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21 Aug 2009, 05:07
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
61% (02:07) correct
39%
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wrong
based on 229
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How many odd three-digit integers greater than 800 are there such that all their digits are different?
A. 40
B. 60
C. 72
D. 81
E. 104
A. 40
B. 60
C. 72
D. 81
E. 104
If the number begins with 8, there are possibilities (5 possibilities for the last digit and 8 possibilities for the middle digit). If the number begins with 9, there are possibilities (4 possibilities for the last digit and 8 possibilities for the middle digit). In all, there are numbers that satisfy the constraints.
The correct answer is C.
GMAC25 : i think there are 9 possiblities for middle digit, 0-9 and excluding 8, now how can we say that there are 5 possibilties for last digit when middle digit is 1 then possible last digit values are 3,5,7,9.
i am not able to understand this.
The correct answer is C.
GMAC25 : i think there are 9 possiblities for middle digit, 0-9 and excluding 8, now how can we say that there are 5 possibilties for last digit when middle digit is 1 then possible last digit values are 3,5,7,9.
i am not able to understand this.
21 Aug 2009, 05:27
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Last digit is always ODD:
case1: first digit is 8, then last digit has 5 options(1,3,5,7,9) and middle digit has 8 options as we have already chosen two digits for the first and the last positions.
total numbers = 1*8*5 = 40
case2: first digit is 9, then last digit has 4 options(1,3,5,7) and middle digit has 8 options again.
total numbers = 1*8*4 = 32
Ans:72
NOTE: last digit should always be one of 1,3,5,7,9
case1: first digit is 8, then last digit has 5 options(1,3,5,7,9) and middle digit has 8 options as we have already chosen two digits for the first and the last positions.
total numbers = 1*8*5 = 40
case2: first digit is 9, then last digit has 4 options(1,3,5,7) and middle digit has 8 options again.
total numbers = 1*8*4 = 32
Ans:72
NOTE: last digit should always be one of 1,3,5,7,9
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