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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1053
GPA: 3.39
Of the three-digit positive integers whose three digits are all differ
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29 Nov 2021, 07:31
29 Nov 2021, 07:31
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Question Stats:
27% (03:28) correct
72% (03:10) wrong based on 29 sessions
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Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?
(A) 84
(B) 91
(C) 100
(D) 105
(E) 243
(A) 84
(B) 91
(C) 100
(D) 105
(E) 243
Re: Of the three-digit positive integers whose three digits are all differ
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01 Dec 2021, 02:16
01 Dec 2021, 02:16
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As for the question, you need to find the number of combinations possible such that:
1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9
Based on this, the numbers can be of the following 3 types:
Type 1: 7AB
Type 2: 8EF
Type 3: 9CD
For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).
Number of combinations for type 1 : 1*7*4 = 28
Number of combinations for type 2: 1*7*5 = 35
Number of combinations for type 3 : 1*7*4 = 28
Thus, total numbers possible = 28+35+28 = 91.
B is thus the correct answer.
1. It is a 3 digit number > 700 (or between 701-999, inclusive).
2. All digits are different and NON ZERO.
3. The numbers must be ODD --> the last digit can be 1 of 1,3,5,7,9
Based on this, the numbers can be of the following 3 types:
Type 1: 7AB
Type 2: 8EF
Type 3: 9CD
For type 1 and type 3, be very careful that the digits must be different. So if it is 7AB, then B can NOT be 7. Similarly for type 3, 9CD, D can NOT be 9. There is no such restriction when you find numbers of type 2 (8EF).
Number of combinations for type 1 : 1*7*4 = 28
Number of combinations for type 2: 1*7*5 = 35
Number of combinations for type 3 : 1*7*4 = 28
Thus, total numbers possible = 28+35+28 = 91.
B is thus the correct answer.
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Re: Of the three-digit positive integers whose three digits are all differ
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18 Apr 2023, 16:33
18 Apr 2023, 16:33
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GeminiHeat wrote:
Of the three-digit positive integers whose three digits are all different and nonzero, how many are odd integers greater than 700?
(A) 84
(B) 91
(C) 100
(D) 105
(E) 243
(A) 84
(B) 91
(C) 100
(D) 105
(E) 243
We know the integers must be greater than 700 and they must be odd; we can divide them into 6 groups: 7EO, 7OO, 8EO, 8OO, 9EO and 9OO where E denotes an even digit and O an odd digit. Let’s determine the number of integers in each group.
7EO:
The hundreds digit is 7, so it’s only 1 choice. Since all the digits are nonzero, E can be 4 numbers: 2, 4, 6 or 8. Since the digits are all different, O can be 4 numbers: 1, 3, 5 or 9. Therefore, there are 1 x 4 x 4 = 16 such integers.
7OO:
Since all the digits are different, O can only be 4 numbers: 1, 3, 5 or 9. However, the second O can only be 3 numbers since it cannot be the same as the first O. Therefore, there are 1 x 4 x 3 = 12 such integers.
8EO:
The hundreds digit is 8, so there is only 1 option. Since all the digits are different and nonzero, E can only be 3 numbers: 2, 4, or 6. Since the last digit is O, an odd digit, O can any of the 5 odd digits: 1, 3, 5, 7 or 9. Therefore, there are 1 x 3 x 5 = 15 such integers.
8OO:
The first O can be any of the 5 odd digits: 1, 3, 5, 7 or 9. However, the second O can only be 4 numbers since it cannot be the same as the first O. Therefore, there are 1 x 5 x 4 = 20 such integers.
9EO:
This is analogous to 7EO, so there should be 16 such integers.
9OO:
This is analogous to 9EO, so there should be 12 such integers.
Therefore, there are 16 + 12 + 15 + 20 + 16 + 12 = 91 odd integers that are greater than 700 where all digits are different and nonzero.
Answer: B
