AIME - Number Theory

font style="font-weight: bold;">Two positive integers differ by The sum of their square roots is the square root of an integer. What is the maximum possible sum of the two integers?

I got rid of one of the conditions because it just made more busy work, but it's really still the same problem, just with a different answer.

Call the integers x and x + 60. Then:



At this point, it was rather intuitive to me that $x(x+60)$ must be a perfect square for this to be possible, since I had already done a lot of work with adding radicals. usamts??? I've said too much ;) . But you can show this by squaring both sides of the equation.







Now let z = x+30:



I'm telling you, that difference of squares is key. One of the 2006 IMO problems was a tough looking number theory problem - until you see the difference of squares. Even Alex Zhai didn't see it immediately! And he's like...omnipotent, when it comes to math. It's like he's THERE when they're writing the problems. Just goes to show you that you can never overlook simple tools.



Since we will maximize x by maximizing z, we'll go through the cases, starting with where z is the largest:



Solving for z yields a number that is not an integer, so we throw it out.



Solving for z yields z = 226, so x = 196 .Thus, our desired sum is:



If you're curious, the condition that I excluded was that our root sum cannot be an integer, which this answer doesn't fit. I took it out because I just would've had to type up another case, and I'm sure you got the idea.

"Dying is the day worth living for"

So I finally started watching At World's End. I don't know why everybody said it was awful. I mean, granted, it was no Dark Knight (which was absolutely mind blowing), but it was worth a watch.

I also started watching a new anime called Monster. The premise is that this doctor saved the life of a small boy who grew up to be a serial killer. So the killer in a way idealizes the doctor, while the doctor is trying to track him down. It reminded me of Death Note a bit, only without the magic notebooks or flying shinigami.

I'm also officially enrolled in WOOT now. I'm wondering if it'll be enough to keep me competitive. After all, all of the MOSP participants automatically get into it. Then again, I think I'm progressing rather quickly. WOOT should help me out with that. We'll see.

Made up number theory problem

Prove that, for all positive integers x and r, the remainder when is divided by is 1.

This is pretty easy if you interpret it as:



Which, according to the binomial theorem, is:



If we expand this out, then every term except for the first will be divisble by (x-1). The first term is just 1, so the remainder is 1. Nifty, right? I thought so.

WOOT

No, that is NOT the douchalicious phrase that people say online when something causes them to orgasm all over their keyboard. Sorry, I'm a bit angry right now. Anyway, it stands for Worldwide Olympiad Online Training. And I've managed to convince one of my parents (guess which one) to sign me up for it. I'm REALLY happy, even if you can't tell from my constant unintentionally unsatisfied mood. It's basically a USAMO prep class.

I've been wondering, would it be possible for me, somebody who had never heard of the AMC less than a year ago, to make it to the IMO? It's certainly a HUGE stretch. I was looking at the winning USAMO scores from this year, they were all in the 33 - 37 range (two people scored 37, the highest). That would mean that I would probably have to solve EVERY problem at least somewhat. Several of them would have to be perfect. I guess I just have to ask myself one thing: Do I think I have what it takes to practice, learn what needs to be learned, and, when the time comes, shit in the mouth of 6 math problems? Of course I can. I'm going for it.

Wish me luck.

AIME - Trigonometry

Find a positive integer n such that:



Keep in mind that, since this is a non calculator test, frantically entering this into your calculator wont help you, as its miles away in the Mojave Desert being cocooned by military spiders. So you best prepare to figure it out on your own.

First off, here's a tip:

Whenever you see inverse trig functions, try to get rid of them by applying trig functions to the equation.

With that said, lets take the tangent of both sides:




The LHS looks like it can be simplified using our tangent sum formula. If you don't know it (and you should), it's:

However, we don't have two terms, we have four. So the best thing to do is split the sum of the four into a sum of two. Just add some parenthesis in to see what I mean:



This can start to get a little messy, so lets split it into two variables:




Calculate each using the sum formula:
















Now plug these back in:






(Here I made the substitution right away, rather than type out that whole ugly expression. Resubstitute in our original expressions for x and y if you want to see what I was avoiding.)



The rest is simply the algebra of solving for n:














Tadaa. If you're practicing outside of the AIME, you can plug that into your calculator and check your answer.

Now, what would've we done if we didn't get a positive integer for an answer? Well, it's the AIME, so if they say solve for n, you can be sure it's a positive integer. Not to mention they right out say it, too. You should go back and check your work for answers. Work your away up instead of starting at the top, checking for arithmetic errors first. If you still cant find it, then check your logic for errors too. If you STILL cant find it, well then, I'd say skip that one for a bit and attempt it again later.

Coming soon to a desolate math blog near you

Geometry problems! I've shied away from geometry problems here because if I don't provide a diagram, then it's just confusing and hard to follow. However, I recently downloaded a neat free geometry program, so perhaps I could get some going. And no, it's not geometer's sketchpad. Not only is that not free, but for some reason I've always found it to be a bitch to use. Perhaps I'll give it another go though (I own it).

And the sun just rose. I really need to get my sleep in order. I keep nagging my parents to buy some red bull so I can even it out, but they wont. I think it's time to ask Nick to procure some for me.

Number Theory and problem solving

I'll try to take you through the steps of solving a basic number theory problem:

Find the smallest positive integer whose cube ends in 888.


How do we begin? And keep in mind that this is a non-calculator question, although you should be expected to be able to pump out some basic arithmetic without one.

Perhaps we can start by looking at an easier problem: whats the smallest integer whose cube ends in 8? Hopefully it didn't take you more than a few seconds to see that it's 2. Next you should ask yourself, does this mean that any integer whose cube ends in 8 has to end in 2? No other single digit cubed ends in 8 (you can check). Also, the unit digit in a multiplication will only be affected by the unit digit of the factors. So this should mean that it has to end in 2.

Now lets work our way up - whats the smallest integer whose cube ends in 88? At first this seems a lot harder, but we can use what we already figured out to make it easier. Realize that the units and tens digit will only be affected by the units and tens digit of the products. So technically, we should be able to actually do out the multiplication of cubing "x2", where x is our unknown digit. You'll have to keep track of the tens digits while adding by considering them mod 10 only. I'll try to draw it out as best as I can here:

x 2
* x 2
------------
2x 4
2x
------------
4x 4
x 2
-------------
8x 8
4x
-------------
12x 8

Now our problem is just, what is the smallest digit x such that 12x is congruent to 8 mod 10. Which is just 4. So the smallest integer cubed who ends in 88 is 42.

Now lets try applying this method to 888. We can't assume that the tens digit is 4, but we can assume that the ones digit is still 2. I won't draw out everything here, but letting the number be x y 2 yields:

(12x + 6y^2) (12 y) (8)

This time, when we find a number for y, we can't just take it mod 10- we'll have to consider what is carried over and add it to the left.

Let's deal with y first. If y is 4 or 9, then 12y is congruent to 8 mod 10. Lets try 4 first:

(12x + 6*16) (48) (8)
(12x + 6*16 + 4) (8) (8)
(12x + 100) (8) (8)

So the smallest x such that 12x + 100 is 8 mod 10 is 4. That yields 442 as a possible number. However, trying out 9 might yield a smaller hundreds digit, so lets try it.

(12x + 6*81) (108) (8)
(12x + 486 +10)(8) (8)

(12x + 496) (8) (8)

So the smallest x such that 496 is 8 mod 10 is 1, yielding 192. Certainly much smaller than 442. Can we say for sure that these are the smallest possible numbers? We showed that any number whose cube ends in 888 must end in these numbers. We also explored the different cases, and picked the smallest from each, so there are no other possibilities for a smaller number. Our final answer is thus 192.

Keep in mind that this isn't the only way to do this. We could also have analyzed the expansion of (100a + 10b + 2)^3. Personally, I preferred the previous method because I'm not exceptional at expanding anything past binomials, but it would be good practice. In fact, I'll try it later tonight.

If anybody happens to be reading this, heres a problem for you - Find the smallest positive integer whose cube ends in 69. And do it WITHOUT brute forcing! (Hint, expand it, and take it mod 100).

By the way, it really isn't worth "memorizing" this trick. The most important thing is being able to solve it on your own, without having seen a very similar problem. That doesn't mean you shouldn't try to relate to elementary problems or other ones that you HAVE memorized, though!

What I'm trying to say is, that if you had never seen a problem like this, and after looking at it for a minute, you thought "eh, this is too hard, whatever." then you're NOT TRYING HARD ENOUGH. Here's a prime example: If you've ever looked at the USAMTS problems, they're hard. Maybe not quite USAMO level, or if you don't look at each one for more than an hour, depending on its difficulty, you probably won't solve it. I have a friend, who is very smart and good at math, and I asked him to try USAMTS this year. His response was:

"Nah, I looked at the problems last night and I couldn't get any, so it's probably too hard for me."

I'm going to try to convince him again, but a key part of problem solving is to not back down from a problem. You have to take the time to explore, play around with, and change the problem until you have a good idea of what is going on, before you can actually solve it. Just because you don't see the solution right away doesn't mean it's not there.