https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Piis3141592653&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T21:18:04ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_I_2015_Problems/Problem_9&diff=118619Mock AIME I 2015 Problems/Problem 92020-03-02T01:27:08Z<p>Piis3141592653: </p>
<hr />
<div>Since <math>a</math> is a multiple of <math>b</math>, let <math>a=kb</math>.<br />
<br />
We can rewrite the first and second conditions as:<br />
<br />
(a) <math>(bk)bc</math> is a perfect square, or <math>ck</math> is a perfect square.<br />
<br />
(b) <math>b(k+7)c</math> is a power of <math>2</math>, so it follows that <math>b</math>, <math>c</math>, and <math>k+7</math> are all powers of <math>2</math>.<br />
<br />
Now we use casework on <math>k</math>. Since <math>k+7</math> is a power of <math>2</math>, <math>k</math> is <math>1, 9, 25, 57, 121,</math> or <math>249</math> or <math>k>500</math>.<br />
<br />
If <math>k>500</math>, then no value of <math>b</math> makes <math>1\leq a, b\leq 500</math>.<br />
<br />
If <math>k=57</math> or <math>k=249</math>, then no value of <math>c</math> that is a power of <math>2</math> makes <math>ck</math> a perfect square.<br />
<br />
If <math>k=1</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16, 32, 64, 128, 256</math> for <math>5\cdot 9=45</math> solutions.<br />
<br />
If <math>k=9</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16, 32</math> for <math>5\cdot 6=30</math> solutions.<br />
<br />
If <math>k=25</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16</math> for <math>5\cdot 5=25</math> solutions.<br />
<br />
If <math>k=121</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4</math> for <math>5\cdot 3=15</math> solutions.<br />
<br />
This is a total of <math>\fbox{115}</math> solutions.</div>Piis3141592653https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_I_2015_Problems/Problem_9&diff=118618Mock AIME I 2015 Problems/Problem 92020-03-02T01:26:05Z<p>Piis3141592653: original solution was incorrect, updated with correct answers</p>
<hr />
<div>Since <math>a</math> is a multiple of <math>b</math>, let <math>a=kb</math>.<br />
<br />
We can rewrite the first and second conditions as:<br />
<br />
(a) <math>(bk)bc</math> is a perfect square, or <math>ck</math> is a perfect square.<br />
<br />
(b) <math>b(k+7)c</math> is a power of <math>2</math>, so it follows that <math>b</math>, <math>c</math>, and <math>k+7</math> are all powers of <math>2</math>.<br />
<br />
Now we use casework on <math>k</math>. Since <math>k+7</math> is a power of <math>2</math>, <math>k</math> is <math>1, 9, 25, 57, 121,</math> or <math>249</math> or <math>k>500</math>.<br />
<br />
If <math>k>500</math>, then no value of <math>b</math> makes <math>1\leq a, b\leq 500</math>.<br />
<br />
If <math>k=57</math> or <math>k=249</math>, then no value of <math>c</math> that is a power of <math>2</math> makes <math>ck</math> a perfect square.<br />
<br />
If <math>k=1</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16, 32, 64, 128, 256</math> for <math>5\cdot 9=45</math> solutions.<br />
<br />
If <math>k=9</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16, 32</math> for <math>5\dot 6=30</math> solutions.<br />
<br />
If <math>k=25</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16</math> for <math>5\dot 5=25</math> solutions.<br />
<br />
If <math>k=121</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4</math> for <math>5\dot 3=15</math> solutions.<br />
<br />
This is a total of <math>\fbox{115}</math> solutions.</div>Piis3141592653https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_I_Problems/Problem_1&diff=1081542002 AIME I Problems/Problem 12019-08-02T18:58:26Z<p>Piis3141592653: /* Problem */</p>
<hr />
<div>== Problem ==<br />
Many states use a sequence of three letters followed by a sequence of three digits as their standard license-plate pattern. Given that each three-letter three-digit arrangement is equally likely, the probability that such a license plate will contain at least one palindrome (a three-letter arrangement or a three-digit arrangement that reads the same left-to-right as it does right-to-left) is <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math><br />
<br />
== Solution ==<br />
=== Solution 1 ===<br />
Consider the three-digit arrangement, <math>\overline{aba}</math>. There are <math>10</math> choices for <math>a</math> and <math>10</math> choices for <math>b</math> (since it is possible for <math>a=b</math>), and so the probability of picking the palindrome is <math>\frac{10 \times 10}{10^3} = \frac 1{10}</math>. Similarly, there is a <math>\frac 1{26}</math> probability of picking the three-letter palindrome. <br />
<br />
By the [[Principle of Inclusion-Exclusion]], the total probability is<br />
<center><math>\frac{1}{26}+\frac{1}{10}-\frac{1}{260}=\frac{35}{260}=\frac{7}{52}\quad\Longrightarrow\quad7+52=\boxed{059}</math></center><br />
=== Solution 2 ===<br />
Using complementary counting, we count all of the license plates that do not have the desired property. In order to not be a palindrome, the first and third characters of each string must be different. Therefore, there are <math>10\cdot 10\cdot 9</math> three digit non-palindromes, and there are <math>26\cdot 26\cdot 25</math> three letter non palindromes. As there are <math>10^3\cdot 26^3</math> total three-letter three-digit arrangements, the probability that a license plate does not have the desired property is <math>\frac{10\cdot 10\cdot 9\cdot 26\cdot 26\cdot 25}{10^3\cdot 26^3}=\frac{45}{52}</math>. We subtract this from 1 to get <math>1-\frac{45}{52}=\frac{7}{52}</math> as our probability. Therefore, our answer is <math>7+52=\boxed{059}</math>.<br />
== See also ==<br />
{{AIME box|year=2002|n=I|before=First Question|num-a=2}}<br />
<br />
[[Category:Intermediate Combinatorics Problems]]<br />
[[Category:Intermediate Probability Problems]]<br />
{{MAA Notice}}</div>Piis3141592653https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=1018462018 AMC 12A Problems/Problem 192019-02-11T19:37:40Z<p>Piis3141592653: /* Solution */</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots</cmath>of the reciprocals of the elements of <math>A</math> 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)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math><br />
<br />
== Solution ==<br />
It's just <cmath><br />
\sum_{a=0}^\infty\frac1{2^a}\sum_{b=0}^\infty\frac1{3^b}\sum_{c=0}^\infty\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br />
</cmath> since this represents all the numbers in the denominator.<br />
(ayushk)<br />
<br />
== Solution 2==<br />
Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math>a_{1}</math>, the second as <math>a_{2}</math>, etc. We know <math>a_{1}=1</math>, <math>a_{2}=1/2</math>, <math>a_{3}=1/4</math>, let us find <math>a_{4}</math>. factoring out <math>1/6</math> from the terms in this subsequence, we would get <math>a_{4}=1/6(1+a_{1}+a_{2}+a_{4})</math>. Knowing <math>a_{1}</math> and <math>a_{2}</math>, we can substitute and solve for <math>a_{4}</math>, and we get <math>1/2</math>. If we do the similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them <math>1/4</math> and <math>1/8</math>. Finally, for the seventh sequence, we see <math>a_{7}=1/30(a_{8})</math>, where <math>a_{8}</math> is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get <math>1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, but when we separated the sequence into its parts, we ignored the <math>1/1</math>, so adding in the <math>1</math>, we get <math>1+1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, which when we solve for, we get <math>29/8=29/30(a_{8})</math>, <math>1/8=1/30(a_{8})</math>, <math>30/8=(a_{8})</math>, <math>15/4=(a_{8})</math>. So our answer is 15/4, but we are asked to add the numerator and denominator, which sums up to 19, which is the answer.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Piis3141592653https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_19&diff=1018452018 AMC 12A Problems/Problem 192019-02-11T19:36:58Z<p>Piis3141592653: Solution 3 is superfluous, it's just a restatement of Solution 1</p>
<hr />
<div>== Problem ==<br />
<br />
Let <math>A</math> be the set of positive integers that have no prime factors other than <math>2</math>, <math>3</math>, or <math>5</math>. The infinite sum <cmath>\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{12} + \frac{1}{15} + \frac{1}{16} + \frac{1}{18} + \frac{1}{20} + \cdots</cmath>of the reciprocals of the elements of <math>A</math> 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)} \text{ 16} \qquad \textbf{(B)} \text{ 17} \qquad \textbf{(C)} \text{ 19} \qquad \textbf{(D)} \text{ 23} \qquad \textbf{(E)} \text{ 36}</math><br />
<br />
== Solution ==<br />
It's just <cmath><br />
\sum_{a\ge 0}\frac1{2^a}\sum_{b\ge 0}\frac1{3^b}\sum_{c\ge 0}\frac{1}{5^c} = 2 \cdot \frac32 \cdot \frac54 = \frac{15}{4}\Rightarrow\textbf{(C)}.<br />
</cmath> since this represents all the numbers in the denominator.<br />
(ayushk)<br />
== Solution 2==<br />
Separate into 7 separate infinite series's so we can calculate each and find the original sum. The first infinite sequence shall be all the reciprocals of the powers of 2, the second shall be reciprocals of the powers of 3, and the third is reciprocals of the powers of 5. We can easily calculate these to be <math>1, 1/2, 1/4</math> respectively. The fourth infinite series shall be all real numbers in the form <math> 1/(2^a3^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The fifth is all real numbers in the form <math> 1/(2^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The sixth is all real numbers in the form <math> 1/(3^a5^b)</math>, where <math>a</math> and <math>b</math> are greater than or equal to 1. The seventh infinite series is all real numbers in the form <math> 1/(2^a3^b5^c)</math>, where <math>a</math> and <math>b</math> and <math>c</math> are greater than or equal to 1. Let us denote the first sequence as <math>a_{1}</math>, the second as <math>a_{2}</math>, etc. We know <math>a_{1}=1</math>, <math>a_{2}=1/2</math>, <math>a_{3}=1/4</math>, let us find <math>a_{4}</math>. factoring out <math>1/6</math> from the terms in this subsequence, we would get <math>a_{4}=1/6(1+a_{1}+a_{2}+a_{4})</math>. Knowing <math>a_{1}</math> and <math>a_{2}</math>, we can substitute and solve for <math>a_{4}</math>, and we get <math>1/2</math>. If we do the similar procedures for the fifth and sixth sequences, we can solve for them too, and we get after solving them <math>1/4</math> and <math>1/8</math>. Finally, for the seventh sequence, we see <math>a_{7}=1/30(a_{8})</math>, where <math>a_{8}</math> is the infinite series the problem is asking us to solve for. The sum of all seven subsequences will equal the one we are looking for, so solving, we get <math>1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, but when we separated the sequence into its parts, we ignored the <math>1/1</math>, so adding in the <math>1</math>, we get <math>1+1+1/2+1/4+1/2+1/4+1/8+1/30(a_{8})=a_{8}</math>, which when we solve for, we get <math>29/8=29/30(a_{8})</math>, <math>1/8=1/30(a_{8})</math>, <math>30/8=(a_{8})</math>, <math>15/4=(a_{8})</math>. So our answer is 15/4, but we are asked to add the numerator and denominator, which sums up to 19, which is the answer.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2018|ab=A|num-b=18|num-a=20}}<br />
{{MAA Notice}}</div>Piis3141592653