GMAT 数学题（5）

sdcar2010 · 发表于 2011-1-1 23:21:32

在 0 和 1000 之间的正整数中，有多少个奇数其每个数位上的数字都不一样？
1) 310
2) 330
3) 365
4) 400
5) 415

And why?

Rebeca · 发表于 2011-1-2 14:11:28

这个我是分段，重复的奇书减一，最后相加的笨方法....
考的时候不会出这么变态的题吧...好歹一百以内的都可以数出来的啊....

smpowe · 发表于 2011-1-2 14:41:36

答案是C

奇数为1位数: 1,3,5,7,9

奇数为2位数且数字不同: C51*C91-C51=40

奇数为3位数且数字不同: C51*C91*C81-C51C81=320

TOTAL: 320+40+5=365

ludanthu · 发表于 2011-1-2 16:52:35

C,

5 choices for unit digit, left 9 for tens digit and 8 for hundreds digit (the suquence of tens hundreds does not matter), makes 5*9*8=360

Exceptions are those having 0 at both tens and hundreds digit: 1, 3, 5, 7, 9. Thus 360+5=365

Actually, I forgot those exceptions at first.....fortunately there is not an option of 360....lol

sdcar2010 · 发表于 2011-1-2 21:26:12

365 is the right answer!

#1: you are on the right track. Just need to figure out a way to quickly calculate the possibilities than to count them by hand.

#2 and #3 are using different methods to solve this question. Brilliant!

GMAT 数学题（5）

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