View source for 2015 AIME II Problems/Problem 1

==Problem==

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

==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 <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>.

==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>.



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

~MathProblemSolvingSkills.com


== See also ==
{{AIME box|year=2015|n=II|before=First Problem|num-a=2}}
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