Difference between revisions of "2020 AMC 12A Problems/Problem 13"
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==See Also== | ==See Also== |
Revision as of 21:32, 17 February 2020
Contents
[hide]Problem
There are integers and
each greater than
such that
for all . What is
?
Solution
can be simplified to
The equation is then which implies that
has to be
since
.
is the result when
and
are
and
being
will make the fraction
which is close to
.
Finally, with being
, the fraction becomes
. In this case
and
work, which means that
must equal
~lopkiloinm
Solution 2
As above, notice that you get
Now, combine the fractions to get .
Assume that and
.
From the first equation we get . Note also that from the second equation,
and
must both be factors of 36.
After some casework we find that and
works, with
. So our answer is
~Silverdragon
Solution 3
Collapsed, . Comparing this to
, observe that
and
. The first can be rewritten as
. Then,
has to factor into 24 while 1 less than that also must factor into 36. The prime factorizations are as follows
and
. Then,
, as only 4 and 3 factor into 36 and 24 while being 1 apart.
(b=1 technically works but I don't care. a,b,c>1 as in the question)
~~BJHHar
edited by - annabelle0913
See Also
2020 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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