Difference between revisions of "2006 AMC 12B Problems/Problem 9"
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Let the integer have digits <math>a</math>, <math>b</math>, and <math>c</math>, read left to right. Because <math>1 \leq a<b<c</math>, none of the digits can be zero and <math>c</math> cannot be 2. If <math>c=4</math>, then <math>a</math> and <math>b</math> must each be chosen from the digits 1, 2, and 3. Therefore there are <math>\binom{3}{2}=3</math> choices for <math>a</math> and <math>b</math>, and for each choice there is one acceptable order. Similarly, for <math>c=6</math> and <math>c=8</math> there are, respectively, <math>\binom{5}{2}=10</math> and <math>\binom{7}{2}=21</math> choices for <math>a</math> and <math>b</math>. Thus there are altogether <math>3+10+21=\boxed{34}</math> such integers. | Let the integer have digits <math>a</math>, <math>b</math>, and <math>c</math>, read left to right. Because <math>1 \leq a<b<c</math>, none of the digits can be zero and <math>c</math> cannot be 2. If <math>c=4</math>, then <math>a</math> and <math>b</math> must each be chosen from the digits 1, 2, and 3. Therefore there are <math>\binom{3}{2}=3</math> choices for <math>a</math> and <math>b</math>, and for each choice there is one acceptable order. Similarly, for <math>c=6</math> and <math>c=8</math> there are, respectively, <math>\binom{5}{2}=10</math> and <math>\binom{7}{2}=21</math> choices for <math>a</math> and <math>b</math>. Thus there are altogether <math>3+10+21=\boxed{34}</math> such integers. | ||
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+ | |||
+ | |||
+ | (Edited by HMSSONI82) | ||
==Solution 2== | ==Solution 2== | ||
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The answer must be half of a triangular number (evens and decreasing/increasing) so <math>\boxed{34}</math> or the letter B. | The answer must be half of a triangular number (evens and decreasing/increasing) so <math>\boxed{34}</math> or the letter B. | ||
- | - | ||
+ | |||
+ | ==Solution 5 (Forward Casework + Listing)== | ||
+ | Casework: | ||
+ | |||
+ | For the sake of simplicity, we are going to call the number <math>\overline{abc}</math>. | ||
+ | |||
+ | 1. If <math>a=1</math>: | ||
+ | |||
+ | a. <math>c=2</math>. No such number exists. | ||
+ | |||
+ | b. <math>c=4</math>. <math>b=2, 3</math>. | ||
+ | |||
+ | c. <math>c=6</math>. <math>b=2, 3, 4, 5</math>. | ||
+ | |||
+ | d. <math>c=8</math>. <math>b=2, 3, 4, 5, 6, 7</math>. | ||
+ | |||
+ | 2. If <math>a=2</math>: continue as above. | ||
+ | |||
+ | We can count up that there are 34 such integers, so select <math>\boxed{B}</math>. | ||
+ | |||
+ | ~hastapasta | ||
== See also == | == See also == | ||
{{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}} | {{AMC12 box|year=2006|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 17:10, 30 April 2024
Contents
Problem
How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
Solution 1 (Alcumus Edition)
Let the integer have digits ,
, and
, read left to right. Because
, none of the digits can be zero and
cannot be 2. If
, then
and
must each be chosen from the digits 1, 2, and 3. Therefore there are
choices for
and
, and for each choice there is one acceptable order. Similarly, for
and
there are, respectively,
and
choices for
and
. Thus there are altogether
such integers.
(Edited by HMSSONI82)
Solution 2
Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).
If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit. If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit. If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,
and so on.
So, the answer is .
Solution 3
The last digit is 4, 6, or 8.
If the last digit is , the possibilities for the first two digits correspond to 2-element subsets of
.
Thus the answer is .
Solution 4
The answer must be half of a triangular number (evens and decreasing/increasing) so or the letter B.
-
Solution 5 (Forward Casework + Listing)
Casework:
For the sake of simplicity, we are going to call the number .
1. If :
a. . No such number exists.
b. .
.
c. .
.
d. .
.
2. If : continue as above.
We can count up that there are 34 such integers, so select .
~hastapasta
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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 |
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