Difference between revisions of "1993 AIME Problems/Problem 1"
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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>. |
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+ | 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}} | ||
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+ | [[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 |