Difference between revisions of "2015 AIME II Problems/Problem 1"
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− | 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>. |
Revision as of 13:18, 28 June 2023
Problem
Let be the least positive integer that is both percent less than one integer and percent greater than another integer. Find the remainder when is divided by .
Solution 1
If is percent less than one integer , then . In addition, is percent greater than another integer , so . Therefore, is divisible by and is divisible by . Setting these two equal, we have . Multiplying by on both sides, we get .
The smallest integers and that satisfy this are and , so . The answer is .
Solution 2
Continuing from Solution 1, we have and . It follows that and . Both and have to be integers, so, in order for that to be true, has to cancel the denominators of both and . In other words, is a multiple of both and . That makes . The answer is .
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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