Thread: Anyone know how to solve this problem?

can you show the work to so i can understand it. thanks

how many digits does the number 8^41* 5^116 have. justify your answer algebraically

Originally Posted by mcdaking84

can you show the work to so i can understand it. thanks

how many digits does the number 8^41* 5^116 have. justify your answer algebraically

A number N has n digits where n = floor(log_10(N))+1 (floor the greatest integer less than or equal to its argument)

8^41* 5^116 = [10^(log_10(8))]^41 * [10^(log_10(5))]^116

.................. = 10^(41*log_10(8)+log_10(5)*116) ~= 10^(118.107)

so floor(log_10(8^41* 5^116))+1 = 119, so 8^41* 5^116 has 119 digits.

RonL

Originally Posted by CaptainBlack

A number N has n digits where n = floor(log_10(N))+1 (floor the greatest integer less than or equal to its argument)

8^41* 5^116 = [10^(log_10(8))]^41 * [10^(log_10(5))]^116

.................. = 10^(41*log_10(8)+log_10(5)*116) ~= 10^(118.107)

so floor(log_10(8^41* 5^116))+1 = 119, so 8^41* 5^116 has 119 digits.

RonL

Can you prove that theorem?

Originally Posted by ecMathGeek

Can you prove that theorem?

Is the Pope Jewish?

If N has 1 digit 1=<N<10, so 0<=log_{10}(N)<1 so floor(log_{10}(N))=0
and n=floor(log_{10}(N))+1 is the number of digits of N.

If N has n digits then 10^{n-1}<=N<10^n, so n-1<=log_{10}(N)<n, so
floor(log_{10}(N))=n-1, and hence n=floor(log_{10}(N))+1.

RonL

Note that (8^41)=(2^123). Hence (8^41)(5^116) has 116 trailing zeros.
123-116=7 and 2^7=128, so it does indeed have 119 digits.

Hello, mcdaking84!

It's easier than you think . . .

How many digits does the number N .= .(8^41)(5^116) have?

Note that: .8^41 .= . (2^3)^41 .= .2^123 .= .(2^7)(2^116)


We have: .N .= .(2^7)(2^116)(5^116) .= .(2^7)(10^116) .= .128 x 10^116


Therefore, N has 119 digits.

Oops . . . exactly what Plato said . . .

Originally Posted by Soroban

Hello, mcdaking84!

It's easier than you think . . .

It depends on what you know. Anyone who was brought up with log books
would know the relationship between the number of digits in front of a
decimal point and the integer part of the log of a number >=1 virtualy
instinctivly.

Also the log_{10} method is more universal (it does not depend on
special numbers).

RonL