Difference between revisions of "2001 AMC 8 Problems/Problem 4"
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==Solution 2 (Faster)== | ==Solution 2 (Faster)== | ||
| − | We know that the smallest possible number is <math>12349</math>. However, <math>12349</math> is an odd number. Thus, by shifting the 4 (the largest even number) into the unit's place and rearranging the rest of the digits from least to greatest | + | We know that the smallest possible number is <math>12349</math>. However, <math>12349</math> is an odd number. Thus, by shifting the 4 (the largest even number) into the unit's place and rearranging the rest of the digits from least to greatest, we arrive at <math>12394</math>. Thus, the digit in the tens place is <math>\boxed{\text{E}}</math> <math>9</math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=3|num-a=5}} | {{AMC8 box|year=2001|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 03:17, 14 November 2024
Problem
The digits 1, 2, 3, 4 and 9 are each used once to form the smallest possible even five-digit number. The digit in the tens place is
Solution 1
Since the number is even, the last digit must be 
or 
. To make the smallest possible number, the ten-thousands digit must be as small as possible, so the ten-thousands digit is 
. Simillarly, the thousands digit has second priority, so it must also be as small as possible once the ten-thousands digit is decided, so the thousands digit is . Similarly, the hundreds digit needs to be the next smallest number, so it is 
. However, for the tens digit, we can't use , since we already used and the number must be even, so the units digit must be and the tens digit is 
(The number is 
.)
Solution 2 (Faster)
We know that the smallest possible number is 
. However, is an odd number. Thus, by shifting the 4 (the largest even number) into the unit's place and rearranging the rest of the digits from least to greatest, we arrive at . Thus, the digit in the tens place is 
.
See Also
| 2001 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
