How Many Digits in 13^18 ?Date: 05/15/2003 at 01:58:33 From: Priya Subject: Number of digits in 13 to the power of 18 (13^18)) Hi, Find the exact number of digits in 13 raised to the 18th power. I tried doing it using the binomial expansion method, but it became too cumbersome. Since this question is from a popular entrance test where the test taker is supposed to take only 45 secs for one question, I guess there should be an easier way to do it. Date: 05/15/2003 at 15:28:54 From: Doctor Douglas Subject: Re: Number of digits in 13 to the power of 18 (13^18)) Hi Priya, How much time it should take depends on whether or not you have access to a calculator (even one without the exponentiation function). With a logarithm, you can use log (base 10) to obtain log(13^18) = 18 log 13 = 18*1.114 = 20.05 So there are twenty-one digits in the number. Now suppose you have no calculator at all (other than your brain), and want to achieve this result in under one minute. If you know your squares and square roots, and the rules for logarithms and exponents, you can estimate it as follows: 1. We need the number log(13), to approximately two or three digits accuracy, since we will multiply it by 18 as in the formula above to obtain the number of digits in 13^18, which is somewhere around twenty digits or so (it must be more than 10^18, which has 19 digits). 13^2 = 169, which is approximately 170. 170^2 = 28900, which is around 30000, 30000^2 = 9 x 10^8 = 10^9, finally we arrive at a number whose logarithm is easy to evaluate, namely log(10^9) = 9. log(13) = (1/8) log(10^9) = 9/8 = 1.125, approximately The actual value of log(13) is 1.1139, so our estimate is in fact good to two and almost three digits. 2. The rest is easy: we multiply 18 by 1.125 to obtain 18 x 9/8 = 9 x 9/4 = 81/4 = 20.25 So we conclude that there are twenty-one digits. Remark: the logarithm is very forgiving - it takes numbers that are off by factors and converts those factors to offsets. So we may have reasonable confidence in our estimate. Although the errors are hard to track without a calculator, note that we rounded up twice (169 to 170 and 28900 to 30000), so that we may expect that the method above gave a slight overestimate of log(13). I was talking to a friend about this very interesting problem, and we developed an even quicker route to the answer: 1. We estimate 13 as approximately equal to 14.142... = sqrt(2)*10. 2. 13^18 is approximately [sqrt(2)*10]^18 = 2^9 * 10^18 = 512 * 10^18 (easy without a calculator if you remember your powers of 2) = 5.12 x 10^20 which has twenty-one digits. Now, we can even estimate the error made in step 1: 14.14/13 is approximately 1.1, so it is a 10% overestimate. This overestimate will propagate when we raise it to the 18th power (the next set of equations are approximations): (1.1)^18 = 1.21^9 = 1.44^4 x 1.21 = sqrt(2)^4 * 1.21 = 4*1.21 = 5 This says that we have overestimated the final result by a factor of 5, which puts the actual value at approximately 13^18 = 1 x 10^20 (probably good to within 10% or so) The actual value is 13^18 = 1.125 x 10^20. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 05/16/2003 at 00:44:12 From: Priya Subject: Thank you (Number of digits in 13 to the power of 18 (13^18)) Thanks a lot for the prompt reply. Really appreciate it. Both the methods were quite easy, though they used different approaches. Thanks again. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/