Difference between revisions of "2014 AIME I Problems/Problem 3"
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+ | == Solution 6 == | ||
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+ | Our fraction can be written in the form <math>\frac{1000 - a}{a} = \frac{1000}{a} - 1.</math> Thus the fraction is reducible when <math>a</math> divides <math>1000.</math> We also want <math>500 < a < 1000.</math> By PIE, the total values of <math>a</math> that make the fraction reducible is, | ||
+ | <cmath>249 + 99 - 49 = 299.</cmath> | ||
+ | By complementary counting, the answer we want is <math>499 - 299 = \boxed{200}.</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=2|num-a=4}} | {{AIME box|year=2014|n=I|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:39, 20 July 2021
Contents
Problem 3
Find the number of rational numbers
such that when
is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000.
Solution 1
We have that the set of these rational numbers is from to
where each each element
has
and
is irreducible.
We note that .
Hence,
is irreducible if
is irreducible, and
is irreducible if
is not divisible by 2 or 5. Thus, the answer to the question is the number of integers between 999 and 501 inclusive that are not divisible by 2 or 5.
We note there are 499 numbers between 501 and 999, and
- 249 numbers are divisible by 2
- 99 numbers are divisible by 5
- 49 numbers are divisible by 10
Using the Principle of Inclusion and Exclusion, we get that there are numbers between
and
are not divisible by either
or
, so our answer is
.
Euler's Totient Function can also be used to arrive at 400 numbers relatively prime to 1000, meaning 200 possible fractions satisfying the necessary conditions.
Solution 2
If the initial manipulation is not obvious, instead ,consider the euclidean algorithm. Instead of using as the fraction to use the euclidean algorithm on, we can rewrite this as
. Thus, we want
. You can either proceed as solution 1, or consider that no even numbers work, limiting us to
choices of numbers and restricting
to be odd. If
is odd,
is odd, so the only possible common factors
and
can share are multiples of
. Thus, we want to avoid these. There are
multiples of
less than
, so the answer is
.
Solution 3
Say ; then
. If this fraction is reducible, then the modulus of some number for
is the same as the modulus for
. Since
, that modulus can only be
or
. This implies that if
or
, the fraction is reducible. There are 249 cases where
, 99 where
, and 49 where
, so by PIE, the number of fails is 299, so our answer is
.
Solution 4
We know that the numerator of the fraction cannot be even, because the denominator would also be even. We also know that the numerator cannot be a multiple of 5 because the denominator would also be a multiple of 5. Proceed by listing out all the other possible fractions and we realize that the numerator and denominator are always relatively prime. We have 499 fractions to start with, and 250 with odd numerators. Subtract 50 to account for the multiples of 5, and we get .
Solution 5
Let the numerator and denominator with
and
Now if
then
Therefore any pair that works satisfies
There are
pairs
such that
and
However, exactly half of them work because of the condition
Therefore the desired answer is
Solution 6 (sort of cheese)
We notice that there are a total of fractions that are in simplest form and the numerator and denominator add up to
because the numerator and denominator have to be relatively prime so there are
. Half of these are greater than
so the answer is
- bedwarsnoob
Solution 6
Our fraction can be written in the form Thus the fraction is reducible when
divides
We also want
By PIE, the total values of
that make the fraction reducible is,
By complementary counting, the answer we want is
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.