问一提 PREP的数学题

Jennywang2 · 发表于 2018-1-20 16:11:37

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 · 发表于 2018-1-20 16:20:04

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 · 发表于 2018-1-20 16:32:23

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！！

问一提 PREP的数学题

所属分类: GMAT考试

正在浏览此版块的会员 ()