262for any number x, no. of digits in x = [log x] +1 ...[] denotes greatest integer function
therefore,
log 530 = 30 log 5 =30(log10 - log2) =30(1-0.3010)=30(0.699)=20.97
hence no. of digits= [20.97]+1 = 21
4 Answers
262
1is there any proof for " no. of digits in x = [log x] +1 " ?
btw right answer. thanks alot
262it is just basic logic, like suppose we have a 2 digit no. say n,
then 10≤n<100
taking log on all sides ,
1≤log n <2
hence for any n such that ,
10m < n < 10m+1
then we can say n has m+1 digits
now taking log,
m < log n <m+1
hence [log n] = m
now
no. of digits = m+1 = [log n] +1
49Nice one Aditya..
Learnt something new! :)
Any more proves, anyone?