Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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==Solution 3== | ==Solution 3== | ||
− | We know that <math>9</math> and <math>11</math> are both factors of <math>19!</math>. Furthermore, we know that H is 0 because <math>19!</math> ends in three zeroes. We can simply use the divisibility rules for <math>9</math> and <math>11</math> for this problem to find T and M. For <math>19!</math> to be divisible by <math>9</math>, the sum of digits must be divisible by <math>9</math>. Summing the digits, we get that T + M + <math>33</math> must be divisible by <math>9</math>. This leaves either A or C as our answer choice. Now we test for divisibility by <math>11</math>. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. <math>2728</math>, <math>2</math>-<math>7</math>+<math>2</math>-<math>8</math> = -11 so <math>2728</math> is divisible by <math>11</math>). 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 <math>4</math> and M is <math>8</math>. The sum is <math>8</math> + <math>4</math> + <math>0</math> = 12 or (C). | + | We know that <math>9</math> and <math>11</math> are both factors of <math>19!</math>. Furthermore, we know that H is 0 because <math>19!</math> ends in three zeroes. We can simply use the divisibility rules for <math>9</math> and <math>11</math> for this problem to find T and M. For <math>19!</math> to be divisible by <math>9</math>, the sum of digits must be divisible by <math>9</math>. Summing the digits, we get that T + M + <math>33</math> must be divisible by <math>9</math>. This leaves either A or C as our answer choice. Now we test for divisibility by <math>11</math>. For a number to be divisible by eleven, the alternating sum must be divisible by 11(ex. <math>2728</math>, <math>2</math>-<math>7</math>+<math>2</math>-<math>8</math> = -11 so <math>2728</math> is divisible by <math>11</math>). 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 <math>4</math> and M is <math>8</math>. The sum is <math>8</math> + <math>4</math> + <math>0</math> = 12 or (C). -krishdhar |
==See Also== | ==See Also== |
Revision as of 16:16, 15 February 2019
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 (C).
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 (C).
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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