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If the number of trailing zeros in n! is three more than the number of trailing zeros in (n - 1)!, how many 3 digit values can "n" assume ?
A. 6
B. 5
C. 4
D. 3
E. 2
A. 6
B. 5
C. 4
D. 3
E. 2
17 Mar 2020, 03:44
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Raksat
If the number of trailing zeros in n! is there more than the number of trailing zeros in (n-1)!, How many 3 digit values can "n" assume ?
A. 6
B. 5
C. 4
D. 3
E. 2
Posted from my mobile device
A. 6
B. 5
C. 4
D. 3
E. 2
Posted from my mobile device
Value of n as a multiple of 5 will give an extra trailing 0 for sure.
REASON :- We will always have more number of 2s as compared to number of 5s, and the trailing 0s would depend on number of 2s and 5s.
so n as 105, 110...990,995 will satisfy the condition..
Total multiple of 5 = 1000-100/5=900/5=180
So answer is 180.
Ofcourse the question is missing something ..
If the number of trailing zeros in n! is there more than the number of trailing zeros in (n-1)!...It must be THREE..
So where would there be a difference of THREE trailing 0s...whereever n gives 3 extra 5s..
hence where ever n is a multiple of 125..
so 125, 250, 375, 500, 625, 750, 875 = 7 but 625 would lead to 4 extra 0s, as \(625 = 5^4\)
A
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25 Apr 2024, 23:27
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Hi chetan2u, can you please explain your reasoning more? How do you conclude that n! has 3 more zeroes than (n-1)! whenever n! is a multiple of 5!, and further, how do you know that 625 must be eliminated, but not the two multiples of 125 that are greater than 625 but still 3 digit integers? Thoroughly confused on how to start this one. I know that the trailing zeroes depend on the number of 5s in an expression, since the number of 2s is always greater, but I'm not sure how to apply that in this case. A more thorough explanation could help me visualize how to solve this one easier.
