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Difference between revisions of "2015 AIME II Problems/Problem 1"

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==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>.
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 +
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==Video Solution==
 +
https://www.youtube.com/watch?v=9re2qLzOKWk&t=7s
 +
 +
~MathProblemSolvingSkills.com
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== 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}}

Revision as of 12:59, 14 February 2023

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 $\boxed{131}$.


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

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