Difference between revisions of "2015 AIME II Problems/Problem 1"
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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>. | ||
+ | |||
+ | 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== | ==Video Solution== |
Revision as of 02:29, 13 November 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.