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27 Mar 2011, 20:26
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If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which [N][/1o]m is an integer?here m is 10 to the number of times m(likex.x)
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27 Mar 2011, 21:15
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Let me write all the multiples of 10 and see how many zeroes we get.
200 ---> two zeroes
210 ---> one zeroes
220 ---> one zeroes
230 ---> one zeroes
240 ---> one zeroes
250 ---> three zeroes. 250 = 5*5 *10. Every 5 delivers one zero.
260 ---> one zeroes
270 ---> one zeroes
280 ---> one zeroes
290 ---> one zeroes
300 ---> two zeroes
Adding we get 2+1+1+1+1+3+1+1+1+1+2 = 15
200 ---> two zeroes
210 ---> one zeroes
220 ---> one zeroes
230 ---> one zeroes
240 ---> one zeroes
250 ---> three zeroes. 250 = 5*5 *10. Every 5 delivers one zero.
260 ---> one zeroes
270 ---> one zeroes
280 ---> one zeroes
290 ---> one zeroes
300 ---> two zeroes
Adding we get 2+1+1+1+1+3+1+1+1+1+2 = 15
TomB
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28 Mar 2011, 00:18
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Basically we have to start by counting how many zeroes does this product have.
Each multiple of 10 will by efault give one zero --> How many multiples of 10 between 199 and 301? (300-200)/10 + 1= 11
So 11 zeroes counted
Now what are the other ways of 0 coming in this product? Any combination of 5 and 2 will yield another 0. This should be over and above the zero that we have counted in default zeroes - e.g. 210 = (2*5 *3 *7) We have already counted this 2 and this 5
Which means the number we are counting will atleast have 2 5's or will be a multiple of 25 along with being a multiple of 10. These are:
200 = 10 * 5*4---> 1 more zero
250 = 10 * 5 * 5
300 = 10 *5 *2 *3 --> 1 more zero
We have counted 13 zeros.
The 2 5's of 250 will pick 2 2's from other number of the product. e.g. (220 = 10*11*2) and give two more zeroes
Hence 15 zeroes.
Each multiple of 10 will by efault give one zero --> How many multiples of 10 between 199 and 301? (300-200)/10 + 1= 11
So 11 zeroes counted
Now what are the other ways of 0 coming in this product? Any combination of 5 and 2 will yield another 0. This should be over and above the zero that we have counted in default zeroes - e.g. 210 = (2*5 *3 *7) We have already counted this 2 and this 5
Which means the number we are counting will atleast have 2 5's or will be a multiple of 25 along with being a multiple of 10. These are:
200 = 10 * 5*4---> 1 more zero
250 = 10 * 5 * 5
300 = 10 *5 *2 *3 --> 1 more zero
We have counted 13 zeros.
The 2 5's of 250 will pick 2 2's from other number of the product. e.g. (220 = 10*11*2) and give two more zeroes
Hence 15 zeroes.
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