https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mshell214&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T09:02:22ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_7&diff=1245362020 AMC 10A Problems/Problem 72020-06-08T23:48:53Z<p>Mshell214: /* Solution 4 */</p>
<hr />
<div>{{duplicate|[[2020 AMC 12A Problems|2020 AMC 12A #5]] and [[2020 AMC 10A Problems|2020 AMC 10A #7]]}}<br />
<br />
==Problem==<br />
The <math>25</math> integers from <math>-10</math> to <math>14,</math> inclusive, can be arranged to form a <math>5</math>-by-<math>5</math> square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?<br />
<math>\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50</math><br />
<br />
== Solution == <br />
<br />
Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by <math>5</math> is the total value per row. The sum of the <math>25</math> integers is <math>-10+9+...+14=11+12+13+14=50</math>, and the common sum is <math>\frac{50}{5}=\boxed{\text{(C) }10}</math>.<br />
<br />
<br />
===Solution 2===<br />
<br />
Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get <math>0 + 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}</math> as our answer.<br />
~Baolan<br />
<br />
<br />
===Solution 3===<br />
<br />
Taking the average of the first and last terms, <math>-10</math> and <math>14</math>, we have that the mean of the set is <math>2</math>. There are 5 values in each row, column or diagonal, so the value of the common sum is <math>5\cdot2</math>, or <math>\boxed{\textbf{(C) } 10}</math>. <br />
~Arctic_Bunny, edited by KINGLOGIC<br />
<br />
===Solution 4===<br />
<br />
Let us consider the horizontal rows. Since there are five of them, each with constant sum <math>x</math>, we can add up the 25 numbers in 5 rows for a sum of <math>5x</math>. Since the sum of the 25 numbers used is <math>-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50</math>, <math>5x=50</math> and <math>x=\boxed{\textbf{(C) }10}</math>. <br />
~cw357<br />
<br />
==Video Solution==<br />
https://youtu.be/JEjib74EmiY<br />
<br />
~IceMatrix<br />
<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=6|num-a=8}}<br />
{{AMC12 box|year=2020|ab=A|num-b=4|num-a=6}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_12&diff=1160732020 AMC 10A Problems/Problem 122020-02-01T03:15:05Z<p>Mshell214: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Triangle <math>AMC</math> is isoceles with <math>AM = AC</math>. Medians <math>\overline{MV}</math> and <math>\overline{CU}</math> are perpendicular to each other, and <math>MV=CU=12</math>. What is the area of <math>\triangle AMC?</math><br />
<br />
<math>\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192</math><br />
<br />
== Solution ==<br />
Since quadrilateral <math>UVCM</math> has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that <math>\triangle AUV</math> has <math>\frac 14</math> the area of triangle <math>AMC</math> by similarity, so <math>[UVCM]=\frac 34\cdot [AMC].</math> Thus,<br />
<cmath>\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]</cmath><br />
<cmath>72=\frac 34\cdot [AMC]</cmath><br />
<cmath>[AMC]=96\rightarrow \boxed{\textbf{(C)}}.</cmath><br />
<br />
<br />
==Solution 2 (CHEATING)==<br />
Draw a to-scale diagram with your graph paper and straightedge. Measure the height and approximate the area.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=11|num-a=13}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160622020 AMC 10A Problems/Problem 152020-02-01T03:12:38Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.)<br />
The probability that the divisor chosen is a perfect square is <math>\frac{1}{22}</math>. <math>m + n = 1 + 22 = 23</math> <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
<br />
~mshell214<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160612020 AMC 10A Problems/Problem 152020-02-01T03:12:22Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
<br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.)<br />
<br />
The probability that the divisor chosen is a perfect square is <math>\frac{1}{22}</math>. <math>m + n = 1 + 22 = 23</math> <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
<br />
~mshell214<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160582020 AMC 10A Problems/Problem 152020-02-01T03:11:40Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.)<br />
The probability that the divisor chosen is a perfect square is <math>\frac{1}{22}</math>. <math>m + n = 1 + 22 = 23</math> <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
~mshell214<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160562020 AMC 10A Problems/Problem 152020-02-01T03:10:48Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.)<br />
The probability that the divisor chosen is a perfect square is <math>\frac{1}{22}</math>. <math>m + n = 1 + 22 = 23</math> <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160532020 AMC 10A Problems/Problem 152020-02-01T03:10:11Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.)<br />
The probability that the divisor chosen is a perfect square is <math>\frac{1}{22}</math>. <math>m + n = 1 + 22 = </math>23<math> </math>\implies \boxed{\textbf{(E) } 23 }$<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160522020 AMC 10A Problems/Problem 152020-02-01T03:09:47Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.)<br />
The probability that the divisor chosen is a perfect square is <math>\frac{1}{22}</math>. m + n = <math>1 + 22</math> = <math>23</math> <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160482020 AMC 10A Problems/Problem 152020-02-01T03:09:20Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!.</math> Thus, there are <math>6 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc.)<br />
The probability that the divisor chosen is a perfect square is <math>\frac{1}{22}</math>. m + n = 1 + 22 = 23 <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160372020 AMC 10A Problems/Problem 152020-02-01T03:07:18Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are <math>5 \cdot 3 \cdot 2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!</math>.)<br />
The probability that the divisor chosen is a perfect square is 1/22. m + n = 1 + 22 = 23 <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160342020 AMC 10A Problems/Problem 152020-02-01T03:06:27Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are <math>5</math> * <math>3</math> * <math>2</math> perfect squares. (For <math>2</math>, you can have <math>0</math>, <math>2</math>, <math>4</math>, <math>6</math>, <math>8</math>, or <math>1</math>0 <math>2</math>s, etc. Note that <math>7</math> and <math>11</math> can not be in the prime factorization of a perfect square because there is only one of each in <math>12!</math>.)<br />
The probability that the divisor chosen is a perfect square is 1/22. m + n = 1 + 22 = 23 <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160332020 AMC 10A Problems/Problem 152020-02-01T03:04:27Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of <math>12!</math> is <math>2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11</math>. <br />
This yields a total of <math>11 \cdot 6 \cdot 3 \cdot 2 \cdot 2</math> divisors of <math>12!.</math><br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are 5 * 3 * 2 perfect squares. (For 2, you can have 0, 2, 4, 6, 8, or 10 2s, etc. Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in 12!.)<br />
The probability that the divisor chosen is a perfect square is 1/22. m + n = 23 <math>\implies \boxed{\textbf{(E) } 23 }</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160282020 AMC 10A Problems/Problem 152020-02-01T03:01:33Z<p>Mshell214: /* See Also */</p>
<hr />
<div>==Problem==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==Solution==<br />
The prime factorization of 12! is <math>2^{10}</math> * <math>3^5</math> * <math>5^2</math> * <math>7</math> * <math>11</math>. <br />
This yields a total of <math>11</math> * <math>6</math> * <math>3</math> * <math>2</math> * <math>2</math> divisors of 12!.<br />
In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Thus, there are 5 * 3 * 2 perfect squares. (For 2, you can have 0, 2, 4, 6, 8, or 10 2s, etc. Note that 7 and 11 can not be in the prime factorization of a perfect square because there is only one of each in 12!.)<br />
The probability that the divisor chosen is a perfect square is 1/22. m + n = 23.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160072020 AMC 10A Problems/Problem 152020-02-01T02:49:01Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem 15==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10A_Problems/Problem_15&diff=1160042020 AMC 10A Problems/Problem 152020-02-01T02:46:55Z<p>Mshell214: /* Solution */</p>
<hr />
<div>==Problem 15==<br />
<br />
A positive integer divisor of <math>12!</math> is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is <math>m+n</math>?<br />
<br />
<math>\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23</math><br />
<br />
== Solution ==<br />
The prime factorization of 12! is <math>2^{10}</math>.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2020|ab=A|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_11&diff=1054022018 AMC 10B Problems/Problem 112019-04-20T03:47:47Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>Which of the following expressions is never a prime number when <math>p</math> is a prime number?<br />
<br />
<math>\textbf{(A) } p^2+16 \qquad \textbf{(B) } p^2+24 \qquad \textbf{(C) } p^2+26 \qquad \textbf{(D) } p^2+46 \qquad \textbf{(E) } p^2+96</math><br />
<br />
==Solution 1==<br />
<br />
Because squares of a non-multiple of 3 is always <math>1\mod 3</math>, the only expression is always a multiple of <math>3</math> is <math>\boxed{\textbf{(C) } p^2+26} </math>. This is excluding when <math>p=0\mod3</math>, which only occurs when <math>p=3</math>, then <math>p^2+26=35</math> which is still composite.<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2018|ab=B|num-b=10|num-a=12}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_8&diff=1028952019 AMC 10B Problems/Problem 82019-02-15T13:42:27Z<p>Mshell214: /* Problem */</p>
<hr />
<div>==Problem==<br />
The figure below shows a square and four equilateral triangles, with each triangle having a side lying on a side of the square, such that each triangle has side length 2 and the third vertices of the triangles meet at the center of the square. The region inside the square but outside the triangles is shaded. What is the area of the shaded region? <br />
<br />
<math>(A)</math> <math>4</math> <math>(B)</math> <math>12 - 4\sqrt{3}</math> <math>(C)</math> <math>3\sqrt{3}</math> <math>(D)</math> <math>4\sqrt{3}</math> <math>(E)</math> <math>16 - \sqrt{3}</math><br />
<br />
==Solution==<br />
We notice that the square can be split into <math>4</math> congruent smaller squares with the altitude of the equilateral triangle being the side of the square. Therefore, the area of each shaded part that resides within a square is the total area of the square subtracted from each triangle (Note that it has already been split in half). <br />
<br />
When we split an equilateral triangle in half, we get <math>2</math> triangles with a <math>30-60-90</math> relationship. Therefore, we get that the altitude and a side length of a square is <math>\sqrt{3}</math>. <br />
<br />
We can then compute the area of the two triangles using the base-height-area relationship and get <math>\frac{2 \cdot 1 \cdot \sqrt{3}}{2} = \sqrt{3}</math>.<br />
<br />
The area of the small squares is the altitude squared which is <math>(\sqrt{3})^2 = 3</math>. Therefore, the area of the shaded region in each of the four squares is <math>3 - \sqrt{3}</math><br />
<br />
Since there are four of these squares, we multiply this by <math>4</math> to get <math>4(3 - \sqrt{3}) = 12 - 4 \sqrt{3}</math> as our answer. This is choice <math>\boxed{ B) 12 - 4 \sqrt{3}}</math>.<br />
<br />
<br />
~ Awesome2.1<br />
<br />
Edited by greersc<br />
<br />
==See Also==<br />
<br />
{{AMC10 box|year=2019|ab=B|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028942019 AMC 10B Problems/Problem 192019-02-15T13:41:06Z<p>Mshell214: /* Solution 1 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <math>2^{10}</math> and <math>5^{10}</math> do not work since the factors chosen must be distinct, and those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again and then added on by mshell214)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5</math> or <math>5^5</math>. Since the factors chosen must be distinct, the last two numbers cannot be created because those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94 (edited by mshell214)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028922019 AMC 10B Problems/Problem 192019-02-15T13:38:15Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5</math> or <math>5^5</math>. Since the factors chosen must be distinct, the last two numbers cannot be created because those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94 (edited by mshell214)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028912019 AMC 10B Problems/Problem 192019-02-15T13:37:24Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5 or 5^5</math>. Since the factors chosen must be distinct, the last two numbers cannot be created because those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94 (edited by mshell214)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028902019 AMC 10B Problems/Problem 192019-02-15T13:36:46Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5 or 5^5</math>. Since the factors chosen must be distinct, the last two numbers cannot be created because those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94 (edited by mshell214)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028892019 AMC 10B Problems/Problem 192019-02-15T13:36:24Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5p or </math>5^5. Since the factors chosen must be distinct, the last two numbers cannot be created because those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94 (edited by mshell214)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028882019 AMC 10B Problems/Problem 192019-02-15T13:36:11Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5</math> or <math>5^5. Since the factors chosen must be distinct, the last two numbers cannot be created because those require </math>2^5 \cdot 2^5<math> or </math>5^5 \cdot 5^5<math>. This gives </math>121-4 = 117<math> candidate numbers. It is not too hard to show that every number of the form </math>2^p5^q<math> where </math>0 \le p, q \le 10<math>, and </math>p,q<math> are not both 0 or 10, can be written as a product of two distinct elements in </math>S<math>. Hence the answer is </math>\boxed{\textbf{(C) } 117}$.<br />
<br />
-scrabbler94 (edited by mshell214)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028872019 AMC 10B Problems/Problem 192019-02-15T13:35:37Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5 or </math>5^5. Since the factors chosen must be distinct, the last two numbers cannot be created because those require <math>2^5 \cdot 2^5</math> or <math>5^5 \cdot 5^5</math>. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94 (edited by mshell214)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028862019 AMC 10B Problems/Problem 192019-02-15T13:34:10Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5 or </math>5^5. Since the factors chosen must be distinct, the last two numbers cannot be created because those require 2 or 5 to the fifth power two times. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94 (edited by mshell214)<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028852019 AMC 10B Problems/Problem 192019-02-15T13:33:36Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5 or </math>5^5. Since the factors chosen must be distinct, the last two numbers cannot be created because those require 2 or 5 to the fifth power two times. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10B_Problems/Problem_19&diff=1028842019 AMC 10B Problems/Problem 192019-02-15T13:31:47Z<p>Mshell214: /* Solution 2 */</p>
<hr />
<div>{{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #19]] and [[2019 AMC 12B Problems|2019 AMC 12B #14]]}}<br />
<br />
==Problem==<br />
Let <math>S</math> be the set of all positive integer divisors of <math>100,000.</math> How many numbers are the product of two distinct elements of <math>S?</math><br />
<br />
<math>\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121</math><br />
<br />
==Solution 1==<br />
To find the number of numbers that are the product of two distinct elements of <math>S</math>, we first square <math>100,000</math> and factor it. Factoring, we find <math>100,000^2 = 2^{10} \cdot 5^{10}</math>. Therefore, <math>100,000^2</math> has <math>(10 + 1)(10 + 1) = 121</math> distinct factors. Each of these can be achieved by multiplying two factors of <math>S</math>. However, the factors must be distinct, so we eliminate <math>1</math> and <math>100,000^2</math>, as well as <math>2^{10}</math> and <math>5^{10}</math>, so the answer is <math>121 - 4 = 117</math>. <br />
<br />
Solution by greersc. (Edited by AZAZ12345 and then by greersc once again)<br />
<br />
==Solution 2==<br />
<br />
The prime factorization of 100,000 is <math>2^5 \cdot 5^5</math>. Thus, we choose two numbers <math>2^a5^b</math> and <math>2^c5^d</math> where <math>0 \le a,b,c,d \le 5</math> and <math>(a,b) \neq (c,d)</math>, whose product is <math>2^{a+c}5^{b+d}</math>, where <math>0 \le a+c \le 10</math> and <math>0 \le b+d \le 10</math>.<br />
<br />
Consider <math>100000^2 = 2^{10}5^{10}</math>. The number of divisors is <math>(10+1)(10+1) = 121</math>. However, some of the divisors of <math>2^{10}5^{10}</math> cannot be written as a product of two distinct divisors of <math>2^5 \cdot 5^5</math>, namely: <math>1 = 2^05^0</math>, <math>2^{10}5^{10}</math>, <math>2^{10}</math>, and <math>5^{10}</math>. The last two can not be created because the maximum factor of 100,000 involving only 2s or 5s is only <math>2^5 or </math>5^5. Since the factors chosen must be distinct, <math>2^10 and </math>5^10 cannot be created because those require 2 or 5 to the fifth power two times. This gives <math>121-4 = 117</math> candidate numbers. It is not too hard to show that every number of the form <math>2^p5^q</math> where <math>0 \le p, q \le 10</math>, and <math>p,q</math> are not both 0 or 10, can be written as a product of two distinct elements in <math>S</math>. Hence the answer is <math>\boxed{\textbf{(C) } 117}</math>.<br />
<br />
-scrabbler94<br />
<br />
==See Also==<br />
{{AMC10 box|year=2019|ab=B|num-b=18|num-a=20}}<br />
{{AMC12 box|year=2019|ab=B|num-b=13|num-a=15}}<br />
{{MAA Notice}}</div>Mshell214