n! has x number of zeroes at the end and (n + 1)! has (x + 3) zeroes at the end. If n is a positive integer and 1 ≤ n ≤ 1000. How many values of n are possible ?
A. 8
B. 7
C. 6
D. 4
E. 1
As 1 ≤ n ≤ 1000, number of 0's in n! are decided by \(5^x\) where 1≤x≤4(since 5^5 = 3125 would always result in 0 number of 0's on dividing 1000 by it). However, 5^4 have to be checked since it has an additional factor which results in 4 extra 0's wherein we are asked about only 3.
So, total number of 0's can be found as follows"
\(\frac{n!}{5^1}\)
\(\frac{n!}{5^2}\)
\(\frac{n!}{5^3}\)
\(\frac{n!}{5^4}\)
where each of the additional power of 5 would contribute an additional '0' in case of \((n+1)!\)
It has to be noted that a difference of 3 0's would only be possible if \(divisor ≥ 5^3\) is counted i.e. n has to be of 3 digits so \(n+1\) precisely should be a multiple of \(5^3\) or \(5^4\).
Hence multiples of \(5^3\) are 125, 250, 375 ..... 1000, a total of 8(\(\frac{1000}{125}\); including a common multiple of \(5^4\) i.e. 625).
For example -
n + 1 = 125 so n = 124
Number of 0's in 124! = \(\frac{124}{5^1} + \frac{124}{5^2} + \frac{124}{5^3} + \frac{124}{5^4}\)
= 24 + 4 + 0 + 0 = 28
&
Number of 0's in 125! = \(\frac{125}{5^1} + \frac{125}{5^2} + \frac{125}{5^3} + \frac{125}{5^4}\)
= 25 + 5 + 1 + 0 = 31
Similarly, for n = 999 the number of 0's are
Number of 0's in 999! = \(\frac{999}{5^1} + \frac{999}{5^2} + \frac{999}{5^3} + \frac{999}{5^4}\)
= 199 + 39 + 7 + 1 = 246
&
Number of 0's in 1000! = \(\frac{1000}{5^1} + \frac{1000}{5^2} + \frac{1000}{5^3} + \frac{1000}{5^4}\)
= 200 + 40 + 8 + 1 = 249
Checking for n = 624 the number of 0's are
Number of 0's in 624! = \(\frac{624}{5^1} + \frac{624}{5^2} + \frac{624}{5^3} + \frac{624}{5^4}\)
= 124 + 24 + 4 + 0 = 152
&
Number of 0's in 625! = \(\frac{625}{5^1} + \frac{625}{5^2} + \frac{625}{5^3} + \frac{625}{5^4}\)
= 125 + 25 + 5 + 1 = 156
So there is a difference of 4 0's so this is not an applicable case here.
So, possible number of values that n can take = 8 - 1 = 7
Answer B.
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