find the value of <math>x</math> straightforwardly: | ||
+ | |||
+ | <cmath>\frac{2016}{(\frac{6}{x})}=100 \implies 2016=\frac{600}{x} \implies x=600/2016 = \frac{25}{84}.</cmath> | ||
+ | |||
+ | Therefore, <math>a+b=25+84=\boxed{\textbf{A)} 109}</math> | ||
+ | |||
+ | -Benedict T (countmath1) | ||
== Video Solution 1== | == Video Solution 1== |
Revision as of 22:36, 6 October 2022
Contents
Problem
A binary operation has the properties that
and that
for all nonzero real numbers
and
. (Here
represents multiplication). The solution to the equation
can be written as
, where
and
are relatively prime positive integers. What is
Solutions
Solution 1
We see that , and think of division. Testing, we see that the first condition
is satisfied, because
. Therefore, division can be the operation
. Solving the equation,
so the answer is
.
Solution 2
We can manipulate the given identities to arrive at a conclusion about the binary operator . Substituting
into the first identity yields
Hence,
or, dividing both sides of the equation by
Hence, the given equation becomes . Solving yields
so the answer is
Solution 3
One way to eliminate the in this equation is to make
so that
. In this case, we can make
.
By multiplying both sides by , we get:
Because
Therefore, , so the answer is
Solution 4 (Proving division)
The given conditions hold for all nonzero numbers and
, even when they're not distinct.
Let From the first two givens, this implies that
From this equation simply becomes
Let Substituting this into the first two conditions, we see that
Substituting , the second equation becomes
Since and
are nonzero, we can divide by
which yields,
Now we can find the value of straightforwardly:
Therefore,
-Benedict T (countmath1)
Video Solution 1
https://www.youtube.com/watch?v=8GULAMwu5oE
Video Solution 2(Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1632
~ pi_is_3.14
See Also
2016 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
2016 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.