Difference between revisions of "2002 AMC 12B Problems/Problem 15"
Ironicninja (talk | contribs) |
Lajiaoxushen (talk | contribs) m (→Solution 2 - Longer but might make more sense) |
||
Line 30: | Line 30: | ||
<cmath>125a = 100b + 10c + d</cmath> | <cmath>125a = 100b + 10c + d</cmath> | ||
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 a there is only one combination of b, c, and d that can make the equation equal. Just think about the RHS as a three digit number bcd. 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 = 975</math> which is the largest a value, then a can be <math>1</math> through <math>7</math>, giving us the answer of <math>\boxed {D) 7}</math> | Thus, we test for how many as are in the domain set by the RHS. Since <math>125\cdot7 = 975</math> which is the largest a value, then a can be <math>1</math> through <math>7</math>, giving us the answer of <math>\boxed {D) 7}</math> | ||
IronicNinja~ | IronicNinja~ | ||
+ | |||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=B|num-b=14|num-a=16}} | {{AMC12 box|year=2002|ab=B|num-b=14|num-a=16}} |
Revision as of 00:37, 3 May 2019
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 - Longer but might make more sense
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 a value, then a 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.