Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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Revision as of 21:22, 13 August 2020
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
. Their divisibility rules (see Solution 2) tell us that
and that
. By guess and checking, we see that
is a valid solution. Therefore the answer is
.
Solution 2 (similar to Solution 1)
We know that , because
ends in three zeroes (see Solution 1). Furthermore, we know that
and
are both factors of
. We can simply use the divisibility rules for
and
for this problem to find
and
. For
to be divisible by
, the sum of digits must simply be divisible by
. Summing the digits, we get that
must be divisible by
. This leaves either
or
as our answer choice. Now we test for divisibility by
. For a number to be divisible by
, the alternating sum must be divisible by
(for example, with the number
,
, so
is divisible by
). Applying the alternating sum test to this problem, we see that
must be divisible by 11. By inspection, we can see that this holds if
and
. The sum is
.
Video Solution
~IceMatrix
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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