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1974 AHSME Problems/Problem 28 - Revision history
2024-03-29T15:33:12Z
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Nathan wailes at 16:44, 5 July 2013
2013-07-05T16:44:27Z
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Nathan wailes
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1=2: /* See Also */
2012-05-30T14:21:24Z
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1=2
https://artofproblemsolving.com/wiki/index.php?title=1974_AHSME_Problems/Problem_28&diff=47165&oldid=prev
Admin25: Created page with "==Problem== Which of the following is satisfied by all numbers <math> x </math> of the form <cmath> x=\frac{a_1}{3}+\frac{a_2}{3^2}+\cdots+\frac{a_{25}}{3^{25}} </cmath> where ..."
2012-05-27T02:41:49Z
<p>Created page with "==Problem== Which of the following is satisfied by all numbers <math> x </math> of the form <cmath> x=\frac{a_1}{3}+\frac{a_2}{3^2}+\cdots+\frac{a_{25}}{3^{25}} </cmath> where ..."</p>
<p><b>New page</b></p><div>==Problem==<br />
Which of the following is satisfied by all numbers <math> x </math> of the form<br />
<br />
<cmath> x=\frac{a_1}{3}+\frac{a_2}{3^2}+\cdots+\frac{a_{25}}{3^{25}} </cmath><br />
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where <math> a_1 </math> is <math> 0 </math> or <math> 2 </math>, <math> a_2 </math> is <math> 0 </math> or <math> 2 </math>,...,<math> a_{25} </math> is <math> 0 </math> or <math> 2 </math>?<br />
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<math> \mathrm{(A)\ } 0\le x<1/3 \qquad \mathrm{(B) \ } 1/3\le x<2/3 \qquad \mathrm{(C) \ } 2/3\le x<1 \qquad </math> <br />
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<math> \mathrm{(D) \ } 0\le x<1/3 \text{ or }2/3\le x<1 \qquad \mathrm{(E) \ }1/2\le x\le 3/4 </math><br />
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==Solution==<br />
We can separate this into two cases.<br />
<br />
'''Case 1: <math> a_1=0 </math>''' Here, clearly the minimum sum is when all of the <math> a_i </math> are <math> 0 </math>, so <math> 0\le x </math>. The maximum sum is when the rest of the <math> a_i </math> are all <math> 2 </math>. In this case, <cmath> \sum_{n=2}^{25}\frac{2}{3^n}<2\sum_{n=2}^{\infty}\frac{1}{3^n}=2\left(\frac{\frac{1}{9}}{1-\frac{1}{3}}\right)=\frac{1}{3}. </cmath> Therefore, in this case, <math> 0\le x<1/3 </math>.<br />
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'''Case 2: <math> a_1=2 </math>''' Here, again, the minimum sum is when the rest of the <math> a_i </math> are all <math> 0 </math>, in which case the minimum sum is <math> \frac{2}{3} </math>. The maximum sum is attained when all of the <math> a_i </math> are <math> 2 </math>. In that case, <cmath> \sum_{n=1}^{25}\frac{2}{3^n}<2\sum_{n=1}^{\infty}\frac{1}{3^n}=2\left(\frac{\frac{1}{3}}{1-\frac{1}{3}}\right)=1. </cmath> Therefore, in this case, <math> 2/3\le x<1 </math>.<br />
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Thus, we must have <math> 0\le x<1/3 </math> or <math> 2/3\le x<1 </math>, <math> \boxed{\text{D}} </math>.<br />
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==See Also==<br />
{{AHSME box|year=1974|num-b=27|num-a=29}}</div>
Admin25