ka May Highlights and 2025 AoPS Online Class Information
jlacosta0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.
Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.
Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!
Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
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Hey! People are starting to get their swag packages from Jane Street for qualifying for USA(J)MO, and after some initial discussion on what we got, people are getting different things. Out of curiosity, I was wondering how they decide who gets what.
Please enter the following info:
- USAMO or USAJMO
- Grade
- Score
- Award/Medal/HM
- MOP (yes or no, if yes then color)
- List of items you got in your package
UNR -> Nevada
St Anselm -> New Hampshire
PSU -> Pennsylvania
WCU -> North Carolina
Put your USERNAME in the list ONLY IF YOU WANT TO!!!!!!!!!
last time i made this poll it got deleted by aops admins cuz they were malding about putting usernames in text boxes and saying the username is their real name or somthing but obviously your username is not ur real name bruh whats wrong with those people
q(x) to be the product of all primes less than p(x)
orl19
Nan hour ago
by ihategeo_1969
Source: IMO Shortlist 1995, S3
For an integer , let be the least prime that does not divide , and define to be the product of all primes less than . In particular, For having , define . Consider the sequence defined by and for . Find all such that .
Let be a cyclic quadrilateral. Let be the midpoint of the arc of its circumcircle which does not contain or . Let the lines and meet at and the lines and meet at . Prove that the lines and are parallel.
Let be an acute triangle with circumcenter and orthocenter . Let denote the circumcircle of triangle , and the midpoint of . The tangents to at and , and the line through perpendicular to line , determine a triangle whose circumcircle we denote by . Define and similarly.
Prove that the common chords of , and are concurrent on line .
Suppose we have distinct positive integers , and an odd prime not dividing any of them, and an integer such that if one considers the infinite sequence and looks at the highest power of that divides each of them, these powers are not all zero, and are all at most . Prove that there exists some (which may depend on ) such that whenever divides an element of this sequence, the maximum power of that divides that element is exactly .
Let ,, be an infinite sequence of positive integers. Suppose that there is an integer such that, for each , the number is an integer. Prove that there is a positive integer such that for all .
Let be a right triangle with . Point lies on the line such that is between and . Let be the midpoint of and let be the seconf intersection point of the circumcircle of and the circumcircle of . Prove that as varies, the line passes through a fixed point.
Converting from base to base 10, we get for a postive integer because its a divisor. Then you can simplify this to and since n or b cant be negative that means n has to be from 1 to 9 exclusive. Then casework from 2-8, and u get n=7 then b=21 and n=8 then b=40 so its 70
First, note that so it follows that must be a divisor of The only divisors of greater than are and so it follows that the possible values of are and yielding an answer of
9b+7 is divisible by b+7, so 9b+7 - 9(b+7) will still be divisible by b+7, so -56 is divisible by b+7. we first try -56-7=-49, so one b possibility could be 49. 49-7=42, 42/2 = 21, 21 is another possibility. 21-7=14, 14/2 = 7, which is less than 9, so we only have 21 and 49, giving us 70.
By the definition of a divisor, for an integer . Clearly does not work. So now we can just try every single value.
no gives which doesn't work gives which doesn't work gives which doesn't work gives which doesn't work gives which doesn't work gives which does work gives which does work
By the definition of a divisor, for an integer . Clearly does not work. So now we can just try every single value.
no gives which doesn't work gives which doesn't work gives which doesn't work gives which doesn't work gives which doesn't work gives which does work gives which does work
I was stuck on this problem for some reason I don't know why
The best way to tackle this problem is to convert everything to variables so first 17 base b = b + 7 and 97 base b = 9b+7 so we can just make a variable when multiplied by it it equals 97 base b
So first to do that we can simplify it like 9b+7 = x(b+7) for some value of x then when we multiply that out we get 9b+7=bx+7x. Because we want to solve for xb we have to subtract that on both sides to get 9b+7-bx= 7x then in any situation like this we have to factor out the b, but first we can subtract 7 on both sides.
When we do that we get b(9-x)= 7(x-1) which when we divide both sides we get
b= (7(x-1))/(9-x) we get this easy equation to solve because we know that x has to be a single digit number because anything greater than 9 won't work so we get that after guess and check x= 7 and 8 so when we plug that in we get 21 and 49, so when we add those together you get
This post has been edited 2 times. Last edited by sadas123, Feb 9, 2025, 4:44 PM
b+7 is divisible by 9b+7, but because b+7 is divisible by 9b+63, then b+7 is divisible by 56. And B+7 is greater than 7, so b=21 or 49[Click][sounds stupid but I got b+7 is divisible by 8b, and did a lot of stupid stuff to get -6+-5+-3+0+1+7+21+49=64 cause I forgot b is greater than 9, but luckily I realized this at the end.]
Find the sum of all integer bases for which is a divisor of
We need , so . Let , do casework and bounding.
If then so , doesn't work.
If then so , doesn't work.
If then so , which does work.
If then so , which does work.
If then so , doesn't work.
Our answer is .
(97)b =9b+7
(17)b=b+7
So b+7 divides 9b+7
gcd(9b+7,b+7)=b+7
gcd(-56,b+7)=b+7
So we get b+7 divides -56
Now , 56=2*2*2*7 and factors greater than 9+7 are 28 and 56 itself
So, b=21 or 49
Sum =70(answer)