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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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=9re2qLzOKWk&t=7s | ||
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+ | ~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 13:59, 14 February 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 50 and
is divisible by 25. 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.