Doubling Pennies for a YearDate: 06/03/2001 at 10:11:08 From: L. Epperson Subject: Giant math compound problem How many pennies will we have at the end of 365 days if we begin on January 1 and double our money each day? We would like to see an algebra formula and see the actual answer to verify that we worked it right. We have been multiplying by 2 and are only at day 40 and it's getting scary. We tried the calculator but it only goes to ten places. We are curious about how many places are in the value without using exponential numbers and just how we would even say such a number, again without using exponential numbers. Thank you. Date: 06/05/2001 at 16:12:42 From: Doctor Twe Subject: Re: Giant math compound problem Hi - thanks for writing to Dr. Math. The number you seek is 1 x 2 x 2 x 2 x ... x 2 a total of 364 times (because you have 1 penny on January 1 and then double it 364 times.) This can be written as 2^364, or 2 to the 364th power. A handy approximation to remember (especially when working with computers) is 2^10 ~= 10^3. If you calculate these values, they actually come to 1024 and 1000, but that's pretty close. (That's why, for example, 1 kilobyte is really 1024 bytes instead of 1000.) Two useful equivalents when dealing with exponents are: a^(b+c) = a^b * a^c and a^(b*c) = (a^b)^c So let's rewrite 2^364 = 2^(4+360) = 2^4 * 2^360 = 16 * 2^(10*36) = 16 * (2^10)^36 =~ 16 * (10^3)^36 ..........first approximation = 16 * 10^108 =~ 10^109 That's a 1 followed by 109 zeroes. That gives you the general order of magnitude of the number you will end up with. To get a more exact value, you can determine the "error factor" in the first approximation by taking the ratio (1024/1000) and raising it to the 36th power (because we're substituting 10^3 for 2^10 thirty-six tmes). (1024/1000)^36 = 1.024^36 = 2.3485... so we can multiply: 16 * 10^108 * (1.204)^36 = 16 * (1.204)^36 * 10^108 = 37.577... * 10^108 = 3.7577... * 10^109 I hope this helps. If you have any more questions, write back. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/ |
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