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If N is the product of all multiples of 10 between 199 and 301, what i
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26 Aug 2015, 12:23
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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 \(\frac{N}{10^{m}}\) is an integer? A. 10 B. 11 C. 13 D. 14 E. 15 What is wrong with my approach below?
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Re: If N is the product of all multiples of 10 between 199 and 301, what i
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26 Aug 2015, 13:40
reto wrote: My approach: We need to find the number of trailing 0 of N, right? Ok, as learned it goes like this: Between 199 and 301 there are 11 multiples of 10 ((200300)/10 +1). So far so good. So N is the product of all multiples of 10 between 199 and 301. Basically this reads as follows: N=(200*210*220*230...) which can be compressed by extracting a 10(20*21*22*23....). It follows that N=10^11* .... usually one can write down a factorial here but in this example, the multiples are not from 1 to something rather from 199 to ... so we can not use a factorial and calculate the trailing zeros from there. What would you offer? How to solve it in an easy way without counting seperately? VeritasPrepKarishma (Would be glad to see your post). Thanks The product of multiples of 10 from 199 to 301 inclusive can be written as: (10*20)*(10*21)*(10*22)......(10*30). It can be rewritten as (10^11*30!)/19!. 30! has 7 trailing zeroes. 19! has 3 trailing zeroes. The numerator of (10^11*30!)/19! has 18 trailing zeroes and the denominator has 3 trailing zeroes. So, (10^11*30!)/19! will be some number with 15 trailing zeroes.




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Re: If N is the product of all multiples of 10 between 199 and 301, what i
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26 Aug 2015, 12:30
My approach: We need to find the number of trailing 0 of N, right? Ok, as learned it goes like this: Between 199 and 301 there are 11 multiples of 10 ((200300)/10 +1). So far so good. So N is the product of all multiples of 10 between 199 and 301. Basically this reads as follows: N=(200*210*220*230...) which can be compressed by extracting a 10(20*21*22*23....). It follows that N=10^11* .... usually one can write down a factorial here but in this example, the multiples are not from 1 to something rather from 199 to ... so we can not use a factorial and calculate the trailing zeros from there. What would you offer? How to solve it in an easy way without counting seperately? VeritasPrepKarishma (Would be glad to see your post). Thanks
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Re: If N is the product of all multiples of 10 between 199 and 301, what i
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26 Aug 2015, 12:37
reto wrote: If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which is an integer? A. 10 B. 11 C. 13 D. 14 E. 15 What is wrong with my approach below? Here is the offical solution, ... Count how many of the multiples of 10 have more than one 5: 200 = 52×8 250 = 52×10 = 53×2 300 = 52×12 One 5 of each of these is counted in our 11 multiples of 10, but there are 4 additional 5s, which can be used to make 4 additional 10s. All in all, there are 11 + 4 = 15 10's in N. Thus, the greatest possible value of m is 15.
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Re: If N is the product of all multiples of 10 between 199 and 301, what i
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26 Aug 2015, 13:41
reto wrote: If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which is an integer? A. 10 B. 11 C. 13 D. 14 E. 15 What is wrong with my approach below? reto, the question has not been defined completely. "what is the greatest integer m for which ?? is an integer"?? Do you mind checking it again?



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Re: If N is the product of all multiples of 10 between 199 and 301, what i
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28 Aug 2015, 10:19
Engr2012 wrote: reto wrote: If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which is an integer? A. 10 B. 11 C. 13 D. 14 E. 15 What is wrong with my approach below? reto, the question has not been defined completely. "what is the greatest integer m for which ?? is an integer"?? Do you mind checking it again? Of course! Sorry for that. Happy to see your response!
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If N is the product of all multiples of 10 between 199 and 301, what i
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28 Aug 2015, 10:34
reto wrote: Engr2012 wrote: reto wrote: If N is the product of all multiples of 10 between 199 and 301, what is the greatest integer m for which is an integer? A. 10 B. 11 C. 13 D. 14 E. 15 What is wrong with my approach below? reto, the question has not been defined completely. "what is the greatest integer m for which ?? is an integer"?? Do you mind checking it again? Of course! Sorry for that. Happy to see your response! Sure, look below: N = 200*210*220*230*240*250*260*270*280*290*300 , you need to find the value of integer m, such that N/\(10^m\) is an integer. For these questions, whenever you need to find the max. or min m such that N/\(P^m\) = integer, make sure to break P down in the form of its Prime factors. After this, the determining value will be the prime factor with the lowest power. Example, in this case, \(10^m = 2^m*5^m\) and for the given series , 5s will be a lot more scarcer than 2s and thus power of 5 will determine how many \(10^m\) can we get. Once you establish which prime factor will be scarcer, it now comes down to counting how many 5s you have in N. N = 200*210*220*230*240*250*260*270*280*290*300 = \(2^{11}\)*\(5^{11}\)*(20*21*22*23*24*25*26*27*28*29*30) and you have 4 5s in the remaining (20*21*22*23*24*25*26*27*28*29*30), bringing the total count of 5s to 15. Thus N/10^m = integer for m = 15. E is the correct answer. Although the above looks time consuming, it is actually not and I was able to get the answer in 1.2 minutes. Hope this helps.



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Re: If N is the product of all multiples of 10 between 199 and 301, what i
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17 Jan 2016, 20:36
11 tens from 200, 210, 220... 300.
And, 4 tens from 4 fives in 200, 250 and 300. So, 15 is the answer.



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Re: If N is the product of all multiples of 10 between 199 and 301, what i
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20 Aug 2017, 11:40
Easy one..
200...to ..300...= 13 zeores.. + 250== there 2 5's are coming... so two more zeroes...
hence answer is 15.



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