3ok mistaken sir.still confused sir for the reason for grouping as 3 terms
Solve with prooof..... (Not tough again!)
highest power of 3 in 100!
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30 Answers
62see basically if a number is of the form
3n then n out of these 3n numbers (1,2,3,.... 3n )will be divisible by 3
if it is
3n+1 then also only n of these (1,2,3,..... 3n+1) will be divisible by 3
if it is
3n+2 then also only n of these (1,2,3,..... 3n+2) will be divisible by 3
and in each of these 3 cases above, the numbers will be (3,6,...3n)
3ok sir still confused for not considering the 36
i cant understanding the meaning of ur explanation
362well see
1,2,3
4,,5,6
7,8,9
....
31,32,33
each of these have 1 multiple of 3
if the last group has
34,35 and no 36 then it does not have a multiple of 3
does this explain?
3i have the same doubt as philip got.can u explain me in some other easier way
62see philip...
The multiples of 3 will contribute for one 3 each...
The multiples of 3^2 will contribute for one additional 3
and so on....
so the final naswer will be multiples of 3+ 9 + 27.....
1sorry i din get it completely
could you tell how it tells the no. of occurrences of p in n factorial
btw bhaiyya please could you see:
http://targetiit.com/iit_jee_forum/posts/circular_path_1398.html
39for this proof is also not necessary
we have some formula na for exponent of a prime
direct formula substitution
13the formula is [n/p] + [n/p2] + [n/p3] + ...
first term [n/p] gives the number of multiples of p that are ≤ n ... similarly [n/p2] gives the number of multiples of p2 that are ≤ n... and so on...
first term gives the number of terms that contribute the prime p... second term gives the number of terms that contribute p2, since one of these 2 p's is already counted in first case, so it adds one extra value of p to the total number of p's.... and so on till the contribution becomes 0...
let me know if i'm wrong...!!!
1i cant for the life of me think of it presently [2]
i'll bookmark this thread
to come back to it later
62yeah philip i was telling this method to sankara..
but i still want the proof from u guys :)
1yes Nishant bhaiyya this is tthe same method i had used but what about the proof
62see sankara..
I am only giving u the fnal answer..
[] shows the greatest integer..
so for our problem, the answer will be
[100/3]+[100/9]+[100/27]+[100/81]+[100/243]+..............
the last terms after the first 4 will be zero...
so the answer will be
33+11+3+1
Now did u understand the method?
62the method to do this is
[n/3]+[n/32]+[n/33] +............
more generally the highest power of p in n!
[n/p]+[n/p2]+[n/p3] +............
3for inspired is there any formula for exponent of prime.if so can u give post the derivation for that formula
1ans should be 48
interestingly the ans is same for the highest power of 4
can't think at all of how the method came