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 "2015 AIME II Problems/Problem 1"

(→‎Solution 2)
(→‎Solution 3)
 
(4 intermediate revisions by 3 users not shown)
Line 5: Line 5:
 
==Solution 1==
 
==Solution 1==
  
−
If <math>N</math> is <math>22</math> percent less than one integer <math>k</math>, then <math>N=\frac{78}{100}k=\frac{39}{50}k</math>. In addition, <math>N</math> is <math>16</math> percent greater than another integer <math>m</math>, so <math>N=\frac{116}{100}m=\frac{29}{25}m</math>. Therefore, <math>k</math> is divisible by 50 and <math>m</math> is divisible by 25. Setting these two equal, we have <math>\frac{39}{50}k=\frac{29}{25}m</math>. Multiplying by <math>50</math> on both sides, we get <math>39k=58m</math>.
+
If <math>N</math> is <math>22</math> percent less than one integer <math>k</math>, then <math>N=\frac{78}{100}k=\frac{39}{50}k</math>. In addition, <math>N</math> is <math>16</math> percent greater than another integer <math>m</math>, so <math>N=\frac{116}{100}m=\frac{29}{25}m</math>. Therefore, <math>k</math> is divisible by <math>50</math> and <math>m</math> is divisible by <math>25</math>. Setting these two equal, we have <math>\frac{39}{50}k=\frac{29}{25}m</math>. Multiplying by <math>50</math> on both sides, we get <math>39k=58m</math>.
  
 
The smallest integers <math>k</math> and <math>m</math> that satisfy this are <math>k=1450</math> and <math>m=975</math>, so <math>N=1131</math>. The answer is <math>\boxed{131}</math>.
 
The smallest integers <math>k</math> and <math>m</math> that satisfy this are <math>k=1450</math> and <math>m=975</math>, so <math>N=1131</math>. The answer is <math>\boxed{131}</math>.
Line 11: Line 11:
 
==Solution 2==
 
==Solution 2==
 
Continuing from Solution 1, we have <math>N=\frac{39}{50}k</math> and <math>N=\frac{29}{25}m</math>. It follows that <math>k=\frac{50}{39}N</math> and <math>m=\frac{25}{29}N</math>. Both <math>m</math> and <math>k</math> have to be integers, so, in order for that to be true, <math>N</math> has to cancel the denominators of both <math>\frac{50}{39}</math> and <math>\frac{25}{29}</math>. In other words, <math>N</math> is a multiple of both <math>29</math> and <math>39</math>. That makes <math>N=\operatorname{lcm}(29,39)=29\cdot39=1131</math>. The answer is <math>\boxed{131}</math>.
 
Continuing from Solution 1, we have <math>N=\frac{39}{50}k</math> and <math>N=\frac{29}{25}m</math>. It follows that <math>k=\frac{50}{39}N</math> and <math>m=\frac{25}{29}N</math>. Both <math>m</math> and <math>k</math> have to be integers, so, in order for that to be true, <math>N</math> has to cancel the denominators of both <math>\frac{50}{39}</math> and <math>\frac{25}{29}</math>. In other words, <math>N</math> is a multiple of both <math>29</math> and <math>39</math>. That makes <math>N=\operatorname{lcm}(29,39)=29\cdot39=1131</math>. The answer is <math>\boxed{131}</math>.
 +
 +
==Solution 3==
 +
We can express <math>N</math> as <math>0.78a</math> and <math>1.16b</math>, where <math>a</math> and <math>b</math> are some positive integers. <math>N=0.78a=1.22b\implies100N=78a=116b.</math> Let us try to find the smallest possible value of <math>100N</math>, first ignoring the integral constraint.
 +
 +
Obviously, we are just trying to find the LCM of <math>78</math> and <math>116.</math> They have no common factors but <math>2</math>, so we multiply <math>78</math> and <math>116</math> and divide by <math>2</math> to get <math>4524.</math> This is obviously not divisible by <math>100</math>, so this is not possible, as it would imply that <math>N=\dfrac{4524}{100},</math> which is not an integer. This can be simplified to <math>\dfrac{1131}{25}</math>.
 +
 +
We know that any possible value of <math>N</math> will be an integral multiple of this value; the smallest such <math>N</math> is achieved by multiplying this value by <math>25.</math> We arrive at <math>1131</math>, which is <math>\boxed{131}\mod1000.</math>
 +
 +
~Technodoggo
 +
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=9re2qLzOKWk&t=7s
 +
 +
~MathProblemSolvingSkills.com
 +
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2015|n=II|before=First Problem|num-a=2}}
 
{{AIME box|year=2015|n=II|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:40, 24 November 2024

Problem

Let $N$ be the least positive integer that is both $22$ percent less than one integer and $16$ percent greater than another integer. Find the remainder when $N$ is divided by $1000$.

Solution 1

If $N$ is $22$ percent less than one integer $k$, then $N=\frac{78}{100}k=\frac{39}{50}k$. In addition, $N$ is $16$ percent greater than another integer $m$, so $N=\frac{116}{100}m=\frac{29}{25}m$. Therefore, $k$ is divisible by $50$ and $m$ is divisible by $25$. Setting these two equal, we have $\frac{39}{50}k=\frac{29}{25}m$. Multiplying by $50$ on both sides, we get $39k=58m$.

The smallest integers $k$ and $m$ that satisfy this are $k=1450$ and $m=975$, so $N=1131$. The answer is $\boxed{131}$.

Solution 2

Continuing from Solution 1, we have $N=\frac{39}{50}k$ and $N=\frac{29}{25}m$. It follows that $k=\frac{50}{39}N$ and $m=\frac{25}{29}N$. Both $m$ and $k$ have to be integers, so, in order for that to be true, $N$ has to cancel the denominators of both $\frac{50}{39}$ and $\frac{25}{29}$. In other words, $N$ is a multiple of both $29$ and $39$. That makes $N=\operatorname{lcm}(29,39)=29\cdot39=1131$. The answer is $\boxed{131}$.

Solution 3

We can express $N$ as $0.78a$ and $1.16b$, where $a$ and $b$ are some positive integers. $N=0.78a=1.22b\implies100N=78a=116b.$ Let us try to find the smallest possible value of $100N$, first ignoring the integral constraint.

Obviously, we are just trying to find the LCM of $78$ and $116.$ They have no common factors but $2$, so we multiply $78$ and $116$ and divide by $2$ to get $4524.$ This is obviously not divisible by $100$, so this is not possible, as it would imply that $N=\dfrac{4524}{100},$ which is not an integer. This can be simplified to $\dfrac{1131}{25}$.

We know that any possible value of $N$ will be an integral multiple of this value; the smallest such $N$ is achieved by multiplying this value by $25.$ We arrive at $1131$, which is $\boxed{131}\mod1000.$

~Technodoggo

Video Solution

https://www.youtube.com/watch?v=9re2qLzOKWk&t=7s

~MathProblemSolvingSkills.com


See also

2015 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=2015_AIME_II_Problems/Problem_1&oldid=235541"