Difference between revisions of "1993 AIME Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | The thousands digit is <math>\in \{4,5,6\} | + | The thousands digit is <math>\in \{4,5,6\}</math>. |
| − | + | Case <math>1</math>: Thousands digit is even | |
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| + | <math>4, 6</math>, two 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 8 \cdot 7 \cdot 4 = 448</math>. | ||
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| + | Case <math>2</math>: Thousands digit is odd | ||
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| + | <math>5</math>, one possibility, then there are <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 8 \cdot 7 \cdot 5= 280</math> possibilities. | ||
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| + | Together, the solution is <math>448 + 280 = \boxed{728}</math>. | ||
== See also == | == See also == | ||
Revision as of 19:47, 9 August 2020
Problem
How many even integers between 4000 and 7000 have four different digits?
Solution
The thousands digit is 
.
Case 
: Thousands digit is even
, two 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 
.
Case 
: Thousands digit is odd
, one possibility, then there are 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
