Difference between revisions of "2018 AMC 10A Problems/Problem 22"
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Revision as of 13:12, 14 April 2018
Let and
be positive integers such that
,
,
, and
. Which of the following must be a divisor of
?
Solution 1
We can say that and
'have'
, that
and
have
, and that
and
have
. Combining
and
yields
has (at a minimum)
, and thus
has
(and no more powers of
because otherwise
would be different). In addition,
has
, and thus
has
(similar to
, we see that
cannot have any other powers of
). We now assume the simplest scenario, where
and
. According to this base case, we have
. We want an extra factor between the two such that this number is between
and
, and this new factor cannot be divisible by
or
. Checking through, we see that
is the only one that works. Therefore the answer is
Solution by JohnHankock
Solution 2 (Better notation)
First off, note that ,
, and
are all of the form
. The prime factorizations are
,
and
, respectively. Now, let
and
be the number of times
and
go into
,respectively. Define
,
,
, and
similiarly. Now, translate the
s into the following:
.
(Unfinished) ~Rowechen Zhong
See Also
2018 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 10 Problems and Solutions |
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