2019 AMC 10B Problems/Problem 14
Contents
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 factors of
in its prime factorization, so there would be 3 powers of 10 meaning it will end in 3 zeros. 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
.
Solution 3 (Brute force)
Multiplying it out, we get . Evidently,
,
, and
. The sum is
.
Do not do this in a real contest.
Solution 4 (1001?)
7, 11, 13 are < 19 and 1001 = 7 * 11 * 13. Check the alternating sum of block 3: H00 - 832 + 40M - 100 + 6T5 - 121 and it is divisible by 1001. HTM + 5 - 53 = 0 (mod 1001) => HTM = 48.
The answer is . ~ AliciaWu
Do this in a real contest.
Video Solution by OmegaLearn
https://youtu.be/p5f1u44-pvQ?t=760
~ pi_is_3.14
Video Solution
~IceMatrix
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AMC 10 Problems and Solutions |
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