Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Problem)
(→‎Solution)
Line 5: Line 5:
 
== Solution ==
 
== Solution ==
  
−
Note that the product of the first <math>\displaystyle 100</math> positive odd integers can be written as <math>\displaystyle 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>
+
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.
−
 
 
−
Hence, we seek the number of threes in <math>\displaystyle 200!</math> decreased by the number of threes in <math>\displaystyle 100!.</math>
 
−
 
 
−
There are
 
−
 
 
−
<math>\displaystyle \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>
 
−
 
 
−
threes in <math>\displaystyle 200!</math> and
 
−
 
 
−
<math>\displaystyle \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>
 
−
 
 
−
threes in <math>\displaystyle 100!</math>
 
−
 
 
−
Therefore, we have a total of <math>\displaystyle 97-48=049</math> threes.
 
−
 
 
−
For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]].
 
  
 
== See also ==
 
== See also ==

Revision as of 13:00, 25 September 2007

Problem

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_A_Problems/Problem_3&oldid=16803"