1995 AMC 8 Problem 23
Attempt Problem 23 of the 1995 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1995 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
23.
How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?
Answer: B
Small hint:
Count choices for each digit position in turn: first, second, third, fourth
Big hint:
First digit: odd options; second: even options; then and for the distinct remaining digits
Solution:
The first digit is odd: choices. The second is even: choices (none of which repeats the odd first digit).
The third digit is any of the unused digits, and the fourth is any of the remaining. In total,
Thus, the correct answer is B .
Problem 23 in other years
- 1985 AMC 8
- 1986 AMC 8
- 1987 AMC 8
- 1988 AMC 8
- 1989 AMC 8
- 1990 AMC 8
- 1991 AMC 8
- 1992 AMC 8
- 1993 AMC 8
- 1994 AMC 8
- 1996 AMC 8
- 1997 AMC 8
- 1998 AMC 8
- 1999 AMC 8
- 2000 AMC 8
- 2001 AMC 8
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