2018 AMC 10A Problems/Problem 22
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 1.1
Elaborating on to what Solution 1 stated, we are not able to add any extra factor of 2 or 3 to because doing so would later the
of
and
. This is why:
The is
and the
of
is
. However, the
of
(meaning both are divisible by 36). Therefore,
is only divisible by
(and no higher power of 3), while
is divisible by only
(and no higher power of 2).
Thus, the of
can be expressed in the form
for which k is a number not divisible by 2 or 3. The only answer choice that satisfies this is
.
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 | ||
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