https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Akhil0422&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T07:25:09ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_2_2006-2007_Problems/Problem_15&diff=29821Mock AIME 2 2006-2007 Problems/Problem 152009-01-30T00:06:31Z<p>Akhil0422: /* Problem */</p>
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<div>== Problem ==<br />
A <math>4\times4\times4</math> [[cube (geometry) | cube]] is composed of <math>64</math> unit cubes. The faces of <math>16</math> unit cubes are colored red. An arrangement of the cubes is "intriguing" if there is exactly <math>1</math> red unit cube in every <math>1\times1\times4</math> rectangular box composed of <math>4</math> unit cubes. Determine the number of "intriguing" colorings.<br />
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[[Image:CubeArt.jpg]]<br />
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==Solution==<br />
{{solution}}<br />
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*[[Mock AIME 2 2006-2007/Problem 14 | Previous Problem]]<br />
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*[[Mock AIME 2 2006-2007]]<br />
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== Problem Source ==</div>Akhil0422https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_5&diff=298201990 AIME Problems/Problem 52009-01-30T00:05:02Z<p>Akhil0422: /* Problem */</p>
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<div>== Problem ==<br />
Let <math>n^{}_{}</math> be the smallest positive [[integer]] that is a multiple of <math>75_{}^{}</math> and has exactly <math>75_{}^{}</math> positive integral divisors, including <math>1_{}^{}</math> and itself. Find <math>\frac{n}{75}</math>.<br />
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== Solution ==<br />
The [[prime factorization]] of <math>75 = 3^15^2 = (2+1)(4+1)(4+1)</math>. For <math>n</math> to have exactly <math>75</math> integral divisors, we need to have <math>n = p_1^{e_1-1}p_2^{e_2-1}\cdots</math> such that <math>e_1e_2 \cdots = 75</math>. Since <math>75|n</math>, two of the [[prime]] [[factor]]s must be <math>3</math> and <math>5</math>. To minimize <math>n</math>, we can introduce a third prime factor, <math>2</math>. Also to minimize <math>n</math>, we want <math>5</math>, the greatest of all the factors, to be raised to the least power. Therefore, <math>n = 2^43^45^2</math> and <math>\frac{n}{75} = \frac{2^43^45^2}{3 \cdot 5^2} = 16 \cdot 27 = \boxed{432}</math>.<br />
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== See also ==<br />
{{AIME box|year=1990|num-b=4|num-a=6}}<br />
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[[Category:Intermediate Number Theory Problems]]</div>Akhil0422https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12B_Problems/Problem_7&diff=298192006 AMC 12B Problems/Problem 72009-01-30T00:03:23Z<p>Akhil0422: Aesthetics</p>
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<div>== Problem ==<br />
Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seating arrangements are possible?<br />
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<math><br />
\text {(A) } 4 \qquad \text {(B) } 12 \qquad \text {(C) } 16 \qquad \text {(D) } 24 \qquad \text {(E) } 48<br />
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== Solution ==<br />
First, we seat the children.<br />
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The first child can be seated in <math>3</math> spaces.<br />
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The second child can be seated in <math>2</math> spaces.<br />
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Now there are <math>2 \times 1</math> ways to seat the adults.<br />
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<math>3 \times 2 \times 2 = 12 \Rightarrow \text{(B)}</math><br />
== See also ==<br />
{{AMC12 box|year=2006|ab=B|num-b=6|num-a=8}}</div>Akhil0422