Difference between revisions of "2012 AIME II Problems/Problem 1"

(Solution)
(Solution 2)
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==Solution 2==
 
==Solution 2==
  
Dividing by <math>4</math> gives us <math>5m + 3n = 503</math>. Solving for <math>n</math> gives <math>n \equiv 1 \pmod 5</math>. The solutions are the numbers <math> 253437
+
Dividing by <math>4</math> gives us <math>5m + 3n = 503</math>. Solving for <math>n</math> gives <math>n \equiv 1 \pmod 5</math>. The solutions are the numbers <math>n
 
  = 1, 6, 11, ... , 166</math>. There are <math>\boxed{34}</math> solutions.
 
  = 1, 6, 11, ... , 166</math>. There are <math>\boxed{34}</math> solutions.
  

Revision as of 14:16, 17 July 2017

Problem 1

Find the number of ordered pairs of positive integer solutions $(m, n)$ to the equation $20m + 12n = 2012$.

Solution

Solution 1

Solving for $m$ gives us $m = \frac{503-3n}{5},$ so in order for $m$ to be an integer, we must have $3n \equiv 503 \mod 5 \longrightarrow n \equiv 1 \mod 5.$ The smallest possible value of $n$ is obviously $1,$ and the greatest is $\frac{503 - 5}{3} = 166,$ so the total number of solutions is $\frac{166-1}{5}+1 = \boxed{034}$

Solution 2

Dividing by $4$ gives us $5m + 3n = 503$. Solving for $n$ gives $n \equiv 1 \pmod 5$. The solutions are the numbers $n   = 1, 6, 11, ... , 166$. There are $\boxed{34}$ solutions.

See Also

2012 AIME II (ProblemsAnswer KeyResources)
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

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