https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Easyas+pi&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T10:09:05ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_10B_Problems/Problem_16&diff=911752015 AMC 10B Problems/Problem 162018-02-13T18:31:40Z<p>Easyas pi: /* Problem */</p>
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<div>==Problem==<br />
Al, Bill, and Cal will each randomly be assigned a whole number from <math>1</math> to <math>10</math>, inclusive, with no two of them getting the same number. What is the probability that Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's?<br />
<math>\textbf{(A) } \dfrac{9}{1000} \qquad\textbf{(B) } \dfrac{1}{90} \qquad\textbf{(C) } \dfrac{1}{80} \qquad\textbf{(D) } \dfrac{1}{72} \qquad\textbf{(E) } \dfrac{2}{121} </math><br />
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==Solution==<br />
We can solve this problem with a brute force approach.<br />
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*If Cal's number is <math>1</math>:<br />
**If Bill's number is <math>2</math>, Al's can be any of <math>4, 6, 8, 10</math>.<br />
**If Bill's number is <math>3</math>, Al's can be any of <math>6, 9</math>.<br />
**If Bill's number is <math>4</math>, Al's can be <math>8</math>.<br />
**If Bill's number is <math>5</math>, Al's can be <math>10</math>.<br />
**Otherwise, Al's number could not be a whole number multiple of Bill's.<br />
*If Cal's number is <math>2</math>:<br />
**If Bill's number is <math>4</math>, Al's can be <math>8</math>.<br />
**Otherwise, Al's number could not be a whole number multiple of Bill's while Bill's number is still a whole number multiple of Cal's.<br />
*Otherwise, Bill's number must be greater than <math>5</math>, i.e. Al's number could not be a whole number multiple of Bill's.<br />
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Clearly, there are exactly <math>9</math> cases where Al's number will be a whole number multiple of Bill's and Bill's number will be a whole number multiple of Cal's. Since there are <math>10*9*8</math> possible permutations of the numbers Al, Bill, and Cal were assigned, the probability that this is true is <math>\frac9{10*9*8}=\boxed{\text{(\textbf C) }\frac1{80}}</math><br />
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==See Also==<br />
{{AMC10 box|year=2015|ab=B|num-b=15|num-a=17}}<br />
{{MAA Notice}}</div>Easyas pi