https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Liant&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T13:46:07ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_25&diff=754902016 AMC 10A Problems/Problem 252016-02-04T22:23:03Z<p>Liant: /* Solution */</p>
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<div>==Problem==<br />
How many ordered triples <math>(x,y,z)</math> of positive integers satisfy <math>\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600</math> and <math>\text{lcm}(y,z)=900</math>?<br />
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<math>\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64</math><br />
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==Solution==<br />
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We prime factorize <math>72,600,</math> and <math>900</math>. The prime factorizations are <math>2^3\times 3^2</math>, <math>2^3\times 3\times 5^2</math> and <math>2^2\times 3^2\times 5^2</math>, respectively. Let <math>x=2^a\times 3^b\times 5^c</math>, <math>y=2^d\times 3^e\times 5^f</math> and <math>z=2^g\times 3^h\times 5^i</math>. We know that <cmath>\max(a,d)=3</cmath> <cmath>\max(b,e)=2</cmath> <cmath>\max(a,g)=3</cmath> <cmath>\max(b,h)=1</cmath> <cmath>\max(c,i)=2</cmath> <cmath>\max(d,g)=2</cmath> <cmath>\max(e,h)=2</cmath> and <math>c=f=0</math> since <math>\text{lcm}(x,y)</math> isn't a multiple of 5. Since <math>\max(d,g)=2</math> we know that <math>a=3</math>. We also know that since <math>\max(b,h)=1</math> that <math>e=2</math>. So now some equations have become useless to us...let's take them out. <cmath>\max(b,h)=1</cmath> <cmath>\max(d,g)=2</cmath> are the only two important ones left. We do casework on each now. If <math>\max(b,h)=1</math> then <math>(b,h)=(1,0),(0,1)</math> or <math>(1,1)</math>. Similarly if <math>\max(d,g)=2</math> then <math>(d,g)=(2,0),(2,1),(2,2),(1,2),(0,2)</math>. Thus our answer is <math>5\times 3=\boxed{15 \text{(A)}}</math>.<br />
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==See Also==<br />
{{AMC10 box|year=2016|ab=A|num-b=24|after=Last Problem}}<br />
{{AMC12 box|year=2016|ab=A|num-b=21|num-a=23}}<br />
{{MAA Notice}}</div>Lianthttps://artofproblemsolving.com/wiki/index.php?title=2013_AMC_8_Problems/Problem_15&diff=648722013 AMC 8 Problems/Problem 152014-10-13T18:53:52Z<p>Liant: /* Solution */</p>
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<div>==Problem==<br />
If <math>3^p + 3^4 = 90</math>, <math>2^r + 44 = 76</math>, and <math>5^3 + 6^s = 1421</math>, what is the product of <math>p</math>, <math>r</math>, and <math>s</math>?<br />
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<math>\textbf{(A)}\ 27 \qquad \textbf{(B)}\ 40 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 70 \qquad \textbf{(E)}\ 90</math><br />
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==Solution==<br />
<math>3^p + 3^4 = 90\\<br />
3^p + 81 = 90\\<br />
3^p = 9</math><br />
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Therefore, <math>p = 2</math>.<br />
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<math>2^r + 44 = 76\\<br />
2^r = 32</math><br />
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Therefore, <math>r = 5</math>.<br />
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<math>5^3 + 6^s = 1421\\<br />
125 + 6^s = 1421\\<br />
6^s=1296</math><br />
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To most people, it would not be immediately evident that <math>6^4 = 1296</math>, so we can multiply 6's until we get the desired number:<br />
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<math>6\cdot6=36</math><br />
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<math>6\cdot36=216</math><br />
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<math>6\cdot216=1296=6^4</math>, so <math>s=4</math>.<br />
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Therefore the answer is <math>2\cdot5\cdot4=\boxed{\textbf{(B)}\ 40}</math>.<br />
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==See Also==<br />
{{AMC8 box|year=2013|num-b=14|num-a=16}}<br />
{{MAA Notice}}</div>Liant