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Putnam/Olympiad Problem of the Day

Anonymous Tue 05 Jan 07:59:49 2021 No.12546876 View Reply Original Report

Quoted By: >>12547230 >>12547245 >>12547388 >>12548918 >>12549048 >>12549490 >>12549503 >>12549525 >>12550105 >>12551118

Let $N$ be the positive integer with $\mathrm{1998}$ decimal digits, all of them 1; that is, $N=1111...11.$ Find the thousandth digit after the decimal point of $\sqrt{N}$

Anonymous

Anonymous Tue 05 Jan 2021 08:01:04 No.12546879 Report

Quoted By:

Previous thread: >>12543236

Anonymous

Anonymous Tue 05 Jan 2021 10:54:48 No.12547230 Report

Quoted By:

>>12546876

Suppose there is a cyclic pattern in the decimal. If that is true, then you can split off all the decimals before the repetition point, and express that quantity as the ratio of an integer and a power of 10. For the repetitive remainder, the repeating decimal sequence can be factored out, leaving a geometric series behind. Because the ratio of the terms of the geometric sequence is a power of 10, it is a rational number, so the geometric series closed-form is a rational number manipulation of a rational number, hence the repeating portion is rational. The sum of the two rational parts is rational, therefore, if there is some sort of cyclic pattern I can follow to find the 1000th decimal place, it must be because the square root of N is some rational number.

Anonymous

Anonymous Tue 05 Jan 2021 11:05:12 No.12547245 Report

Quoted By:

>>12546876
If the square root of N is a rational number, then the ratio in lowest terms must have denominator 1, otherwise N can’t be expressed in lowest terms with denominator 1. So, if square root of N is a rational number, then it must be an integer.

If the square root of N is a rational number, then N must be a square integer. 9 is a square integer, so when we cast out the 9s of N, we find that it is a multiple of 9.

Anonymous

Anonymous Tue 05 Jan 2021 12:27:11 No.12547388 Report

Quoted By: >>12547409 >>12548864

>>12546876
Ok, so here is what I did:
Define the sequence
$a_{n+1}=10^n+a_n$
Obviously we have:
$a_1=1$
$a_2=11$
$a_3=111$
$a_4=1111$
and so on.
We can show that:
$a_n= 1+10+100+1000+...+10^{n-1} = \frac{10^n-1}{10-1} = \frac{10^n-1}{9}$ (sum of a geometric series)
We then define $b_n= \sqrt{a_n}$ since we are interested in $\sqrt{111...1}$
Then we have that $b_n= \sqrt{ \frac{10^n-1}{9} }$ so we know it grows proportionally to $\sqrt{10^n}$
We then divide with it and take the limit:
$\frac{b_n}{ \sqrt{10^n} } = \sqrt{ \frac{10^n-1}{9*10^n} } = \frac{1}{ \sqrt{9} } \sqrt{ 1 - \frac{1}{10^n} } \to frac{1}{3}$
This means that $b_n ~ \frac{1}{3} \sqrt{10^n}$ for large $n$
Since the $n=1998$ we are interested is even and kinda big $b_{1998} ~ \frac{1}{3} 10^999$ so a large number of its first digits are $3$ (note that this wouldn't work if we had an odd number, pretty cool!)
To determine whether the thousand digit after the decimal is 3 or not we just need to check weather $| \frac{1}{3}-\frac{ b_{1998} }{ \sqrt{ 10^{1998} } } | < \frac{1}{ 10^{1001} }$
We have $\frac{1}{3}-b_{1998} = \frac{1}{3}-sqrt{ \frac{10^{1998}-1}{ 9*10^{1998} } } = \frac{1}{3}-\frac{1}{3} \sqrt{ \frac{10^{1998}-1}{ 10^{1998} } } = \frac{1}{3} (1- \sqrt{1-\frac{1}{10^{1998} } }$
Now since $\sqrt{1-x}<1-\frac{x}{2}$ (can be proven with many ways for example Taylor series or just tangent at the point x=0 and the function has a negative second derivative)
We have $\frac{1}{3} (1- \sqrt{1-\frac{1}{10^{1998} } } < \frac{1}{3} (1-1+\frac{1}{ 2*10^{1998} } = \frac{1}{3} \frac{1}{ 2*10^{1998} } < \frac{1}{ 10^{1001} }$
Also we need to show that the thing whithin the absolute value is positive, which is easy since $\sqrt{1-x}<1$ for $1>x>0$
So 3

Anonymous

Anonymous Tue 05 Jan 2021 12:38:59 No.12547409 Report

Quoted By: >>12547412

>>12547388
I kinda fucked up latex in some parts, I hope it's still readable. Obviously frac13 is 1/3 and I meant $b_n \sim \frac{1}{3} \sqrt{10^n}$ for large n and $b_{1998} \sim \frac{1}{3} \sqrt{10^{999}}$
Plus some parenthesis and a sqrt but yeah

Anonymous

Anonymous Tue 05 Jan 2021 12:40:28 No.12547412 Report

Quoted By:

>>12547409
$b_{1998} \sim \frac{1}{3} 10^{999}$

Anonymous

Anonymous Tue 05 Jan 2021 18:13:55 No.12548227 Report

Quoted By: >>12550022

1/9 = 0.111111111

1/9 x 10^X just shifts decimal.

sqrt (1/9) = 1/3.

thus 3 is on the number you ever see.

Anonymous

Anonymous Tue 05 Jan 2021 21:35:08 No.12548864 Report

Quoted By:

>>12547388
For the inequality, both sides need to be divided by $\sqrt{10^{1998}}$ , so is should be along the lines of
$\lvert \frac{1}{3} - \frac{b_{1998}}{\sqrt{10^{1998}}} \rvert < \frac{1}{10^{2000}}$
It's probably easier around this step to work with the Taylor expansion of $\sqrt{x}$ at $\frac{10^{1998}}{9}$ at this point, and it is easy to see that only the constant and linear terms matter towards the thousandth digit after the decimal.
$b_{1998} \cong \frac{1}{3} 10^{999} - \frac{1}{ \frac{2}{3} 10^{999} }$
Multiply through by $10^{1000}$ , and the ones digit is the same as that of
$\frac{100}{3} - \frac{30}{2} = 18.333\dots$

Anonymous

Anonymous Tue 05 Jan 2021 21:54:31 No.12548918 Report

Quoted By:

>>12546876
step 1. $\sqrt{1111\ldots1}$ is
$10^{999}\sqrt{0.1111\ldots1}$

step 2. $\sqrt{0.1111\ldots1}$ is about $\sqrt{1/9} =1/3$

step 3. So the digit is probably 3 I dunno

Anonymous

Anonymous Tue 05 Jan 2021 22:28:38 No.12549014 Report

Quoted By:

>still no correct solution to OP's problem
I laugh at you brainlets.

Anonymous

Anonymous Tue 05 Jan 2021 22:41:59 No.12549048 Report

Quoted By:

>>12546876
0
I just used bigfloat lib

Anonymous

Anonymous Wed 06 Jan 2021 00:50:21 No.12549490 Report

Quoted By:

>>12546876
Try to manipulate sqrt(1/9*10^1998-1/9)

Anonymous

Anonymous Wed 06 Jan 2021 00:51:59 No.12549503 Report

Quoted By:

>>12546876
3

Anonymous

Anonymous Wed 06 Jan 2021 00:56:14 No.12549525 Report

Quoted By:

>>12546876
gotta be 1

Anonymous

Anonymous Wed 06 Jan 2021 03:16:38 No.12550022 Report

Quoted By:

>>12548227
This is a neat way to do it for a specific problem like this

Anonymous

Anonymous Wed 06 Jan 2021 03:54:55 No.12550105 Report

Quoted By: >>12550204 >>12551118

>>12546876
$N\;=\;\sum_{n\;=\;0}^{1997}10^n\;=\;\frac{10^{1998}\,-\,1}9$
$\sqrt N\;=\;\frac{10^{999}}3\,\sqrt{1\,-\,\frac1{10^{1998}}}$
$1000\,\sqrt N\;=\;\frac{10^{1002}}3\,\sqrt{1\,-\,\frac1{10^{1998}}}$
$\frac{10^{1002}}3\,-\,\frac1{6\,\times\,10^{996}}\;=\;\frac{10^{1002}}3\,\left(1\,-\,\frac1{2\,\times\,10^{1998}}\right)\;<\;\frac{10^{1002}}3\,\sqrt{1\,-\,\frac1{10^{1998}}}\;<\;\frac{10^{1002}}3\,\left(1\,-\,\frac1{2\,\times\,10^{1998}}\,+\,\frac1{8\,\times\,10^{3996}}\right)\;=\;\frac{10^{1002}}3\,-\,\frac1{6\,\times\,10^{996}}\,+\,\frac1{24\,\times\,10^{2994}}$
$\sum_{n\;\leq\;1001}3\,\times\,10^n\,-\,\frac1{10^{997}}\,-\,\sum_{n\;\leq\;998}6\,\times\,10^n\;<\;1000\,\sqrt N\;<\;\sum_{n\;\leq\;1001}3\,\times\,10^n\,-\,\frac1{10^{997}}\,-\,\sum_{n\;\leq\;998}6\,\times\,10^n\,+\,\frac1{24\,\times\,10^{2994}}$
$\sum_{n\;=\;0}^{1001}3\,\times\,10^n\;=\;\left\lfloor\sum_{n\;\leq\;1001}3\,\times\,10^n\,-\,\frac1{10^{997}}\,-\,\sum_{n\;\leq\;998}6\,\times\,10^n\right\rfloor\;\leq\;\left\lfloor1000\,\sqrt N\right\rfloor\;\leq\;\left\lfloor\sum_{n\;\leq\;1001}3\,\times\,10^n\,-\,\frac1{10^{997}}\,-\,\sum_{n\;\leq\;998}6\,\times\,10^n\,+\,\frac1{24\,\times\,10^{2994}}\right\rfloor\;=\;\sum_{n\;=\;0}^{1001}3\,\times\,10^n$
$\left\lfloor1000\,\sqrt N\right\rfloor\;=\;\sum_{n\;=\;0}^{1001}3\,\times\,10^n$
It’s 3.

Anonymous

Anonymous Wed 06 Jan 2021 04:38:55 No.12550204 Report

Quoted By:

>>12550105
Erratum: the second-order term is negative and on the left; doesn’t change the result.

Anonymous

Anonymous Wed 06 Jan 2021 12:56:18 No.12551118 Report

Quoted By:

>>12550105
>>>12546876 (OP)
>3
Can you write AES encryption as sum?

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