Difference between revisions of "2002 AMC 12B Problems/Problem 15"
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The upper limit for the right hand side (RHS) is <math>999</math> (when <math>b = 9</math>, <math>c = 9</math>, and <math>d = 9</math>). | The upper limit for the right hand side (RHS) is <math>999</math> (when <math>b = 9</math>, <math>c = 9</math>, and <math>d = 9</math>). | ||
It's easy to prove that for an <math>a</math> there is only one combination of <math>b, c,</math> and <math>d</math> that can make the equation equal. Just think about the RHS as a three digit number <math>bcd</math>. There's one and only one way to create every three digit number with a certain combination of digits. | It's easy to prove that for an <math>a</math> there is only one combination of <math>b, c,</math> and <math>d</math> that can make the equation equal. Just think about the RHS as a three digit number <math>bcd</math>. 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 <math>125\cdot7 = | + | Thus, we test for how many as are in the domain set by the RHS. Since <math>125\cdot7 = 875</math> which is the largest <math>a</math> value, then <math>a</math> can be <math>1</math> through <math>7</math>, giving us the answer of <math>\boxed {D) 7}</math> |
IronicNinja~ | IronicNinja~ |
Latest revision as of 15:50, 3 April 2022
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 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.