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25 Feb 2020, 01:14
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B
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71% (01:24) correct
29%
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If x is a positive integer and \(\frac{50!}{30^x}\) is a positive integer, what is the greatest possible value of x ?
A. 11
B. 12
C. 13
D. 14
E. 15
Are You Up For the Challenge: 700 Level Questions
25 Feb 2020, 01:37
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30=2*3*5
Our bottle neck is 5 so we need to know how many 5 are in 50!
we have for every 5 numbers 1 and on 25 and 50 we have 2 fives
thus 10+2=12
if we have 12 fives is 50! we can have 30^12 as the greatest value
Ans (b)
Our bottle neck is 5 so we need to know how many 5 are in 50!
we have for every 5 numbers 1 and on 25 and 50 we have 2 fives
thus 10+2=12
if we have 12 fives is 50! we can have 30^12 as the greatest value
Ans (b)
25 Feb 2020, 01:58
Kudos
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We need to Prime factorizing for 30 -> 2*3*5
So, the maximum number for 2 will be calculated by:
\(\frac{50}{2}\) + \(\frac{50}{2^2}\) + \(\frac{50}{2^3}\) + \(\frac{50}{2^4}\) + \(\frac{50}{2^5}\) = 25+12+6+3+1= 47
So, the maximum number for 3 will be calculated by:
\(\frac{50}{3}\) + \(\frac{50}{3^2}\) + \(\frac{50}{3^3}\) = 16+5+1= 22
So, the maximum number for 5 will be calculated by:
\(\frac{50}{5}\) + \(\frac{50}{5^2}\) = 10+2 = 12
As 30 is the combination of 2, 3 and 5 - We take the minimum value of them, i.e. - 12. 'B' is the winner
So, the maximum number for 2 will be calculated by:
\(\frac{50}{2}\) + \(\frac{50}{2^2}\) + \(\frac{50}{2^3}\) + \(\frac{50}{2^4}\) + \(\frac{50}{2^5}\) = 25+12+6+3+1= 47
So, the maximum number for 3 will be calculated by:
\(\frac{50}{3}\) + \(\frac{50}{3^2}\) + \(\frac{50}{3^3}\) = 16+5+1= 22
So, the maximum number for 5 will be calculated by:
\(\frac{50}{5}\) + \(\frac{50}{5^2}\) = 10+2 = 12
As 30 is the combination of 2, 3 and 5 - We take the minimum value of them, i.e. - 12. 'B' is the winner
