/sci/ - Science & Math

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Putnam Problem of the Day http://math.harvard.edu/putnam(...) !!84u21X0KjW+ Sun Apr 7 03:08:51 2019 No.10529139 [Reply] [Original]

[math]
\text{Let }N\text{ be the positive integer with 1998 decimal digits,}
\\
\text{all of them 1; that is,}
\\
\qquad \qquad \qquad \qquad N=1111\cdots 11.
\\
\text{Find the thousandth digit after the decimal point of }\sqrt N\text{.}
[/math]

>>	http://math.harvard.edu/putnam(...) !!84u21X0KjW+ Sun Apr 7 03:09:17 2019 No.10529142 Previous Thread >>10526643

Anonymous Sun Apr 7 04:50:24 2019 No.10529258

[eqn]\sqrt{111 \ldots 1} \\
= \sqrt{\sum_{k=0}^{1997} 10^k} \\
= \sqrt{\frac{1 - 10^{1998}}{1 - 10}} \\
= \frac{10^{999}}{3} \sqrt{1 - 10^{-1998}} [/eqn]
Remember that [math]\sqrt{1 - x} = 1 - \frac{1}{2} x [/math] for small x so
[eqn] \frac{10^{999}}{3} \sqrt{1 - 10^{-1998}} \\
= \frac{10^{999}}{3} \left(1 - \frac{1}{2} 10^{-1998} \right) \\
= \frac{10^{999}}{3} - \frac{10^{-999}}{6} \\
= \frac{10^{999} - 10^{-999}}{3} + \frac{10^{-999}}{6}
[/eqn]
The first term is 0 at the thousandth digit after the decimal point while the second term is 1.

>>	Anonymous Sun Apr 7 05:36:14 2019 No.10529318 File: 16 KB, 1850x510, putnam btfo.png [View same] [iqdb] [saucenao] [google] >>10529139 Seems to be 1, by direct calculation.

Anonymous Sun Apr 7 05:59:04 2019 No.10529358 [DELETED]

>>10529258
[math]\frac{10^{999}}{3}-\frac{10^{-999}}{6}[/math]
[math]=3\sum\limits_{k=998}^{-\infty}10^k-\frac{3}{2}\sum\limits_{k=-1000}^{-\infty}10^k[/math]
[math]=3\sum\limits_{k=998}^{-999}10^k+3\sum\limits_{k=-1000}^{-\infty}10^k-\frac{3}{2}\sum\limits_{k=-1000}^{-\infty}10^k[/math]
[math]=3\sum\limits_{k=998}^{-999}10^k+\frac{3}{2}\sum\limits_{k=-1000}^{-\infty}10^k[/math]
[math]=3...3.333...333000...+0.000...000166...[/math]
[math]=3...3.333...333166...[/math]
It's a 1 and a 6.
>√1-x=1−x/2
Approximately.

Anonymous Sun Apr 7 06:07:10 2019 No.10529373

>>10529258
[math]\frac{10^{999}}{3}-\frac{10^{-999}}{6}[/math]
[math]=3\sum\limits_{k=998}^{-\infty}{10^k}-\frac{1}{2}3\sum\limits_{k=-1000}^{-\infty}{10^k}[/math]
[math]=3\sum\limits_{k=998}^{-999}{10^k}+3\sum\limits_{k=-1000}^{-\infty}{10^k}-\frac{1}{2}3\sum\limits_{k=-1000}^{-\infty}{10^k}[/math]
[math]=3\sum\limits_{k=998}^{-999}{10^k}+\frac{3}{2}\sum\limits_{k=-1000}^{-\infty}{10^k}[/math]
[math]=3...3.333...333000... + 0.000...000166...[/math]
[math]=3...3.333...333166...[/math]
It's a 1 and a 6.
>√1-x=1−x/2
Approximately.

>>	Anonymous Sun Apr 7 19:19:41 2019 No.10531333 bump

>>	Anonymous Sun Apr 7 19:32:42 2019 No.10531360 >>10529139 thinly veiled homework thread

>>	Anonymous Sun Apr 7 20:09:10 2019 No.10531456 >>10529318 based

>>	Anonymous Sun Apr 7 21:02:27 2019 No.10531535 my wife flan is so cute

>>	Anonymous Mon Apr 8 08:22:34 2019 No.10532576 >>10529139 What was the solution to one in the last OP? >>10531535 Remi is better

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