2001 AMC 8 Problems/Problem 4
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[hide]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 |
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