Difference between revisions of "2015 AIME II Problems/Problem 1"
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==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>. | ||
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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>1.22b</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>. | ||
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+ | 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> | ||
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+ | ~Technodoggo | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 07:44, 17 December 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 .
Solution 3
We can express as and , where and are some positive integers. Let us try to find the smallest possible value of , first ignoring the integral constraint.
Obviously, we are just trying to find the LCM of and They have no common factors but , so we multiply and and divide by to get This is obviously not divisible by , so this is not possible, as it would imply that which is not an integer. This can be simplified to .
We know that any possible value of will be an integral multiple of this value; the smallest such is achieved by multiplying this value by We arrive at , which is
~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 |
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