GMAT 数学题（7）

yayafl

One more deduction:

n 为正整数。问 n 最大为多少能让 (2^m)! 整除 2ⁿ ？

The answer is : n = 2^m - 1

For 2!, n = 1
For 4!, n = 3
For 8!, n = 7
For 16!, n = 15
For 32!, n = 31
For 64!, n = 63
For 128!, n = 127
. . . . .

-- by 会员 sdcar2010 (2011/1/3 23:03:56)

It is really smart!

Thanks.

小困蛇

One more deduction:

n 为正整数。问 n 最大为多少能让 (2^m)! 整除 2ⁿ ？

The answer is : n = 2^m - 1

For 2!, n = 1
For 4!, n = 3
For 8!, n = 7
For 16!, n = 15
For 32!, n = 31
For 64!, n = 63
For 128!, n = 127
. . . . .

-- by 会员 sdcar2010 (2011/1/3 23:03:56)

得通过你列的那一溜推出等比数列先！厉害！！
我也马克下，这一步很销魂阿
2) If we repeat step (1) on (2⁵)!, we will get

(2⁶)! = (2^{2^5}) * [(2⁵)!] = (2^{2^5}) * (2^{2^4}) * [(2⁴)!] = . . . = (2^{2^5}) * (2^{2^4}) * (2^{2^3}) * (2^{2^2}) (2^{2^1}) * (2^{2^0})

        = 2^{(2^5 + 2^4 + 2^3 + 2^2 + 2^1 + 2^0)}

        = 2ⁿ

sdcar2010

As it turned out, one of the recent questions from JJ is following the same analytical path used here:

(Original wrong info about a GMAT question)209. What is the greatest number for n such that 2^n divides 2!/8! ?
选项大概有：2，3，4，6，8 （哈我给了6个选项了）

参考答案： 待定

思路：题干得出的等式为2!/(8!*2^n)还是2^n*8!/2!？


(Confired right info about the GMAT question) What is the greatest number for n such that 2^n divides 8!/2！ ?
我的最后一道题 很明确