Difference between revisions of "1993 AIME Problems/Problem 1"

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== 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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The thousands digit is <math>\in \{4,5,6\}</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>.
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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 ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Revision as of 20:47, 9 August 2020

Problem

How many even integers between 4000 and 7000 have four different digits?

Solution

The thousands digit is $\in \{4,5,6\}$.

Case $1$: Thousands digit is even

$4, 6$, two possibilities, then there are only $\frac{10}{2} - 1 = 4$ possibilities for the units digit. This leaves $8$ possible digits for the hundreds and $7$ for the tens places, yielding a total of $2 \cdot 8 \cdot 7 \cdot 4 = 448$.


Case $2$: Thousands digit is odd

$5$, one possibility, then there are $5$ choices for the units digit, with $8$ digits for the hundreds and $7$ for the tens place. This gives $1 \cdot 8 \cdot 7 \cdot 5= 280$ possibilities.

Together, the solution is $448 + 280 = \boxed{728}$.

See also

1993 AIME (ProblemsAnswer KeyResources)
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

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