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)
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发表于 2011-7-26 13:06:41
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