Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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==Solution 2 (similar to Solution 1)== | ==Solution 2 (similar to Solution 1)== | ||
We know that <math>H = 0 </math>, because <math>19!</math> ends in three zeroes (see Solution 1). Furthermore, we know that <math>9</math> and <math>11</math> are both factors of <math>19!</math>. We can simply use the divisibility rules for <math>9</math> and <math>11</math> for this problem to find <math>T</math> and <math>M</math>. For <math>19!</math> to be divisible by <math>9</math>, the sum of digits must simply be divisible by <math>9</math>. Summing the digits, we get that <math>T + M + 33</math> must be divisible by <math>9</math>. This leaves either <math>\text{A}</math> or <math>\text{C}</math> as our answer choice. Now we test for divisibility by <math>11</math>. For a number to be divisible by <math>11</math>, the alternating sum must be divisible by <math>11</math> (for example, with the number <math>2728</math>, <math>2-7+2-8 = -11</math>, so <math>2728</math> is divisible by <math>11</math>). Applying the alternating sum test to this problem, we see that <math>T - M - 7</math> must be divisible by 11. By inspection, we can see that this holds if <math>T=4</math> and <math>M=8</math>. The sum is <math>8 + 4 + 0 = \boxed{\textbf{(C) }12}</math>. | We know that <math>H = 0 </math>, because <math>19!</math> ends in three zeroes (see Solution 1). Furthermore, we know that <math>9</math> and <math>11</math> are both factors of <math>19!</math>. We can simply use the divisibility rules for <math>9</math> and <math>11</math> for this problem to find <math>T</math> and <math>M</math>. For <math>19!</math> to be divisible by <math>9</math>, the sum of digits must simply be divisible by <math>9</math>. Summing the digits, we get that <math>T + M + 33</math> must be divisible by <math>9</math>. This leaves either <math>\text{A}</math> or <math>\text{C}</math> as our answer choice. Now we test for divisibility by <math>11</math>. For a number to be divisible by <math>11</math>, the alternating sum must be divisible by <math>11</math> (for example, with the number <math>2728</math>, <math>2-7+2-8 = -11</math>, so <math>2728</math> is divisible by <math>11</math>). Applying the alternating sum test to this problem, we see that <math>T - M - 7</math> must be divisible by 11. By inspection, we can see that this holds if <math>T=4</math> and <math>M=8</math>. The sum is <math>8 + 4 + 0 = \boxed{\textbf{(C) }12}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/mXvetCMMzpU | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Revision as of 05:12, 28 January 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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