在beatthegmat上发现的一道题，求解释！

holyxie

x is the product of each integer from 1 to 50, inclusive and , where k is an integer .   What is the greatest value of k for which y is a factor of x?

(A)   0

(B)   5

(C)   6

(D)   10

(E)    12

老美给出的方法极度繁琐而且不知所云，请哪位NN指教一下！！

luciferzwei

首先，化简Y=（2X2X5X5）^k=2^2k*5^2k
要使Y属于X，就得从2和5下手。
又X=1*2*3*……50
其中2的个数是40,5的个数是12.
必然看小的
故另5^2k小于等于5^12,
得：2k小于等于12，得k小于等于6.
故答案为6.

superbat28

to LZ, 这道题我觉得是这么做哈

Y= 100^k, 简化就是10^2k
因为Y要作为X的商，所以就要看X中有多少个10的倍数，因为1-50所有数里头，2的倍数的个数要多于5的倍数的个数，所以我们就看1-50里头有多少个是5的倍数。从5，10,15,20,25（注意，这里是2次5的倍数！！），30,35,40,45,50（注意，这里也是2次5的倍数！！）
所以要求2k小于等于12，故k最大是6~

holyxie

有没有NN能解释下老美给出的这个方法是什么意思啊？
Now obviously there will be more 2s than 5s in 50! so we will check only for 5s that will give us the no. of (5x2)s pairs which will give us no. of zeroes in 50!.
No. of 5s in 50! = (50/5)+(10/5) = 12
          (to find no. of 5s in 50! : First divide 50 by 5 i.e. 10.
                                             Then divide 10 (from above division) by 5 i.e. 2
                                              So 10+2 = 12)

holyxie

C。
100=2*2*5*5；
1-50里面有12个5，又因为偶数有25个，肯定足够与5相乘。
因此12/2=6.

-- by 会员 kidxun (2011/7/28 11:42:29)

这个地方为什么要考虑25个偶数啊？就是因为偶数里面都有2？

kidxun

C。
100=2*2*5*5；
1-50里面有12个5，又因为偶数有25个，肯定足够与5相乘。
因此12/2=6.

bmwwhz

先说声不好意思，昨天我忽略了2个零。我昨天算的是找2和5相乘的0，但是忽略了3个特殊的数字--25,20和50.25可以拆分为5*5即又出现一个新的5，和任何一个含因子为2的数相乘即可得一个0.20和50相乘得三个0，我昨天忽略掉了千位上的那个。所以一共应该出现12个0是正确的。

再来说老美的方法，比我做的更加有效。我当时考虑的是通过2,5相乘找零，可是忽略了一个问题，就是50个数中决定0的是5，因为在50个数中找到的2的个数远远大于5（there will be more 2s than 5s in 50），去个数最少的就是其决定性的那个【so we will check only for 5s that will give us the no. of (5x2)】。能找到12个5就相应能乘得出12个0.

至于“No. of 5s in 50! = (50/5)+(10/2) = 12”，在下糊涂，不太明白。

机经里有一道题跟这个差不多。
69. K是正整数，5^K是99到199所有奇数乘积的factor，问K最大可能是多少。我记得是13。
关键考点：在99-199中带5的奇数里面找到质因子5的个数；
99-199中带5的奇数为：105,115,125……；总共13个5；K=13；

老美给出的官方答案是C啊...
Question:
x is the product of each integer from 1 to 50, inclusive
and ,y=100^k  where k is an integer .  
What is the greatest value of k for which y is a factor of x?
Answer:
x is product of each integer from 1 to 50 i.e. x=50!
and y=100^k.
For "y" to be factor of "x" : x=zy
where z should be an integer.
So, x=z . 100^k
=> 50! = z . 100^k
=> z = 50!/(100^k)
Now "z" will be an integer if 100^k divides 50! without leaving any remainder.
So we have to find no. of zeroes i.e. 10s or (5x2)s in 50!.
Now obviously there will be more 2s than 5s in 50! so we will check only for 5s that will give us the no. of (5x2)s pairs which will give us no. of zeroes in 50!.
No. of 5s in 50! = (50/5)+(10/2) = 12
           (to find no. of 5s in 50! : First divide 50 by 5 i.e. 10.
                                              Then divide 10 (from above division) by 5 i.e. 2
                                               So 10+2 = 12)
So back to our equation above : z = 50!/100^k
Since there are 12 zeroes or 10s in 50! so there will be six (6) 100s.
So k=6 for "z" to be an integer.
So Option C is the answer.
但是我还是看不懂是什么意思

-- by 会员 holyxie (2011/7/27 22:52:48)

holyxie

老美给出的官方答案是C啊...
Question:
x is the product of each integer from 1 to 50, inclusive
and ,y=100^k  where k is an integer .  
What is the greatest value of k for which y is a factor of x?
Answer:
x is product of each integer from 1 to 50 i.e. x=50!
and y=100^k.
For "y" to be factor of "x" : x=zy
where z should be an integer.
So, x=z . 100^k
=> 50! = z . 100^k
=> z = 50!/(100^k)
Now "z" will be an integer if 100^k divides 50! without leaving any remainder.
So we have to find no. of zeroes i.e. 10s or (5x2)s in 50!.
Now obviously there will be more 2s than 5s in 50! so we will check only for 5s that will give us the no. of (5x2)s pairs which will give us no. of zeroes in 50!.
No. of 5s in 50! = (50/5)+(10/2) = 12
           (to find no. of 5s in 50! : First divide 50 by 5 i.e. 10.
                                              Then divide 10 (from above division) by 5 i.e. 2
                                               So 10+2 = 12)
So back to our equation above : z = 50!/100^k
Since there are 12 zeroes or 10s in 50! so there will be six (6) 100s.
So k=6 for "z" to be an integer.
So Option C is the answer.
但是我还是看不懂是什么意思

bmwwhz

感觉自己没解释清楚，重新解释下。
所有整数，能在个位出现0的只有2*5和10，也就是说10个数相乘，所得结果只能有两个0，也就是个位和十位，就是100了。那1-50个数，有5次这种情况，即100的5次方。所得100的k次方。k=5.

答案是B。1-10十个数相乘，出现2个0，同理1-50五十个数相乘会出现2*5=10个0，被100的k次方整除，k=5.

-- by 会员 bmwwhz (2011/7/27 16:14:05)

bmwwhz

答案是B。1-10十个数相乘，出现2个0，同理1-50五十个数相乘会出现2*5=10个0，被100的k次方整除，k=5.