## math problem - driving me crazy

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### math problem - driving me crazy

I've been working on this for 2 days, i don't feel like I'm getting any closer.

What is the 100th digit after the decimal point in the following expression:

( \sqrt{5} + \sqrt {7} )^{2011}

If any one can help me regain my sanity I'll really appreciate it.
joekz

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### Re: math problem - driving me crazy

joekz wrote:I've been working on this for 2 days, i don't feel like I'm getting any closer.

What is the 100th digit after the decimal point in the following expression:

( \sqrt{5} + \sqrt {7} )^{2011}

If any one can help me regain my sanity I'll really appreciate it.

Was this on the Putnam exam (or some other math contest)? It seems like it since they tend to incorporate the current year into their tests.

Here are my thoughts:

Spoiler:
When we expand it, we can ignore the integer part of the expansion (for example,
( \sqrt{5} + \sqrt {7} )^{2} = 12 + 2\sqrt{35}
in which case we can ignore the 12.

Looking at the number 2011, it's binary representation is 11111011001, so
( \sqrt{5} + \sqrt {7} )^{2011} = ( \sqrt{5} + \sqrt {7} )^{1024}( \sqrt{5} + \sqrt {7} )^{512}( \sqrt{5} + \sqrt {7} )^{256}( \sqrt{5} + \sqrt {7} )^{128}( \sqrt{5} + \sqrt {7} )^{64}( \sqrt{5} + \sqrt {7} )^{16}( \sqrt{5} + \sqrt {7} )^{8}( \sqrt{5} + \sqrt {7} )

It's cute that all of them have the form a+b\sqrt{35} except for the last one, which can be rewritten as
\frac{5\sqrt{7} + 7\sqrt{5}}{\sqrt{35}}

So, once we multiply all terms and then remove the integers, the part we have to look at will be of the form
m( 5\sqrt{7} + 7\sqrt{5} )

where m is the coefficient of the \sqrt{35} term in ( \sqrt{5} + \sqrt {7} )^{2010} Of course, finding m by brute force would be extremely tiresome, so hopefully there is an easy pattern we can see.

So, let's look at the main question: What is the 100th digit after the decimal point?

My first thought would be to find the 100th digit of ( 5\sqrt{7} + 7\sqrt{5} ) and see what happens as we change m, but that would be tiresome. Instead, let's look at what makes the 100th digit. If you multiply two numbers with some arbitrary number of digits after the decimal point, the number of digits in the new number is equal to the sum of the number of digits of both of your original numbers.

So perhaps if we find a way to break m( 5\sqrt{7} + 7\sqrt{5} ) into smaller factors that are easier to handle, we can piece together what the 100th digit would be. I'd be willing to bet that this has something to do with pascal's triangle, which I'm not terribly familiar with.

I have no idea if I'm on the right track, this is just stream of consciousness.
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Xias

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### Re: math problem - driving me crazy

Hint:

Spoiler:
Look at
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n.
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### Re: math problem - driving me crazy

Xias thanks a lot! I'm at work right now but i will definitively try that path when ill get home.

Macbi wrote:Hint:

Spoiler:
Look at
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n.

Spoiler:
I tried going in that path... but that only gave me results when n was even (then the expression is an integer)... maybe there is something i dont see?
joekz

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### Re: math problem - driving me crazy

Spoiler:
6. Proof by Maple.

Proginoskes

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### Re: math problem - driving me crazy

joekz wrote:Xias thanks a lot! I'm at work right now but i will definitively try that path when ill get home.

Macbi wrote:Hint:

Spoiler:
Look at
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n.

Spoiler:
I tried going in that path... but that only gave me results when n was even (then the expression is an integer)... maybe there is something i dont see?

Ah, for some reason I thought we were in 2010. I now have no idea.

(Wolfram Alpha says the answer is 4.)
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Macbi

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### Re: math problem - driving me crazy

I got 6 from W|A like the above maple poster, 4 is the 99th digit?

http://www.wolframalpha.com/input/?i=fr ... +sqrt+7%29^2011%29*10^99%29

A few things I would consider, (bearing in mind I relatively suck at this type of problem), would be to

a) look at the Taylor series of things like (sqrt (4+x) + sqrt (9-2x))^2011
b) 2011 = 2048 - 37, look at the binary decomposition that way, in a similar way to what Xias is saying
c) Can we find the Nth digit of sqrt(5) + sqrt(7) on its own? How about its square? If we can do those, can we generalize?
d) Write it as
\frac{2^{2011}}{(\sqrt 7 - \sqrt 5)^{2011}}
and use b above to get the annoying double radical term in the numerator only.

No clue if these lead anywhere.
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mike-l

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### Re: math problem - driving me crazy

We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.

In balanced trinary 2011 is:
2457 - 819 + 273 +81 + 27 -9 + 1
Or +-+++-0+
In trinary 2011 is:
2110201

sqrt(5)+sqrt(7) cubed is (12 + 2 sqrt(35)) * (sqrt(5) + sqrt(7)) = 12 sqrt(5) + 12 sqrt(7) + 10 sqrt(7) + 14 sqrt(5) = 26 sqrt(5) + 22 sqrt (7)
a sqrt(5) + b sqrt(7) cubed is:
= (5 a^3 + 21 ab^2) sqrt(5) + (15 b a^2 + 7 b^3) sqrt(7)
hmm, that doesn't look all that useful.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Yakk
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### Re: math problem - driving me crazy

Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.

Right, but at the end, we'll have a 1s part and a non-1s part, and only the latter matters. It's for that reason though that actually calculating the non-1s part by hand would be difficult.

Macbi wrote:Hint:

Spoiler:
Look at
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n.

I'm curious now. For n=>even, you end up with double the integer part. What exactly happens when n=>odd, and how can we use this to our advantage?
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Xias

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### Re: math problem - driving me crazy

Xias wrote:
Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.

Right, but at the end, we'll have a 1s part and a non-1s part, and only the latter matters. It's for that reason though that actually calculating the non-1s part by hand would be difficult.

Macbi wrote:Hint:

Spoiler:
Look at
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n.

I'm curious now. For n=>even, you end up with double the integer part. What exactly happens when n=>odd, and how can we use this to our advantage?

Spoiler:
If n is even
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n.
will be an integer (the odd powers will cancel themselves out and the even powers will give integers). so any digit after the decimal point of
(\sqrt{7}-\sqrt{5})^n
will be 9 less the digit after the decimal point in the same place of
(\sqrt{7}-\sqrt{5})^n
. And it isn't hard to show that
(\sqrt{7}-\sqrt{5})^{2011}< \frac{1}{2}^{2011} < 10^{-100}
which means the first 100 digits after the decimal point of
(\sqrt{7}-\sqrt{5})^{2011}
are 0, so the 100 digit after the decimal point of
(\sqrt{7}+\sqrt{5})^{2011}
is 9. I'm not sure if this helps at all with the original problem.
joekz

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### Re: math problem - driving me crazy

2011 isn't even...
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Yakk
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### Re: math problem - driving me crazy

Wait...

Wouldn't it just be:

Then under distributive property:

5+7

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.
sundalius

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### Re: math problem - driving me crazy

sundalius wrote:Wait...

Wouldn't it just be:

Then under distributive property:

5+7

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.

http://en.wikipedia.org/wiki/Freshmans_dream
Quite relevant. Basically, following your train of thought
(2+3)^2 = 2^2 + 3^2
5^2 = 4+9
25 = 13 and that doesnt quite sound right to me.. A search for "expanding brackets" might help you as well.

t1mm01994

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### Re: math problem - driving me crazy

@sundalius: the distributive property means that a(b+c) = ab+ac. From this you can prove that (a+b)(c+d) = ac + ad + bc + bd. You should be able to work out where you're going wrong from that.
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jestingrabbit

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### Re: math problem - driving me crazy

Xias wrote:
Yakk wrote:We cannot completely ignore the 1s part until the very end, because those factors end up infecting our non-ones parts when we multiply.

Right, but at the end, we'll have a 1s part and a non-1s part, and only the latter matters. It's for that reason though that actually calculating the non-1s part by hand would be difficult.

Macbi wrote:Hint:

Spoiler:
Look at
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n.

I'm curious now. For n=>even, you end up with double the integer part. What exactly happens when n=>odd, and how can we use this to our advantage?

Spoiler:
everything that has sqrt(7) in an odd power will be canceled out. we will be left with a*sqrt(5) when a is a (very large) integer. This expression will be identical up to 10^-100 to the original one.
I'm still no sure this will get us to the answer, but it seems to be worth a shoot.

now my problems are -
1. expressing a (working on it...).
2. find a smart way to find the 100th digit after the decimal point in a*sqrt(5) (might be easier after expressing a)

Another thought - i can do the exect same thing with
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n

which will result in a*sqrt(7) - if that makes life any easier.
joekz

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### Re: math problem - driving me crazy

So (\sqrt 5 + \sqrt 7)^{2010} is close enough to an integer that, for the purpose of finding the 100th digit, we can just assume that it is an integer (the difference is smaller than 10^(-500)). So we essentially have a huge 1000 digit integer times \sqrt 5 + \sqrt 7. (And we know what the integer is, it's 2*(5^1005+7^1005)+2*(2010 choose 2)*35*(5^1004 + 7^1004) + ...., so maybe something can be said for digits of (ab)^n(a^m+b^m)(\sqrt a + \sqrt b)?
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mike-l

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### Re: math problem - driving me crazy

Well, only the first 100 digits of that number matter.

So what if we looked at it mod 10^100?

It might end up having a pattern in its first 100 digits. So we could also look at it mod 10 and see what we get.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Yakk
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### Re: math problem - driving me crazy

Pretty sure every digit in the big integer matters, root 5 + root 7 being irrational.
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mike-l

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### Re: math problem - driving me crazy

... ya, that was dumb.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Yakk
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### Re: math problem - driving me crazy

sundalius wrote:Wait...

Wouldn't it just be:

Then under distributive property:

5+7

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.

Even if the first part worked, you'd actually get

(5 + 7)^(2010/2) * (sqrt 5 + sqrt 7) = 12^1005 * (sqrt 5 + sqrt 7),

or maybe (sqrt 12)^(2011),

and I'm not sure you're any better off.

Proginoskes

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### Re: math problem - driving me crazy

I don’t know if this helps, but we can rewrite (√7 + √5) = 2/(√7 - √5).
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Qaanol

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### Re: math problem - driving me crazy

I think you need a series that converges more rapidly than the binomial theorem.

Proginoskes

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### Re: math problem - driving me crazy

Macbi wrote:
joekz wrote:Xias thanks a lot! I'm at work right now but i will definitively try that path when ill get home.

Macbi wrote:Hint:

Spoiler:
Look at
(\sqrt{7}+\sqrt{5})^n + (\sqrt{7}-\sqrt{5})^n.

Spoiler:
I tried going in that path... but that only gave me results when n was even (then the expression is an integer)... maybe there is something i dont see?

Ah, for some reason I thought we were in 2010. I now have no idea.

(Wolfram Alpha says the answer is 4.)

Yes, W|A has the 100th digit as being 4, but is that not just the 100th digit and not the 100th digit after the decimal place?

EDIT: I am no mathematician, I just use W|A a lot...
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Kick

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### Re: math problem - driving me crazy

Kick wrote:Yes, W|A has the 100th digit as being 4, but is that not just the 100th digit and not the 100th digit after the decimal place?

EDIT: I am no mathematician, I just use W|A a lot...

Ah, yes. That was my mistake.
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### Re: math problem - driving me crazy

After working on this problem for a while, I'm wondering whether the OP is remembering it incorrectly. All the approaches seem to eventually lead to knowing the 100th digit after the decimal place of \sqrt5, or some other equally (in)accessable data.

Late last night, I had an inspiration. Since there are a lot of symbols, and I don't really want to mess with the math interface for that, I wrote it up and put in a PDF; the link is http://math.la.asu.edu/~checkman/extras/sqrt5sqrt7.pdf .

Proginoskes

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### Re: math problem - driving me crazy

Spoiler:
[list=*][*]In your last expression, the right side inside the "[ ]" is much smaller than 10^(-100), so it can be neglected for the important digit.
[*]You now multiply a very large number with sqrt(5), so you need a lot of digits of sqrt(5).
[*]The expression inside the "[ ]" is ~10^1071, which is a lot (~300 orders of magnitude) smaller than the original number. I wonder where this comes from.
[/list]
mfb

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### Re: math problem - driving me crazy

t1mm01994 wrote:
sundalius wrote:Wait...

Wouldn't it just be:

Then under distributive property:

5+7

5+7=12 then 12^2011

This was my first thought. Please explain how it's wrong or it'll bug me.

http://en.wikipedia.org/wiki/Freshmans_dream
Quite relevant. Basically, following your train of thought
(2+3)^2 = 2^2 + 3^2
5^2 = 4+9
25 = 13 and that doesnt quite sound right to me.. A search for "expanding brackets" might help you as well.

Thanks. Interesting read, will have to show my Algebra II/Geometry teacher tomorrow... I feel so lost in the math parts here, being only a freshman/sophomore/Junior(it's bloody complicated) in High School. But yeah, that doesn't sound right... I'll go study now >.>
sundalius

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### Re: math problem - driving me crazy

mfb wrote:After looking at your pdf:
The expression inside the "[ ]" is ~10^1071, which is a lot (~300 orders of magnitude) smaller than the original number. I wonder where this comes from.

A mistake. The 12 in the auxiliary equation should be a 24. (Also, remember that you're substituting n=1005 into the formula for b_n, not 2011.) This means
b_n = \left[ \left(12 + 2 \sqrt{35}\right)^n + \left(12 - 2 \sqrt{35}\right)^n\right]\cdot 10^{100} \cdot\sqrt 5

Unfortunately, there are still \sqrt7s lurking about. (I wonder about how \sqrt5+\sqrt7 turned into \sqrt5, though ...

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### Re: math problem - driving me crazy

Well, (\sqrt{7}+\sqrt{5})^2 = 12+2\sqrt{35}
Change \sqrt{5} to \sqrt{7}+\sqrt{5} in your bn definition and you have the original problem.

So after your correction, the whole work looks a bit pointless.
mfb

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### Re: math problem - driving me crazy

Cheating with Wolfram Alpha

Spoiler:
Using the input

(((\sqrt{7}+\sqrt{5})^{2011})*10^{100)})mod10

Wnderer

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### Re: math problem - driving me crazy

mfb wrote:Well, (\sqrt{7}+\sqrt{5})^2 = 12+2\sqrt{35}
Change \sqrt{5} to \sqrt{7}+\sqrt{5} in your bn definition and you have the original problem.

So after your correction, the whole work looks a bit pointless.

Of course, it's going to evaluate to the same number (or it should, anyway). It's the rewriting-and-putting-into-a-form-we-can-deal-with that is important; if it's done properly, it will enable us to solve the problem without doing the long calculation directly. (The "simpler" problem is an example of how this can be done.)

BTW, I think I solved the problem of where the \sqrt7 went. b_1 = 52, and I imagine the original formula pops up again. (I realized this last night when I was dropping off to sleep.)

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