Difference between revisions of "2019 AMC 10B Problems/Problem 14"
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==Solution 2== | ==Solution 2== | ||
| − | With investing just a little bit of time, we can manually calculate 19!. If we prime factorize 19!, it becomes <math>2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19</math>. This looks complicated, but we can use elimination methods to make it simpler. <math>2^3 \cdot 5^3 = 1000</math>, and <math>7 \cdot 11 \cdot 13 \cdot = 1001</math>. If we put these aside for a moment, we have <math>2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19</math> left from the original 19!. <math>2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192</math>, and <math>3^8 = (3^4)^2 = 81^2 = 6561</math>. We have the 2's and 3's out of the way, and then we have <math>7 \cdot 17 \cdot 19 = 2261</math>. Now if we multiply all the values calculated, we get <math>1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000</math>. Thus <math>T = 4, M = 8, H = 0</math>, and the answer <math>T + M + H = 12</math>, thus (C). | + | With investing just a little bit of time, we can manually calculate 19!. If we prime factorize 19!, it becomes <math>2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19</math>. This looks complicated, but we can use elimination methods to make it simpler. <math>2^3 \cdot 5^3 = 1000</math>, and <math>7 \cdot 11 \cdot 13 \cdot = 1001</math>. If we put these aside for a moment, we have <math>2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19</math> left from the original 19!. <math>2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192</math>, and <math>3^8 = (3^4)^2 = 81^2 = 6561</math>. We have the 2's and 3's out of the way, and then we have <math>7 \cdot 17 \cdot 19 = 2261</math>. Now if we multiply all the values calculated, we get <math>1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000</math>. Thus <math>T = 4, M = 8, H = 0</math>, and the answer <math>T + M + H = 12</math>, thus <math>\boxed{\textbf{(C) }12}</math>. |
==Solution 3== | ==Solution 3== | ||
Revision as of 20:51, 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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
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