问一提 PREP的数学题

Jennywang2

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

答案是B
求解题思路，谢谢

yracheld

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.
搬运来的 原楼在这里https://gmatclub.com/forum/of-the-three-digit-positive-integers-whose-three-digits-are-all-different-and-nonzero-how-many-are-odd-integers-greater-than-217298.html

Jennywang2

yracheld 发表于 2018-1-20 16:20
As for the question, you need to find the number of combinations possible such that:

1. It is a 3 d ...

谢谢！！