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16 Sep 2014, 01:23
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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:
85%
(hard)
Question Stats:
51% (02:12) correct
49%
(02:22)
wrong
based on 392
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How many odd three-digit integers greater than 800 exist, such that all their digits are distinct?
A. 40
B. 60
C. 72
D. 81
E. 104
A. 40
B. 60
C. 72
D. 81
E. 104
16 Sep 2014, 01:23
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Bookmarks
Official Solution:
How many odd three-digit integers greater than 800 exist, such that all their digits are distinct?
A. 40
B. 60
C. 72
D. 81
E. 104
In the range 800 - 899:
1 choice for the first digit: 8
5 choices for the third digit: 1, 3, 5, 7, 9 (as it must be an odd number)
8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.
Total for this range: \(1*5*8 = 40\).
In the range 900 - 999:
1 choice for the first digit: 9
4 choices for the third digit: 1, 3, 5, 7 (9 is not an option, as it's the first digit)
8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.
Total for this range: \(1*4*8 = 32\).
Combining both ranges, the total number of odd three-digit integers greater than 800 with distinct digits is: \(40 + 32 = 72\).
Answer: C
How many odd three-digit integers greater than 800 exist, such that all their digits are distinct?
A. 40
B. 60
C. 72
D. 81
E. 104
In the range 800 - 899:
1 choice for the first digit: 8
5 choices for the third digit: 1, 3, 5, 7, 9 (as it must be an odd number)
8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.
Total for this range: \(1*5*8 = 40\).
In the range 900 - 999:
1 choice for the first digit: 9
4 choices for the third digit: 1, 3, 5, 7 (9 is not an option, as it's the first digit)
8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.
Total for this range: \(1*4*8 = 32\).
Combining both ranges, the total number of odd three-digit integers greater than 800 with distinct digits is: \(40 + 32 = 72\).
Answer: C
General Discussion
28 Feb 2015, 12:23
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Bunuel
Official Solution:
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.
Total for this range: \(1*5*8 = 40\).
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 the first digit)
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
Total for this range: \(1*4*8 = 32\).
Total: \(40+32 = 72\).
Answer: C
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.
Total for this range: \(1*5*8 = 40\).
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 the first digit)
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
Total for this range: \(1*4*8 = 32\).
Total: \(40+32 = 72\).
Answer: C
Hi Bunuel,
I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit odd integers then why are we including the numbers from 801 - 899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913 - all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you.
