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 | + | 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>. |
== 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 04:37, 12 May 2021
Contents
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
.
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.