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

Revision as of 01:29, 13 November 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 $\boxed{131}$ .

Solution 3

We can express $N$ as $0.78a$ and $1.22b$ , where $a$ and $b$ are some positive integers. $N=0.78a=1.22b\implies100N=78a=122b.$ 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

@@ Line 12: / Line 12: @@
 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>1.22b</math>, where <math>a</math> and <math>b</math> are some positive integers. <math>N=0.78a=1.22b\implies100N=78a=122b.</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==

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

Page

Toolbox

Search