Difference between revisions of "1993 AIME Problems/Problem 1"
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| − | The thousands digit is <math>\in \{4,5,6\}</math>. If the thousands digit is even (<math>4, | + | 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 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 5 \cdot 8 \cdot 7 = 280</math> possibilities. Together, the solution is <math>448 + 280 = \boxed{728}</math>. | If the thousands digit is odd (<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 5 \cdot 8 \cdot 7 = 280</math> possibilities. Together, the solution is <math>448 + 280 = \boxed{728}</math>. | ||
Revision as of 19:39, 9 August 2020
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 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.