Difference between revisions of "2019 AMC 10B Problems/Problem 14"

Revision as of 21:29, 17 February 2019

Problem

The base-ten representation for $19!$ is $121,6T5,100,40M,832,H00$ , where $T$ , $M$ , and $H$ denote digits that are not given. What is $T+M+H$ ?

Solution 1

We can figure out $H = 0$ by noticing that $19!$ will end with $3$ zeroes, as there are three $5$ s in its prime factorization. Next, we use the fact that $19!$ is a multiple of both $11$ and $9$ . Their divisibility rules tell us that $T + M \equiv 3 \;(\bmod\; 9)$ and that $T - M \equiv 7 \;(\bmod\; 11)$ . By inspection, we see that $T = 4, M = 8$ is a valid solution. Therefore the answer is $4 + 8 + 0 = \boxed{\textbf{(C) }12}$ .

Solution 2

By investing just a little bit of time, we can manually calculate $19!$ . If we prime factorize $19!$ , it becomes $2^{16} \cdot 3^8 \cdot 5^3 \cdot 7^2 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ . This looks complicated, but we can use elimination methods to make it simpler. $2^3 \cdot 5^3 = 1000$ , and $7 \cdot 11 \cdot 13 \cdot = 1001$ . If we put these aside for a moment, we have $2^{13} \cdot 3^8 \cdot 7 \cdot 17 \cdot 19$ left. $2^{13} = 2^{10} \cdot 2^3 = 1024 \cdot 8 = 8192$ , and $3^8 = (3^4)^2 = 81^2 = 6561$ . We have the $2$ s and $3$ s out of the way, and then we have $7 \cdot 17 \cdot 19 = 2261$ . Now if we multiply all the values calculated, we get $1000 \cdot 1001 \cdot 8192 \cdot 6561 \cdot 2261 = 121,645,100,408,832,000$ . Thus $T = 4, M = 8, H = 0$ , and the answer is $T + M + H = \boxed{\textbf{(C) }12}$ .

Solution 3 (similar to Solution 1)

We know that $9$ and $11$ are both factors of $19!$ . Furthermore, we know that $H = 0$ , because $19!$ ends in three zeroes (see Solution 1). We can simply use the divisibility rules for $9$ and $11$ for this problem to find $T$ and $M$ . For $19!$ to be divisible by $9$ , the sum of digits must simply be divisible by $9$ . Summing the digits, we get that $T + M + 33$ must be divisible by $9$ . This leaves either $\text{A}$ or $\text{C}$ as our answer choice. Now we test for divisibility by $11$ . For a number to be divisible by $11$ , the alternating sum must be divisible by $11$ (for example, with the number $2728$ , $2-7+2-8 = -11$ , so $2728$ is divisible by $11$ ). Applying the alternating sum test to this problem, we see that $T - M - 7$ must be divisible by 11. By inspection, we can see that this holds if $T=4$ and $M=8$ . The sum is $8 + 4 + 0 = \boxed{\textbf{(C) }12}$ .

@@ Line 3: / Line 3: @@
 ==Solution 1==
-We can figure out <math>H = 0</math> by noticing that <math>19!</math> will end with <math>3</math> zeroes, as there are three <math>5</math>'s in its prime factorization. Next we use the fact that <math>19!</math> is a multiple of both <math>11</math> and <math>9</math>. Since their divisibility rules gives us that <math>T + M</math> is congruent to <math>3</math> mod <math>9</math> and that <math>T - M</math> is congruent to <math>7</math> mod <math>11</math>. By inspection, we see that <math>T = 4, M = 8</math> is a valid solution. Therefore the answer is <math>4 + 8 + 0 = 12</math>, which is <math>\boxed{\textbf{(C) }12}</math>
+We can figure out <math>H = 0</math> by noticing that <math>19!</math> will end with <math>3</math> zeroes, as there are three <math>5</math>s in its prime factorization. Next, we use the fact that <math>19!</math> is a multiple of both <math>11</math> and <math>9</math>. Their divisibility rules tell us that <math>T + M \equiv 3 \;(\bmod\; 9)</math> and that <math>T - M \equiv 7 \;(\bmod\; 11)</math>. By inspection, we see that <math>T = 4, M = 8</math> is a valid solution. Therefore the answer is <math>4 + 8 + 0 = \boxed{\textbf{(C) }12}</math>.
 ==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 <math>\boxed{\textbf{(C) }12}</math>.
+By investing just a little bit of time, we can manually calculate <math>19!</math>. If we prime factorize <math>19!</math>, 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. <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 <math>2</math>s and <math>3</math>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 is <math>T + M + H = \boxed{\textbf{(C) }12}</math>.
-==Solution 3==
+==Solution 3 (similar to Solution 1)==
-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 <math>\boxed{\textbf{(C) }12}</math>. -- krishdhar
+We know that <math>9</math> and <math>11</math> are both factors of <math>19!</math>. Furthermore, we know that <math>H = 0 </math>, because <math>19!</math> ends in three zeroes (see Solution 1). 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>.
 ==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
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