2019 AMC 10B Problems/Problem 14
Problem
The base-ten representation for is
, where
,
, and
denote digits that are not given. What is
?
Solution 1
We can figure out by noticing that
will end with
zeroes, as there are three
's in its prime factorization. Next we use the fact that
is a multiple of both
and
. Since their divisibility rules gives us that
is congruent to
mod
and that
is congruent to
mod
. By inspection, we see that
is a valid solution. Therefore the answer is
, which is
Solution 2
With investing just a little bit of time, we can manually calculate 19!. If we prime factorize 19!, it becomes . This looks complicated, but we can use elimination methods to make it simpler.
, and
. If we put these aside for a moment, we have
left from the original 19!.
, and
. We have the 2's and 3's out of the way, and then we have
. Now if we multiply all the values calculated, we get
. Thus
, and the answer
, thus
.
Solution 3
We know that and
are both factors of
. Furthermore, we know that H is 0 because
ends in three zeroes. We can simply use the divisibility rules for
and
for this problem to find T and M. For
to be divisible by
, the sum of digits must be divisible by
. Summing the digits, we get that T + M +
must be divisible by
. This leaves either A or C as our answer choice. Now we test for divisibility by
. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex.
,
-
+
-
= -11 so
is divisible by
). Applying the alternating sum to this problem, we see that T-M-7 must be divisible by 11. By inspection, we can see that this holds if T is
and M is
. The sum is
+
+
= 12 or (C). -- krishdhar
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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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