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The New Coffee Room

  1. TNCR
  2. General Discussion
  3. Puzzle Time - after the decimal point

Puzzle Time - after the decimal point

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  • J Online
    J Online
    jon-nyc
    wrote on 7 Mar 2021, 19:03 last edited by
    #1

    What is the first digit after the decimal point in the number you get by raising the square root of 2 plus the square root of 3 to the billionth power?

    Only non-witches get due process.

    • Cotton Mather, Salem Massachusetts, 1692
    1 Reply Last reply
    • K Online
      K Online
      Klaus
      wrote on 7 Mar 2021, 22:16 last edited by
      #2

      I guess the point is that if you expand (2^0.5+3^0.5)^(10^9), then almost all terms are whole numbers, right?

      1 Reply Last reply
      • J Online
        J Online
        jon-nyc
        wrote on 7 Mar 2021, 22:26 last edited by
        #3

        I guess that will get you there I found a simple short proof though.

        Only non-witches get due process.

        • Cotton Mather, Salem Massachusetts, 1692
        1 Reply Last reply
        • J Online
          J Online
          jon-nyc
          wrote on 7 Mar 2021, 22:27 last edited by jon-nyc 3 Jul 2021, 22:28
          #4

          I’ll give you a hint - my proof relies on the fact that the difference between the two numbers (sqrt3 and sqrt2) is less than one.

          Only non-witches get due process.

          • Cotton Mather, Salem Massachusetts, 1692
          1 Reply Last reply
          • K Online
            K Online
            Klaus
            wrote on 8 Mar 2021, 09:51 last edited by
            #5

            Hm, using the binomial equation (a+b)(a-b)=a^2-b^2, I can transform sqrt3 + sqrt2 to 1/(sqrt3-sqrt2). Right direction?

            1 Reply Last reply
            • J Online
              J Online
              jon-nyc
              wrote on 8 Mar 2021, 12:26 last edited by jon-nyc 3 Aug 2021, 12:26
              #6

              That’s not where I was headed.

              I can give a bigger hint.

              Let sqrt(3) be a, sqrt(2) be b.

              We’re looking for (a+b)^n for n=10^9

              Here’s the big hint:

              :::

              What happens if you do the binomial expansion of (a+b)^n then add (a-b)^n to it?

              :::

              Only non-witches get due process.

              • Cotton Mather, Salem Massachusetts, 1692
              1 Reply Last reply
              • K Online
                K Online
                Klaus
                wrote on 8 Mar 2021, 13:40 last edited by Klaus 3 Aug 2021, 13:42
                #7

                Too big of a hint 😉 (sqrt3 -sqrt2)^10^9 is too small to influence the first digit after the dot.

                1 Reply Last reply
                • J Online
                  J Online
                  jon-nyc
                  wrote on 8 Mar 2021, 13:56 last edited by
                  #8

                  It is a huge hint

                  Only non-witches get due process.

                  • Cotton Mather, Salem Massachusetts, 1692
                  1 Reply Last reply
                  • K Online
                    K Online
                    Klaus
                    wrote on 8 Mar 2021, 14:16 last edited by Klaus 3 Aug 2021, 14:17
                    #9

                    I'd say the digit must be 9. Or 0 if I screwed up.

                    If you apply the binomial theorem to the term you suggest and do some simplification, you get a whole number minus (sqrt3 - sqrt2)^n, which gets very small.

                    1 Reply Last reply
                    • J Online
                      J Online
                      jon-nyc
                      wrote on 8 Mar 2021, 14:23 last edited by jon-nyc 3 Aug 2021, 14:23
                      #10

                      Right. We know that by adding an infinitesimal positive number to the number we seek, we get an integer.

                      Therefore the first many many digits after the decimal must be 9.

                      Note this works for any even n sufficiently large.

                      Only non-witches get due process.

                      • Cotton Mather, Salem Massachusetts, 1692
                      1 Reply Last reply
                      • J Online
                        J Online
                        jon-nyc
                        wrote on 8 Mar 2021, 14:33 last edited by jon-nyc 3 Aug 2021, 14:34
                        #11

                        You can already see the pattern with quite small n.

                        f5b69dba-a6b9-4d89-a6aa-d89faee3f576-image.png

                        Only non-witches get due process.

                        • Cotton Mather, Salem Massachusetts, 1692
                        K 1 Reply Last reply 8 Mar 2021, 14:39
                        • J jon-nyc
                          8 Mar 2021, 14:33

                          You can already see the pattern with quite small n.

                          f5b69dba-a6b9-4d89-a6aa-d89faee3f576-image.png

                          K Online
                          K Online
                          Klaus
                          wrote on 8 Mar 2021, 14:39 last edited by
                          #12

                          @jon-nyc said in Puzzle Time - after the decimal point:

                          You can already see the pattern with quite small n.

                          f5b69dba-a6b9-4d89-a6aa-d89faee3f576-image.png

                          Yes, but I was confused for a while because that pattern switches when you try larger numbers.

                          I was at first thinking that this was a rounding error and then tried again with a "big float" library, but still, with n >=50 the first digit seems to be 0.

                          8973a130-fec9-4a1f-ba5d-a4847d59ab45-image.png

                          However, I assume that it's just a particularly numerically unstable formula.

                          J 1 Reply Last reply 8 Mar 2021, 14:44
                          • K Klaus
                            8 Mar 2021, 14:39

                            @jon-nyc said in Puzzle Time - after the decimal point:

                            You can already see the pattern with quite small n.

                            f5b69dba-a6b9-4d89-a6aa-d89faee3f576-image.png

                            Yes, but I was confused for a while because that pattern switches when you try larger numbers.

                            I was at first thinking that this was a rounding error and then tried again with a "big float" library, but still, with n >=50 the first digit seems to be 0.

                            8973a130-fec9-4a1f-ba5d-a4847d59ab45-image.png

                            However, I assume that it's just a particularly numerically unstable formula.

                            J Online
                            J Online
                            jon-nyc
                            wrote on 8 Mar 2021, 14:44 last edited by jon-nyc 3 Aug 2021, 14:44
                            #13

                            @Klaus

                            Has to be because (sqrt(3)-sqrt(2))^n stays tiny and positive. Same thing happened on my excel, too.

                            Only non-witches get due process.

                            • Cotton Mather, Salem Massachusetts, 1692
                            1 Reply Last reply
                            • K Online
                              K Online
                              Klaus
                              wrote on 8 Mar 2021, 14:46 last edited by
                              #14

                              Effect of increasing the precision to 100 bits, and then 500 bits:

                              59125c6e-c1ca-4cd3-9f5c-a24fd9fddd88-image.png

                              1 Reply Last reply
                              • J Online
                                J Online
                                jon-nyc
                                wrote on 8 Mar 2021, 15:09 last edited by
                                #15

                                Cool.

                                Only non-witches get due process.

                                • Cotton Mather, Salem Massachusetts, 1692
                                1 Reply Last reply
                                • J Online
                                  J Online
                                  jon-nyc
                                  wrote on 13 Mar 2021, 12:06 last edited by
                                  #16

                                  The ‘official’ answer came out today.

                                  SOLUTION: If you try entering "(sqrt(2) + sqrt(3))^1,000,000,000" into Google or your calculator, you're likely to find that you get only the dozen or so most significant figures; that is, you don't get a nearly accurate enough answer to see what happens after the decimal point.

                                  But you can try smaller powers and see what happens. For example, the decimal expansion of the 10th power of the square root of 2 plus the square root of 3 begins 95049.9999895. A bit of experimentation shows that each even power of the square root of 2 plus the square root of 3 seems to be just a hair below some integer. So the solution to the puzzle seems to be 9. But why?

                                  If we merely square the number "square root of 2 plus square root of 3," we get about 9.9. If we play with 10 minus this number, we discover that it's equal to the square of the square root of 2 minus the square root of 3. Aha!

                                  Yes, (sqrt(3) + sqrt(2))^2n + (sqrt(3) - sqrt(2))^2n is always an integer, because when you expand it, the odd-power terms cancel and the even ones are integers. But (sqrt(3) - sqrt(2))^2n is very small, about 10^-n, so the first roughly n digits of (sqrt(3) + sqrt(2))^2n past the decimal point are all 9's.

                                  [The idea for this puzzle comes from the 29th Annual Virginia Tech Mathematics Contest, 2007

                                  Only non-witches get due process.

                                  • Cotton Mather, Salem Massachusetts, 1692
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