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

Revision as of 14:55, 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.

@@ Line 8: / Line 8: @@
 == See also ==
-*[{{AIME box|year=2006|n=II|num-b=2|num-a=4}}
+*{{AIME box|year=2006|n=II|num-b=2|num-a=4}}
 [[Category:Intermediate Number Theory Problems]]

2006 AIME II (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

Page

Toolbox

Search

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

Revision as of 14:55, 25 September 2007

Problem

Solution

See also