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n! has 13 zeros than wat is the higest and lowest value of n??
Read Solution (Total 9)
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- Here we have to consider only the 5,10,15,20.... these terms..
so form (5,10,15,20,30,35,40,45,55 ) we got 1 no of zero from each.
Now from 25(5*5) and 50(5*5*2) we got 2 no of zeroes form each.
so upto 55 factorial we get (9+4)= 13 zeroes.
so 55! is the smallest value and 59! is the largest value. - 12 years agoHelpfull: Yes(51) No(11)
- @ Aditya ,
pls check.
65! will have 13 +2= 15 zeros.
55! ,56!,.....59! will have 13 zeros.
lowest value of n=55
highest value of n =59 - 12 years agoHelpfull: Yes(12) No(12)
- 10! has 2 zeros.. these zeros hav cam frm 10*5*even no.
so this means 20! will hav 4 zeros
now 25=5*5 ohk.. so 30! has 2 times 5 and 1 time 10
so these two 5 will multiply wd any 2 even no. n giv 2 zeros
so in all 30! will hav 7 zeros
nw 40! hav 9 zeros.
50=5*5*2
again we hav two 5
so 50! has 12 zeros
and till 55! we hav 13 zeros
0n 60! we will hav 14 zeros
so we ended at 59!. - 12 years agoHelpfull: Yes(10) No(3)
- can anyone explain me more clearly
- 12 years agoHelpfull: Yes(9) No(1)
- Ans:- n=65 Because n!=65! has 13 zeros...65/5=13
if we want to check in 200!
how many zeros are there then we will divide it by 5 let see
200/5=40; we will take again 40
40/5=8 ; then
8/5=1 now we will add all the ans 40+8+1=49 zeros possible in 200! - 12 years agoHelpfull: Yes(4) No(23)
- olz expain clearly
- 12 years agoHelpfull: Yes(4) No(1)
- To find the number of zeros of factorial number we use this formula
[n/5]+[n/5^2]+[n/5^3]+-.... so on where [] represents integral part of number..For this problem
[n/5]+[n/5^2]+...........=13
we should get 11+2=13 which exactly fits..To get 11+2 value of n :lowest-55 and highest --59 - 9 years agoHelpfull: Yes(2) No(0)
- 55! is the lowest value ans 59! is the max valu.
go through the arun sharma book, chapter number system.
- 9 years agoHelpfull: Yes(0) No(1)
- 55! is lowest and 59! is highest.
- 8 years agoHelpfull: Yes(0) No(1)
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