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[Christes]

I got it.  Here's a hint...

First show that no perfect square has last digit equal to 3.

Then show that for sufficiently large n, the sum of factorials has last digit 3.

Edit:  As an aside... if you know modular arithmetic, you will note that we were really only considering factorials mod 10.  You could probably use another value here.

[Nivim]

I'll just leave this here.

Beautiful.

Indeed. Thanks for that, Vector.

Indeed, indeed.

EDIT:  ...I disagree with some points he makes though.  He rails about how stupid it is to prove the obvious.  Well, a lot of obvious things aren't true, and if you don't have the correct tools, you can't prove the non-obvious.

Also, it's really annoying not having a common language to talk about stuff with other people.  Artists have common language for things like "vanishing points", after all.

He's talking about terminology glut and intuition-killing proofs.  It's very important to be rigorous, but there's a certain kind of rigorous that really isn't needed.  Mathematicians basically never do two-column proofs.  Ever.  They also have one hell of a lot of terminology, but don't have names for "whole numbers" versus "mixed numbers" versus "improper fractions," because those aren't important terms to anyone.

Ugh, yes. I can personally vouch for how terrible mathematics is in schools. I FAILED early math classes because I couldn't memorize the tables and formulas I was supposed to. The only thing that finally got me to do well in a math class, and ENJOY it, was one teacher who did exactly what he proposes. It was a private school, with it's own curriculum(closed down now...) and her math class involved her giving us interesting problems and just having the class go at it with no help but the occasional nudge in the right direction.

By gods yes! Why hasn't this been publicized thoroughly and spread to all offending parties?

[Sowelu]

I got it.  Here's a hint...

First show that no perfect square has last digit equal to 3.

Then show that for sufficiently large n, the sum of factorials has last digit 3.

Edit:  As an aside... if you know modular arithmetic, you will note that we were really only considering factorials mod 10.  You could probably use another value here.

Oh!  Nice.  Let me take a crack at that, though that hint seriously gives it away.

Spoiler (click to show/hide)

Observe that S4 = 33.  Observe that n! for all n > 4 have both 5 and 2 as factors, so their last digit in base 10 is 0, therefore the last digit of Sn will not change above S4, it will always be 3.

And...uh...As for the last digit always being 3.  Let me look at that.  ...Oh, that's funny.  The last digit of n^2 appears to cycle through the same ten numbers in the same order every time:  1 4 9 6 5 6 9 4 1 0  And I have no idea why.  3 isn't in there, though.  Well, that's not rigorous enough, why does it cycle?  Oh!  Because n^2 - (n-1)^2 = 2n-1, which follows a much more rigorously predictable pattern of five values.  Since adding that value is the only thing that changes the last digit of n^2, nothing can break us out of that observed cycle.

Well that was a terrible stream-of-conscious proof, but it makes sense to me now! 

[Christes]

While I agree with the main thrust of Lockhart's lament, something about it has always rubbed me the wrong way.  Maybe it's because I'm immersed in a collegiate environment where students take my class to fulfill prerequisites, not because they are interested in them.  (And, incedentally, that's not the math department's fault at all!)

[Christes]

I got it.  Here's a hint...

First show that no perfect square has last digit equal to 3.

Then show that for sufficiently large n, the sum of factorials has last digit 3.

Edit:  As an aside... if you know modular arithmetic, you will note that we were really only considering factorials mod 10.  You could probably use another value here.

Oh!  Nice.  Let me take a crack at that, though that hint seriously gives it away.

Spoiler (click to show/hide)

Observe that S4 = 33.  Observe that n! for all n > 4 have both 5 and 2 as factors, so their last digit in base 10 is 0, therefore the last digit of Sn will not change above S4, it will always be 3.

And...uh...As for the last digit always being 3.  Let me look at that.  ...Oh, that's funny.  The last digit of n^2 appears to cycle through the same ten numbers in the same order every time:  1 4 9 6 5 6 9 4 1 0  And I have no idea why.  3 isn't in there, though.  Well, that's not rigorous enough, why does it cycle?  Oh!  Because n^2 - (n-1)^2 = 2n-1, which follows a much more rigorously predictable pattern of five values.  Since adding that value is the only thing that changes the last digit of n^2, nothing can break us out of that observed cycle.

Well that was a terrible stream-of-conscious proof, but it makes sense to me now! 

Sure, that works.  Also note that the last digit of n^2 is purely determined by the last digit of n - that's how I looked at it.

[Vector]

Why hasn't it been spread everywhere?

A mathematician wrote it.

While I agree with the main thrust of Lockhart's lament, something about it has always rubbed me the wrong way.  Maybe it's because I'm immersed in a collegiate environment where students take my class to fulfill prerequisites, not because they are interested in them.  (And, incedentally, that's not the math department's fault at all!)

Yeah, goddamned engineering and physics departments >_>

[Shinziril]

Heh heh heh.  I'm an engineering major at the moment, although I do actually like abstract math, even if I never delve very deeply into it.  I have to admit that having the practical-math reach a level where it's actually USEFUL helps keep interest up (using Matlab to solve systems of linear equations, numerical differentiation/integration, etc, versus boring shit like memorizing times tables that you do back in early school and never really use). 

[Christes]

While I agree with the main thrust of Lockhart's lament, something about it has always rubbed me the wrong way.  Maybe it's because I'm immersed in a collegiate environment where students take my class to fulfill prerequisites, not because they are interested in them.  (And, incedentally, that's not the math department's fault at all!)

Yeah, goddamned engineering and physics departments >_>

Heh, actually I'm teaching college algebra/trig now.  Most engineering and physics students are okay with those already xD  At my school every person who gets a BS needs a full year of math.  There is also a general requirement from several departments (like sociology) that their students take college algebra.  The students usually put this requirement off until senior year.  That doesn't end well.  College algebra is also required for the business calc class. 


It seems like these departments are having us do their dirty work.

[ed boy]

I don't see an obvious way to do it, but I've been wrong before, especially in Number Theory.  Is this for a class?  It might be helpful to know what sort of machinery you have to work with if it is.

My aunt's friend's daughter has university interviews starting on sunday, and I have been helping her. She has a bunch of practice interview questions, and I'm helping her by guiding her through the problems. That one, however, I was unable to manage. I would have liked to spend longer on it to do it myself, but there are a lot of questions to get through before interviews start.

I got it.  Here's a hint...

First show that no perfect square has last digit equal to 3.

Then show that for sufficiently large n, the sum of factorials has last digit 3.

Edit:  As an aside... if you know modular arithmetic, you will note that we were really only considering factorials mod 10.  You could probably use another value here.

Many thanks.

[ed boy]

I have another couple. I think I can work out solutions, but I'm not confident in them and would appreciate criticism of them. Questions are in red, my attempts at solutions are in blue.

Spoiler: the first one (click to show/hide)

Let An be 2²²ⁿ (that is, 2^(2^(2^(n)))) and Bn be 100¹⁰⁰ⁿ (that is, 100^(100^(n))).
As n tends to infinity, which out of An and Bn is larger?

log(An)=2²ⁿlog(2)
log(log(An))=log(log(2))+2ⁿlog(2)
log(Bn)=100ⁿlog(100)
log(log(Bn))=log(log(100))+nlog(100)

Am I right in thinking that since log(log(An)) increases exponentially and log(log(Bn)) increases linearly, An will be greater then Bn as n tends to infinity?

Spoiler: the second one (click to show/hide)

In a regular pentagon ABCDE, C’ is the intersection of AD and BE etc. If the area of the triangle AC’D’ is 1, what is the area of the pentagon A’B’C’D’E’?

I can do this one by expressing the area of the pentagon ABCDE in terms of the pentagon A'B'C'D'E' and some triangles. I can then find simultaneous equations for the area, and use those to find an expression for the area of the pentagon A'B'C'D'E'. However, this method is very long and unpleasant, and I feel that I am missing a simple and easy way to do it. Am I?

[Vector]

Am I right in thinking that since log(log(An)) increases exponentially and log(log(Bn)) increases linearly, An will be greater then Bn as n tends to infinity?

Yes, this is true.

Working on the other one as we speak.

[Sowelu]

Yes on the first one.

On the second one...Well it depends on what you consider to be too long and unpleasant. I mean it's just three triangles, seriously.  Calculate BAC and EAD with SAS, then find the length of AC or AD then calculate ACD with SSA.  Or you can think of it as a triangle and a trapezoid...

[Vector]

Trying to do it without side-lengths.  I think this might work.

Let N = area of ABCDE, q = area of small triangles, n = area of inner pentagon, p the integer such that pn = N.

Observations:

N = 5 + 5q + n  --> 5(1+q) = (p-1)n
1 + 2q = n + 2
p = 3 + n + q

That should be enough.  It's a bit more elegant, I feel.

(If you want to know how I got the last equation, I reflected one of the area-1 triangles over the edge of the inner pentagon)

Spoiler: Potentially helpful diagram (click to show/hide)

[Sowelu]

I realized I didn't read the question nearly enough.  Oops.

On the other hand, I re-derived the general solution for the area of an N-sided polygon, given the length of a side.  Whee.

[ed boy]

N = 5 + 5q + n  --> 5(1+q) = (p-1)n
1 + 2q = n + 2
p = 3 + n + q
(If you want to know how I got the last equation, I reflected one of the area-1 triangles over the edge of the inner pentagon)

I'm sorry, I can see how reflecting the triangle over the inner edge of the pentagon gives the second one, but how does it give the third?