2002 AMC 12B Problems/Problem 15
Contents
Problem
How many four-digit numbers have the property that the three-digit number obtained by removing the leftmost digit is one ninth of ?
Solution
Let , such that . Then . Since , from we have three-digit solutions, and the answer is .
Solution 2
Since N is a four digit number, assume WLOG that , where a is the thousands digit, b is the hundreds digit, c is the tens digit, and d is the ones digit. Then, , so Set these equal to each other: Notice that , thus:
Go back to our first equation, in which we set , Then: The upper limit for the right hand side (RHS) is (when , , and ). It's easy to prove that for an there is only one combination of and that can make the equation equal. Just think about the RHS as a three digit number . There's one and only one way to create every three digit number with a certain combination of digits. Thus, we test for how many as are in the domain set by the RHS. Since which is the largest value, then can be through , giving us the answer of
IronicNinja~
See also
2002 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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All AMC 12 Problems and Solutions |
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