(tracking progress 1) grinding at 12 am

by flec, Nov 5, 2023, 7:19 AM

(just solving problems) i have very limited knowledge on the 500 formulas so i literally use guess + check, logic, creativity, and funny weird stuff that may or may not work for the problems.
this post is basically just to track my progress
to prove that i have been doing crap
besides i think organizing something with latex is really aesthetically pleasing and easier for me to review later
problem 1 - 2022 amc 10b #13
question:
The positive difference between a pair of primes is equal to $2$, and the positive difference between the cubes of the two primes is $31106$. What is the sum of the digits of the least prime that is greater than those two primes?
my solution:
first, i made a system of equations pretending that x=larger prime and y=smaller prime
$x-y=2$
$x^3-y^3=31106$
then, i decided that it would be possible to substitute the x-y equation into the other one, so i tried this:
first factorizing $x^3-y^3$
$(x-y)(x^2+xy+y^2)=31106$
since x-y=2, you get plug that into the equation:
$2(x^2+xy+y^2)=31106$ -> $x^2+xy+y^2=15553$
then, pretending that one prime is x and the other is x+2 (since it says in the problem that the positive difference between the primes is 2), you can substitute into the equation again
$x^2+x(x+2)+(x+2)^2=15553$ -> $x^2+x^2+x^2+2x+4x+4=15553$ -> $3x^2+6x+4=15553$ -> $3x^2+6x=15549$
[divide everything by 3]
$3(x^2+2x)=5183$
[complete the square]
$x^2+2x+1=5183+1$ -> $(x+1)^2=5184$ and 5184 nicely turns out to be a perfect square
[square root both sides]
$|x+1|=72$ which gets you $x=-73$ or $71$ but since its the difference they are both positive so $73$ and $71$ (correct me on my logic flow here)
the next prime number is $79$ so you do $7+9=16$ and you get E (16) as your answer
overall: i'm sure there is a much simpler way to solve this, but then again i doubt my knowledge could support that. i'm using what i already know to just plug it in and although that is more time-consuming i doubt i can learn + comprehend new formulas 3 days before the amc.
problem 2 - 2017 amc 10a #20
question: Let $S(n)$ equal the sum of the digits of positive integer $n$. For example, $S(1507) = 13$. For a particular positive integer $n$, $S(n) = 1274$. Which of the following could be the value of $S(n+1)$?
my solution:
i decided to use kind of the process of elimination for this one
A - for the digits to add up to one $n$ must be 99999... and $1274$ is not a multiple of $9$ so we know this isn't correct
B - an example of a number that would work would be 20999 with the nine repeating because once you add one every nine cancels out and leaves one $1$ in the beginning so $2+1=3$. $(1274-2)/9$ isn't divisible by $9$ so we also know this is wrong
C - example of this would be like $11111111111$ (eleven ones) and then the appropriate amount of 9's in the end. $1274-12=1262$ and this is not divisible by $9$ so it doesn't work
D - example is $1238$ ones and then $09999$ (once you add the one, the nines will cancel out and leave $10000$, and the digit addition difference is 35). $1238+1=1239$ (add one from the nines leaving the one once the is added) so you obtain D. so that's the correct answer
overall: i feel like this is a really wonky way to do it and that there's some super easy way to do it involving modular. i know some mod but i'm not comfy enough to implement it into these scenarios. again, i'm finding ridiculous ways to solve these problems that end up working but take some time. not so good for a fast-paced exam like the amc 10, but i guess it helps me use my brain and have some fun.
questions i'm having trouble on: (i read the solutions but still dont understand so i need someone to dumb down for me)
2011 10b #16
2017 10a #19
2015 10b #17
anyways this is for my practice and for smart people to judge me and be like what the hell is she doing why is she making it 500 times more complicated
the answer to that is i am simply not smart enough to know that my methods are really time consuming, but smart enough to know that the methods are kind of dumb
but if people could help me on the problems i'm stuck on that would be great
i heard c&p and especially geo are really important and unfortunately i suck butt at both of them
This post has been edited 6 times. Last edited by flec, Nov 5, 2023, 7:23 AM

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Quote:
first, i made a system of equations pretending that x=larger prime and y=smaller prime
[...]
then, pretending that one prime is x and the other is x+2 (since it says in the problem that the positive difference between the primes is 2), you can substitute into the equation again

Initially, you assumed that $x$ was the larger prime, but then you make $x$ the smaller prime (as $x$ is the smaller number out of $x$ and $x+2$. Basically you're redefining what $x$ means in the middle of solving the problem, which could be problematic.
Quote:
$|x+1|=72$ which gets you $x=-73$ or $71$ but since its the difference they are both positive so $73$ and $71$ (correct me on my logic flow here)

Maybe I'm being dumb but I don't really get what you're trying to say here

Despite these minor issues, your overall idea is correct. Your way of solving the problem is also quite reasonable. I looked at the contest thread in C&P and this is the method most people used anyway so dw you're not overcomplicating it

by mathlearner2357, Nov 5, 2023, 9:32 AM

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AIME qual is not that hard

iv never qualed but i get 96-103.5 on practice amc 10s and I almost qualed last year

Just answer most of the questions 1-15 and stop at around question 20, skipping the ones between 15 and 20 you dont know. The ones before 15 are mostly answerable with basic intro to number theory/intro to geometry/c and p, basic AoPS stuff. If you know calculus, good for you it will help with optimization and minima and maxima problems but there is always a way to do it without calculus at least before question 15.

Basically after you reach around question 15-16 go to around 20, answer only the questions you know, stare at questions 21-25 and die inside because theres absolutely know way you can solve them even if you know calculus, go back and double check questions 1-15ish as well as any questions 15-20 that you answered again and again until you are 100 percent sure that you have gotten 0 wrong and the time runs out. It might be boring but it works.

by tacowizard, Nov 5, 2023, 2:10 PM

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#16 is basically finding the area of the center region divided by the area of the outside
You can basically let the side length of the regular octagon be 1
then use 45-45-90 triangle facts to get all the rest of the lengths inside the octagon.

#19 is like you think about what happens with Alice is at the end seat, and what happens when not

#17 what's the volume of a pyramid? an octahedron is two pyramids

by awesomeming327., Nov 5, 2023, 4:42 PM

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thank you everyone! also @tacowizard i think aime is somewhat easier for people with experience and who took like aops courses and have experience with competition math but i have never done competition math except for like mathcounts/kangaroo/amc 8 once... also my base is kind of iffy and i silly a LOT

@awesomeming thank you! here is what i did with your advice
#16 - area of each triangle is $0.25$ and the area of each rectangle is $sqrt(0.5)*4=sqrt(2)/2$. the area of the middle square is $1$.
since there are four triangles and four rectangles, just multiply each area by four and add it to the area of the square.
probability = area of the square / area of everything
so
$1/(sqrt(2)/2*4+2)=1/(2sqrt(2)+2)$ (rationalize) $(2sqrt(2)-2)/4=(sqrt(2)-1)/2$ A

woopsies i dont know latex for sqrt so sorry guys
This post has been edited 1 time. Last edited by flec, Nov 5, 2023, 7:22 PM

by flec, Nov 5, 2023, 7:21 PM

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It's $\sqrt$

by awesomeming327., Nov 5, 2023, 8:33 PM

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thanks i suck at latex lol

and bro kms i couldnt solve 2022 amc10a #4 conversions suck

by flec, Nov 5, 2023, 9:41 PM

scream, don't shout, for contrib

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