Difference between revisions of "2012 AIME II Problems/Problem 1"
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== Solution == | == Solution == | ||
− | Solving for <math>m</math> gives us <math>m = \frac{503-3n}{5},</math> so in order for <math>m</math> to be an integer, we must have <math>3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.</math> The smallest possible value of <math>n</math> is obviously <math>1,</math> and the greatest is <math>\frac{503 - 5}{3} = 166,</math> so the total number of solutions is <math>\frac{166-1}{5}+1 = \boxed{034.}</math> | + | |
+ | ==Solution 1== | ||
+ | |||
+ | Solving for <math>m</math> gives us <math>m = \frac{503-3n}{5},</math> so in order for <math>m</math> to be an integer, we must have <math>3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.</math> The smallest possible value of <math>n</math> is obviously <math>1,</math> and the greatest is <math>\frac{503 - 5}{3} = 166,</math> so the total number of solutions is <math>\frac{166-1}{5}+1 = \boxed{34}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Dividing by <math>4</math> gives us <math>5m + 3n = 503</math>. Thus, we have <math>503-5m\equiv 0 \pmod 3</math> since <math>n</math> is an integer. Rearranging then gives <math>m\equiv 1\pmod 3.</math> Since <math>503-5m>0,</math> we know that <math>m<503/5.</math> Because <math>m</math> is an integer, we can rewrite this as <math>m\le 100.</math> Therefore, <math>m</math> ranges from | ||
+ | <cmath>0\cdot 3+1\quad \text{to} \quad 33\cdot 3+1,</cmath> | ||
+ | giving <math>\boxed{034}</math> values. | ||
+ | ~vaporwave | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Because the x-intercept of the equation is <math>\frac{2012}{20}</math>, and the y-intercept is <math>\frac{2012}{12}</math>, the slope is <math>\frac{\frac{-2012}{12}}{\frac{2012}{20}} = \frac{-5}{3}</math>. Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: <math>(100,1), (97,6), (94,11)...</math> Because the solutions are only positive, we can generate only 33 more solutions, so in total we have <math>33+1=\boxed{34}</math> solutions. | ||
+ | |||
+ | == Solution 4 (Quick)== | ||
+ | Note that a positive integer is divisible by 20 if it ends in 0. | ||
+ | |||
+ | Notice that if a multiple of 12 end in 2, 12 must be multiplied by an integer that ends in 1 or 6. | ||
+ | |||
+ | So let's start checking because 2012 ends in 2, same as 12. | ||
+ | |||
+ | When <math>n=1</math>, <math>m=100</math>. | ||
+ | |||
+ | When <math>n=6</math>, <math>m=97</math>. | ||
+ | What is happening? Why doesn't, say, as values of <math>n</math>, do <math>1, 11, ...</math> work, while <math>6, 16, 26</math> work for <math>m</math> to be an integer (as seen in 2022 CEMC Cayley #21 (https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CayleyContest.pdf, | ||
+ | https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CayleySolution.pdf)?) | ||
+ | |||
+ | Notice that we can break <math>1, 6, ...</math> down into <math>0, 5, 10, ...</math>. So <math>20m=2000-12k</math>, where <math>k</math> is a member of the set <math>{0, 5, 10, 15...}</math>. | ||
+ | |||
+ | For a number to be divisible by 20, it must be divisible 10, and the quotient (that number divided by 10) must be divisible by 2. This means that the number must end in 0, and its tens digit must be even. | ||
+ | |||
+ | So notice that the tens digit cycle in: <math>0, 4, 8, 2, 6</math>. Each of this is even (notice that when <math>k=25</math>, <math>2000-12k=1700</math>, it cycles again). | ||
+ | |||
+ | So we know that all values of <math>n</math>, which end in 1 or 6, that make <math>2012-12n\geq20</math> works. | ||
+ | |||
+ | So values of <math>n</math> that belong to <math>{1, 6, ..., 166}</math> work. Clearly, that is 34 integer values of <math>n</math>. Therefore, the answer is <math>\boxed{034}</math>. | ||
+ | |||
+ | ~hastapasta | ||
== See Also == | == See Also == |
Latest revision as of 14:56, 2 June 2023
Contents
Problem 1
Find the number of ordered pairs of positive integer solutions to the equation .
Solution
Solution 1
Solving for gives us so in order for to be an integer, we must have The smallest possible value of is obviously and the greatest is so the total number of solutions is
Solution 2
Dividing by gives us . Thus, we have since is an integer. Rearranging then gives Since we know that Because is an integer, we can rewrite this as Therefore, ranges from giving values. ~vaporwave
Solution 3
Because the x-intercept of the equation is , and the y-intercept is , the slope is . Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: Because the solutions are only positive, we can generate only 33 more solutions, so in total we have solutions.
Solution 4 (Quick)
Note that a positive integer is divisible by 20 if it ends in 0.
Notice that if a multiple of 12 end in 2, 12 must be multiplied by an integer that ends in 1 or 6.
So let's start checking because 2012 ends in 2, same as 12.
When , .
When , . What is happening? Why doesn't, say, as values of , do work, while work for to be an integer (as seen in 2022 CEMC Cayley #21 (https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CayleyContest.pdf, https://www.cemc.uwaterloo.ca/contests/past_contests/2022/2022CayleySolution.pdf)?)
Notice that we can break down into . So , where is a member of the set .
For a number to be divisible by 20, it must be divisible 10, and the quotient (that number divided by 10) must be divisible by 2. This means that the number must end in 0, and its tens digit must be even.
So notice that the tens digit cycle in: . Each of this is even (notice that when , , it cycles again).
So we know that all values of , which end in 1 or 6, that make works.
So values of that belong to work. Clearly, that is 34 integer values of . Therefore, the answer is .
~hastapasta
See Also
2012 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.