1993 AIME Problems/Problem 1: Difference between revisions
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== Problem == | == Problem == | ||
How many even | How many even [[integer]]s between 4000 and 7000 have four different digits? | ||
== Solution == | == Solution == | ||
{{ | The thousands digit is <math>\in \{4,5,6\}</math>. If the thousands digit is even (<math>4,\ 6</math>, 2 possibilities), then there are only <math>\frac{10}{2} - 1 = 4</math> possibilities for the units digit. This leaves <math>8</math> possible digits for the hundreds and <math>7</math> for the tens places, yielding a total of <math>2 \cdot 4 \cdot 8 \cdot 7 = 448</math>. | ||
If the thousands digit is odd (<math>5</math>, one possibility), then there is <math>5</math> choices for the units digit, with <math>8</math> digits for the hundreds and <math>7</math> for the tens place. This gives <math>1 \cdot 5 \cdot 8 \cdot 7 = 280</math> possibilities. Together, the solution is <math>448 + 280 = 728</math>. | |||
== See also == | == See also == | ||
{{AIME box|year=1993|before=First question|num-a=2}} | {{AIME box|year=1993|before=First question|num-a=2}} | ||
[[Category:Intermediate Combinatorics Problems]] | |||
Revision as of 15:21, 26 March 2007
Problem
How many even integers between 4000 and 7000 have four different digits?
Solution
The thousands digit is 
. If the thousands digit is even (
, 2 possibilities), then there are only 
possibilities for the units digit. This leaves 
possible digits for the hundreds and 
for the tens places, yielding a total of 
.
If the thousands digit is odd (
, one possibility), then there is choices for the units digit, with digits for the hundreds and for the tens place. This gives 
possibilities. Together, the solution is 
.
See also
| 1993 AIME (Problems • Answer Key • Resources) | ||
| Preceded by First question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
