Finding Digits in Specific Decimal Places of Large NumbersDate: 03/16/2006 at 13:19:19 From: Doris Subject: Number Theory Find the digits immediately before and after the decimal point in (root2 + root3)^2000. The power is too large to get the answer! Date: 03/16/2006 at 16:49:14 From: Doctor Vogler Subject: Re: Number Theory Hi Doris, Thanks for writing to Dr. Math. Are you familiar with the Fibonacci numbers? They are usually defined using a recurrence relation F(n+2) = F(n+1) + F(n) but they can also be given with an explicit (exponential) formula: 1 1 + sqrt(5) n 1 - sqrt(5) n F(n) = ------- * ( (-----------) - (-----------) ) sqrt(5) 2 2 as also seen in equation (3) of Fibonacci Number http://mathworld.wolfram.com/FibonacciNumber.html Well, your formula *almost* looked like the first term in a similar expression, except that you have two roots instead of just one. So I would recommend squaring the sum and rewriting this as (sqrt(2) + sqrt(3))^2000 = (5 + 2*sqrt(6))^1000. Does that make sense? Then you can define the sequence G(n) = (5 + 2*sqrt(6))^n + (5 - 2*sqrt(6))^n. It turns out that this is a sequence of integers, which you can prove by writing out a recurrence relation that it satisfies: G(n+2) = 10*G(n+1) - G(n). Prove that G is given by this recurrence, and then use the recurrence to show that G(n+2) = -G(n) mod 10 and therefore G(n+4) = G(n) mod 10 and G(4n) = 2 mod 10, so that the last digit of G(4n) is 2 for any integer n (i.e. every fourth term ends in a 2). Since 1000 = 4*250 is divisible by 4, that means that G(1000) = (5 + 2*sqrt(6))^1000 + (5 - 2*sqrt(6))^1000 is an integer whose last digit is a 2. Now you want to know about the first term, so let's look at the second term (5 - 2*sqrt(6))^1000. It turns out that 5 - 2*sqrt(6) = 0.10102... is between 1/10 and 1/9. So what does that tell you about (5 - 2*sqrt(6))^1000 ? Finally, you have (5 + 2*sqrt(6))^1000 = G(1000) - (5 - 2*sqrt(6))^1000, so what would be the digits immediately before and immediately after the decimal point? If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 03/19/2006 at 20:46:49 From: Doris Subject: Number Theory Thanks for the reply. I understand all you wrote except the step to get: G(4n) = 2 mod 10 from G(n+4) = G(n) mod 10. I can show G(n+2) = -G(n) mod 10 to G(n+4) = G(n) mod 10 as below. G(n+4)= -G(n+2)mod 10 = (-10 G(n+1)+ G(n))mod 10 = G(n)mod 10 Can you please show me the step for G(4n) = 2 mod 10? Thanks for your concern and I am looking forward for your further reply. Doris Date: 03/20/2006 at 17:27:43 From: Doctor Vogler Subject: Re: Number Theory Hi Doris, First compute G(0) mod 10. What do you get? Then compute G(4) mod 10 from the relation you just proved. Then compute G(8) mod 10, and notice a pattern. To prove this result, use induction. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 03/27/2006 at 18:05:33 From: Doris Subject: Thank you (Number Theory) Thanks Dr. Math for answering my questions. It is very useful. You gave methods and major hints that I could use to get the final answer. |
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