Don't they increase at a steady rate?
Out of the first 1 0 0 natural numbers,the probability that for a selected number k , k ! has two trailing zeroes less than ( k + 1 ) ! in the end, is?
Source: FIITJEE AITS!
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2 solutions
Moderator note:
Good. What would the answer be if I replace the number 1 0 0 by 1 0 0 0 ?
Thanks, I've added "Trailing zeroes".
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you are most welcome sir.always glad to help.
It is possible when (k+1) =25,50,75,100
That is correct but please mention why that is so,plz.
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@Adarsh Kumar , Can youHelp me in finding the solution?Here it what I did:-
For The Given Criteria To be Fulfilled, (k+1)!/k*k!=100
=(k+1)/k=100. But That Doesn't seem to work. Can you tell Where I am Going Wrong?
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there is a comma after k it is not product!!!
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@Adarsh Kumar – Then that should be mentioned :3 XD Many may Get confused :P
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@Mehul Arora – ok sorry edited!now?
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@Adarsh Kumar – Yeah, That's better :P XD. Good Problem Nonetheless.
You must be knowing that number of zeros at the end of n! is same as number of 5's in n! which is obtained using the formula [n/5]+[n/5^2]+[n/5^3]+.......... till we start getting 0; where [.] stands for floor function(greatest integer function).Because of presence of 5^2 in 25,50,75 & 100 you will observe that extra jump in number of 5's compared to the general trend which in turn affects number of zeros in same manner.
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Yes. I knew that. I just Was Confused after seeing the Question thinking that It was a dot in Place of a comma. :)
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@Mehul Arora – Well , as an exercise to your eye , find a fullstop in these commas: xD xD xD
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@Nihar Mahajan – Well, I am Mad, But Not Psychic -_-
Yeah... thats correct.I gotthis prob by this idea only.
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@Nihar Mahajan – Nice,kinda hit & run question.You cannot afford to exhaust yourself over such questions in JEE paper!
yup exact same strategy here
@Deepak Kumar , Could you Please provide the Solution?
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Try working on finding number of 5's in n!
Since there wasn't any serious solutions I thought I should do it.
Solution:
If k ! has 2 less trailing zeroes than ( k + 1 ) ! .Now assuming that the power of 5 in k ! equals 5 i we can see that the power of 5 in ( k + 1 ) ! has to be 5 i + 2 (To make exactly 2 more trailing zeroes when multiplied by 2).Now we need to find the numbers for which the following is true:
k = A ( 5 2 k ) − 1 → k + 1 = A ( 5 2 k ) →
The only numbers among the interval [ 1 , 1 0 0 ] that at least consist of 2, 5's (Distinct 5's) are 25,50,75 and 100,meaning that P ( S ) = 1 0 0 4 = 2 5 1
P.S:I need to state that the question remains flawed despite author's edits.It really has to mention the fact that the number of "Trailing zeroes" are counted.