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Difference between revisions of "2006 AIME I Problems/Problem 3"

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== Problem ==
 
== Problem ==
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Let <math>P </math> be the product of the first <math>100</math> [[positive integer | positive]] [[odd integer]]s. Find the largest integer <math>k </math> such that <math>P </math> is divisible by <math>3^k .</math>
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Find the least [[positive]] [[integer]] such that when its leftmost [[digit]] is deleted, the resulting integer is <math>\frac{1}{29}</math> of the original integer.
  
 
== Solution ==
 
== Solution ==
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The number can be represented as <math>10^na+b</math>, where <math> a </math> is the leftmost digit, and <math> b </math> is the rest of the number. We know that <math>b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na</math>. Thus <math> a </math> has to be 7 since <math> 10^n </math> can not have 7 as a factor, and the smallest <math> 10^n </math> can be and have a factor of <math> 2^2 </math> is <math> 10^2=100. </math> We find that <math> b </math> is 25, so the number is 725.
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Note that the product of the first <math>100</math> positive odd integers can be written as <math>1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}</math>
 
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Hence, we seek the number of threes in <math>200!</math> decreased by the number of threes in <math>100!.</math>
 
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There are
 
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<math>\left\lfloor \frac{200}{3}\right\rfloor+\left\lfloor\frac{200}{9}\right\rfloor+\left\lfloor \frac{200}{27}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor =66+22+7+2=97</math>
 
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threes in <math>200!</math> and
 
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<math>\left\lfloor \frac{100}{3}\right\rfloor+\left\lfloor\frac{100}{9}\right\rfloor+\left\lfloor \frac{100}{27}\right\rfloor+\left\lfloor\frac{100}{81}\right\rfloor=33+11+3+1=48 </math>
 
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threes in <math>100!</math>
 
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Therefore, we have a total of <math>97-48=049</math> threes.
 
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For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]].
 
  
 
== See also ==
 
== See also ==

Revision as of 18:24, 25 September 2007

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.

Solution

The number can be represented as $10^na+b$, where $a$ is the leftmost digit, and $b$ is the rest of the number. We know that $b=\frac{10^na+b}{29} \implies 28b=2^2\times7b=10^na$. Thus $a$ has to be 7 since $10^n$ can not have 7 as a factor, and the smallest $10^n$ can be and have a factor of $2^2$ is $10^2=100.$ We find that $b$ is 25, so the number is 725.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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