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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.
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
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
In the range 800 - 900: 1 choice for the first digit: 8; 5 choices for the third digit: 1, 3, 5, 7, 9; 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
1*5*8 = 40.
In the range 900 - 999: 1 choice for the first digit: 9; 4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's first digit); 8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
Re: How many odd three-digit integers greater than 800 are there [#permalink]
23 Jan 2015, 02:34
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