Let's do dis
page 2/2
>>7658381
most promising thread i've seen in ages
1) It's useless to pick an a_i to be 4 or more - you can split it into 2 + (a_i - 2) if that's the case and get a better one product. So we're left with choosing from 2 and 3. Two 3's are better than 3 2's (9 > 8) so we have to pick as many 3s as possible. That takes us to 999. Then we have to compare 3 vs 2+2, where the 2's win.
Final answer:
n = 334
Sequence = 2 times 2, 332 times 3,
>>7658381
2) Let's assume a solution exists, x needs to be close to the square root of 379. This is about 19. Try 19 (it's easy to multiply these numbers because it includes 20), doesn't work, too big. Try 18. It works, looks like this:
380 * 378 + 1 = 379^2
>tfw too stupid to do this
>>7658389
>Two 3's are better than 3 2's (9 > 8)
I don't understand where you get the "(9 > 8)" part from, care to explain? Is it a simple typo, i.e. should it be 9 > 6?
>That takes us to 999
What do you mean by this?
How did you get from this reasoning to n = 334 and the sequence?
15) This is a cubic polynomial with a positive leading coefficient, so it has one real root if and only if it is monotonic---that is, if its derivative is everywhere nonnegative.
The derivative of the polynomial is [math]3 x^2 - 2 a x - 2 a.[/math]
This derivative is itself a quadratic, so it is everywhere nonnegative if and only if its minimum is nonnegative. The derivative of this quadratic is given by [math]6 x - 2 a[/math], so the minimum occurs at [math]x = \frac{1}{3} a [/math].
Evaluating the quadratic at this minimum yields
[eqn]
\frac{1}{3} a^2 - \frac{2}{3} a^2 - 2 a = - \frac{1}{3} a^2 - 2 a.
[/eqn]
This expression is nonnegative when [math]a \geq 0, a \leq -6.[/math] These are precisely the values of [math]a[/math] for which the polynomial has exactly one real root.
>>7658389
This is incorrect, I haven't figured out the answer yet, but doing 96×97×98×99×100×101×102×103×104 is gonna give you a much larger product than 332×333×334
>>7658412
Oh sorry I meant $-6 \leq a \leq 0$, not the other way around.
Use AM-GM for the first one:
[math] \prod_{i=1}^n a_i \leq \left(\frac{\sum_{k=1}^n a_k}{n}\right)^n, [/math]
with equality iff [math] a_1=\dots=a_n [/math], so that the largest value attained is in fact
[math] \left(\frac{1000}{n}\right)^n [/math]
>>7658413
Also I've clearly made a mistake because the real answer is a < 3/4
>>7658394
I would say that the point for 2 is to realize that you have
x(x+1)(x+2)(x+3)=379^2-1=(379-1)(379+1)=378·380,
and so now you could try to factor into primes both numbers on the RHS, and notice that the restriction on x is immense because you are asking for divisibility and that the divisors appear 'in a row' (i.e. n, n+1, n+2, n+3), so that should solve the problem quickly
>>7658413
Actually these don't add to a thousand, but the principle is correct that something like 198×199×200×201×202 is gonna have a greater product.
For problem 5 put x-3=y, then expand the quadratics so you get cancellations on the odd terms and you get
(y+2)^4+(y-2)^4+14=0,
y^4+24y^2+23=0 (I might have made mistakes here)
then y^2=z, and you can solve this easily.
>>7658428
i think he meant multiplying 2*2*3*3*3*3*3....*3
where you have two 2s and three hundred and thirty two 3s in the equation
14 is also very easy: 1990^3=(1000+990)^3, so expanding with the binomial formula and cancelling gives you 3.
>>7658435
I misread the problem I assumed it was asking for consecutive integers
2+2=5
For problem 11, let a_0,a_0+d,a_0+2d be the solutions to the equation. Then use cardano-vieta formulas to get the coefficents; for example you know that you need
3(a_0+d)=48/32
a_0/2(d+1/2)=21/32
From this system you can solve for a_0, d and get the original solutions.
>>7658404
a) product-wise
b) use all the 3's you can, you can use 333 before reaching 999 where you can't use any more 3's. then we can greedily replace at most one 3, because replacing 2 is bad (9 > 8). so we try to replace one 3 for two 2's and get a factor of 4 instead of the factor of 3 we had
For problem 13 notice that
(111,111,111,111)(1,000,000,000,005)+1=(10^12-1)(10^12+5)/9+1=((10^12+2)/3)^2,
so the final answer is (10^12+2)/3
>>7658434
Thanks for the help. I initially tried the naive method of just subbing x with a + bi and expanding the whole thing, you get a quadratic with b^2 terms, with a determinant of -7((a-1)^4 + (a-5)^4) +2((a-1)(a-5))^2 - 112
Now this determinant had to be positive as b was real, but I sure as hell wasn't going to solve for a at this point.
is #11 really as stupid as listing out all roots with rational root thm after factoring away a 2?
>>7658458
Yea I think so. There's probably and easy way to factor it tho.
For problem 8, apply difference of squares twice and after a quick computation you get 18 if I didn´t make any algebra mistake
>>7658416
That's odd. If you solve the quadratic, you get x = a±√(a^2 +6a).
The range you provide for a is basically the range where the stationary points of the function are *not* real.
I don't get it, I guess.
>>7658423
(2)(3)(3)(3)(7)*(2)(5)(2)(19)
We have 9 numbers, and we need to turn this into 4 consecutive numbers. Notice that 3*7 is 21, which is already pretty close to 19. So lets find the next easiest 5*2*2=20, and the remaining is 3*3*1=18. so x=18.
Fuck, I tried 3 different questions and they all kicked my ass. I'll try a few more in the morning; in the meantime, does anyone have a few tips on how to solve 17? I can't get my head round it (though I'm sure it's incredibly simple).
>>7658412
>so it has one real root if and only if it is monotonic
x^3 + 2x^2 + 1 has only one real root and it is not monotonic.
>Get your hands dirty: plug in some numbers
>Try some special cases: guess and check
>#3: plug 2002/2003 into high precision online calculator.
>Realize x would be the negative of this value after learning what the floor function does
>multiply by 10^22 to and simplify both to irreducible form
>about to post answer
>realize "highly precise" calculator only goes to 22 digits
>look up million digit calculator
>find a monster that never ends
all of my nope
For problem 9 you want to find the biggest number such that (x2 + 2) - x2 is smaller or equal than 2008 and then you want that (x2 + 2) is the last term of the sequence. This gets you that a = 10042 - 2008 should be the answer.
Problem 15 ask for the probability of picking a connected graph with 5 nodes. The answer of this problem follows from some horrible computations. It is easy to see that the number of all graphs in 5 nodes is 2^(10). Then you have to get the number of connected graphs in 5 nodes. I did this in a recursive way: You calculate the number of connected graphs in 1,2,3 and 4 nodes. The first 3 are easy to calculate, however some works must be done to calculate the number of connected graphs in 4 nodes. Using the same counting technique you get that the probability for picking a connected graph in a set of 5 nodes is 1 - (286/(2^10)). Maybe you can do it easier if you calculate the generating functions for connected graphs.
>>7658603
>a = 1004^2 - 2008
Let's not let this die
10.
[math] 0 < x < 1 \Longrightarrow \forall n \in \mathbb{N} : x \geq x^n [/math]
So,
[math] f(x) = (1+x)(1+x^4)(1+x^{16})(1+x^{64})(1+x^{256}) \dots \leq 2(1+x) [/math]
Is this on the right tracks so far?
1)
n=500
Sequence: 2,2,2,2,2...
Product is 2^500.
What is the time limit on ARML problems?
Also calculators are?...
20.
Distance from rabbit hole converges at
[eqn]\frac{2\sqrt{3}}{3}[/eqn]
don't ask
Problem 4.
Abusing notation, notice that [/math] 111111111 \dots ^2 = 12345678901234567890123\dots [\math]. Then
[/math] 11111.1111\dots^2 = 123456789.01234\dots [\math]. 11111 is clearly the integer part, and the closest fractional part is [\math] \frac{1}{9} [\math], which one can confirm by checking the cases [\math] \frac{1}{8} [\math] and [\math] \frac{1}{10} [\math].
>>7658381
>minimal technical knowledge
Why do they lie to you?
>>7661428
n=367, sequence is 266 3's and 101 2's gives greater product that your 2^500.
You can easily solve the problem for real n and a's (after you realize that with reals all a's have to be equal) by finding the extreme of the product. The integer solution is probably close to the real one.
>>7658381
3) x=-2002/2003
since floor(-2002/2003)=-1
14) Since (a+b)^3=a^3+b^3+3ab(a+b), we have that ((a+b)^3-a^3-b^3)/((a+b)ab)=3. In particular this is true for the given expression.
>>7661719
but how do you show it's the least solution?
>>7661725
i thought an exact solution is the least solution or am i dumb?
19.
Height = 8
One way of doing it, if it helps anyone
5) Let's search a root in the form x=3+ih. The equation becomes
(ih-2)^4+(ih+2)^4 = -14
h^4 + 3h^2 + 9 = 0
-> h^2 = -3/2 +- i*sqrt(3)*3/2 = 3(-1/2+-i*sqrt(3)/2) = 9/2*exp(i*s), with s=2pi/3 or s=4pi/3
-> x1 = 3 + 3sqrt(2)/2*exp(i*pi/3)
-> x2 = 3 + 3sqrt(2)/2*exp(i*4pi/3)
-> x3 = 3 + 3sqrt(2)/2*exp(i*2pi/3)
-> x4 = 3 + 3sqrt(2)/2*exp(i*5pi/3)
>>7661735
wouldn't a larger negative number be a better "least solution"?
>>7661800
You can just multiply the solution x=-2002/2003 by any positive integer smaller than 2003 and you will get same result. Making me think the least solution is then -2002*2002/2003.
>>7658379
>>7658381
fun post OP
since no one seems to have done #6: we know that it is a repeating decimal so we can simply work backwards.
The last digit is 7 and 7*97+1=680 so the remainder of the previous step was 68.
The second to last digit is 6 so 6*97+68=650 so the remainder of the previous step was 65.
The digit we are looking for is A and we know that A*97+65 must be divisible by 10. The only digit that satisfies this condition is 5.
A=5
>>7661699
>with reals all a's have to be equal
Why is this?
>>7661699
n=334
a(1 to 333) = 3
a(334) = 1
My solution seems to be 333 3's. and 1 1 (pic related, assuming all the a's have to be the same).
Yours is 266 3's and 101 2's.
Instead of multiplying by 101 2's I multiplied by 67 3's.
3^67 > 2^101 if you go to see though (I'd be dumbfounded without a calculator here).
Anyone have any tips on trying to understand when a^b >=< c^d? It's something I'm generally curious about and is applicable here.
>>7662474
Pic din't upload myb
>>7662475
>>7662474
>>7662474
>3^67 > 2^101 if you go to see though (I'd be dumbfounded without a calculator here).
Multiply both exponents by (1/10), then compare.
>>7658402
>>tfw too stupid to do this
These problems are meant for people attending HS, how does that make you feel?
>>7662019
>7*97+1=680
Why plus one here?
>>7662506
Nvm got it, we know the remainder has to be the same as the first digit you divided for it to recur, I guess.
>>7662479
>integers
also that is like the staple property of e
>>7662489
Confused.
My high school education was nowhere near as hard as this, I didn't even get to calculus until A level
>inb4 British "education"
20. Represent each leg of the rabbit's path as a complex number and add them all up.
>>7658381
Another easy one is #14:
Let a= 1000, b=990
\frac{(a+b)^3-a^3-b^3}{(a+b)*a*b}
reduces to
\frac{3a^2b+3b^2a}{(a+b)*a*b}
which equals 3
>>7664096
hmm, I can not into LaTeX
>>7664098
You have to do [math] tags.
Either way it seems to be somewhat broken, only working half of the times.
>>7658381
6) For this we can observe the behavior of small repetends.
1/3 = 0.3bar => 3 x 3 = 9
1/7 = 0.142857 bar => 142857 x 7 = 999999
we can deduce from this that multiplying the repetend of 1/97 by 97 gives us 10^97 - 1 (a string of 96 9's.
We can do a pseudo-multiplication with the last three digits of the repetend and 97.
the third to the last digit comes from 7A +4 which should end in 9.
Therefore, A = 5.
I have checked the actual answer and the repetend does indeed end with ...567.
Here's a brief explanation of a solution to 10: by expanding the product, (I hope I get the LaTeX right) [eqn] f(x)f(x^2)=1+x^2+x^3+\dots =\frac{1}{1-x}[\eqn]. Plugging in [math]x=\frac 3 8 [/math] rearranges to 9/64 being the sought answer.
I think #1 is (3^332) * 4. So, n = 333, and our sequence is 332 3's and then one four at the end. Notice that this is larger than (3^333) * 1.
>>7664137
codemonkey try to fix the jsMath crippleware,
codemonkey make it worse
https://www.youtube.com/watch?v=qYodWEKCuGg
>>7663440
>also that is like the staple property of e
Elaborate please?
A staple property of e is what?
>>7663440
Yeap, it just got us the e value, then you take the closest integer.
>>7664306
Or two 2's instead of the 4.
>>7664363
given a finite positive integer, if you were to express it as a sum of reals and make a product of them, e gives you the biggest number, or say multiplying e approaches infinity the fastest if that finite integer goes big
read it on wiki or something, also that's the reason why you see random e's pop up when you're calculating interest rates in highschool for fastest growth
>>7658379
Friday night fun is going out and getting pissed and then having a one night stand with some slut. I don't see any sluts or beer in here, you neckbeard.
>>7658381
15) a+1 is a root of the function (checking the constant term by the Rational Root test and using synthetic division.):
f(x) = (x-a-1)(x^2+x-a+1)
For the quadratic factor to have no real roots, we find values of a for which the discriminant is less than zero:
1-4a-4<0
a < 3/4.
>>7664890
Pic related
>>7658381
19) Let's assume ABCD is a right trapezoid where AD is perpendicular to DC. If the height is equal to the length of AB, any point can be chosen on BD such that E,F and G can be constructed similarly with D coinciding with F. In this case, EFGP is always going to be square.
I'm running into a problem applying this to the general case (by translation of AB maybe?).
>>7663527
They're olympiad problems. Not anything that a typical high-schooler would see.
>>7658379
Problem 8 can also be solved by simple substitution. Let:
x=19991998000
a=1994+3
b=1994
c=1994-3
Our equation is now (x+a)^2-2(x+b)^2+(x+c)^2. Expanding and canceling gives: 2x(a-2b+c)+a^2-2b^2+c^2.
Looking at a-2b+c we have (1994+3)-2*(1994)+(1994-3)=0
Now expanding the a^2-2b^2+c^2 we have 1994^2-2*3*1994+3^2-2*1994^2+1994^2-2*3*1994+3^2=3^2+3^2=18.
Answer is 18.
>>7658381
9) I got a = 110890.
Let's define a number n so that a is the smallest number larger than n^2 (i.e., a-1 = n^2). the next three squares will be spaced at 2n+1, 2n+3 and 2n+5. We then set the sum of these values to be smaller than 2008 and find the largest integer n that will do.
Doing this we find that n<333.5 so the largest n would be 333. Therefore, a = 333^2+1.
Is question 16 easy or am I missing something?
All the computers can communicate with each other if they are all linked to each other, for this to happen there needs to be [math] 5 \choose 2 = 10[\math] links. Links are only created when the coin flip is heads so
[eqn] P(AllConnected) = 2^{-10}[\eqn]
>>7666470
I changed my mind, I think the answers 5 / 16
live pls
