Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 4 (Designed to be very quick))
(→‎Solution)
 
(One intermediate revision by the same user not shown)
(No difference)

Latest revision as of 03:24, 2 November 2024

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 $\frac{166-1}{5}+1 = \boxed{34}$

Solution 2

Dividing by $4$ gives us $5m + 3n = 503$. Thus, we have $503-5m\equiv 0 \pmod 3$ since $n$ is an integer. Rearranging then gives $m\equiv 1\pmod 3.$ Since $503-5m>0,$ we know that $m<503/5.$ Because $m$ is an integer, we can rewrite this as $m\le 100.$ Therefore, $m$ ranges from \[0\cdot 3+1\quad \text{to} \quad 33\cdot 3+1,\] giving $\boxed{034}$ values. ~vaporwave

Solution 3

Because the x-intercept of the equation is $\frac{2012}{20}$, and the y-intercept is $\frac{2012}{12}$, the slope is $\frac{\frac{-2012}{12}}{\frac{2012}{20}} = \frac{-5}{3}$. Now, notice the first obvious solution: (100,1). From it, we derive all the other solutions by applying the slope in reverse, i.e: $(100,1), (97,6), (94,11)...$ Because the solutions are only positive, we can generate only 33 more solutions, so in total we have $33+1=\boxed{34}$ 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 $n=1$, $m=100$.

When $n=6$, $m=97$. What is happening? Why doesn't, say, as values of $n$, do $1, 11, ...$ work, while $6, 16, 26$ work for $m$ 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 $1, 6, ...$ down into $0, 5, 10, ...$. So $20m=2000-12k$, where $k$ is a member of the set ${0, 5, 10, 15...}$.

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: $0, 4, 8, 2, 6$. Each of this is even (notice that when $k=25$, $2000-12k=1700$, it cycles again).

So we know that all values of $n$, which end in 1 or 6, that make $2012-12n\geq20$ works.

So values of $n$ that belong to ${1, 6, ..., 166}$ work. Clearly, that is 34 integer values of $n$. Therefore, the answer is $\boxed{034}$.

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_II_Problems/Problem_1&oldid=231195"