Let $a$ and $b$ be two-digit positive integers. Find the greatest possible value of $a+b$, given that the greatest common factor of $a$ and $b$ is $6$. [i]Proposed by Jacob Xu[/i] [hide=Solution][i]Solution[/i]. $\boxed{186}$ We can write our two numbers as $6x$ and $6y$. Notice that $x$ and $y$ must be relatively prime. Since $6x$ and $6y$ are two digit numbers, we just need to check values of $x$ and $y$ from $2$ through $16$ such that $x$ and $y$ are relatively prime. We maximize the sum when $x = 15$ and $y = 16$, since consecutive numbers are always relatively prime. So the sum is $6 \cdot (15+16) = \boxed{186}$.[/hide]

Let $a$ and $b$ be two-digit positive integers. Find the greatest possible value of $a+b$, given that the greatest common factor of $a$ and $b$ is $6$. [i]Proposed by Jacob Xu[/i] [hide=Solution][i]Solution[/i]. $\boxed{186}$ We can write our two numbers as $6x$ and $6y$. Notice that $x$ and $y$ must be relatively prime. Since $6x$ and $6y$ are two digit numbers, we just need to check values of $x$ and $y$ from $2$ through $16$ such that $x$ and $y$ are relatively prime. We maximize the sum when $x = 15$ and $y = 16$, since consecutive numbers are always relatively prime. So the sum is $6 \cdot (15+16) = \boxed{186}$.[/hide]

The $n$-factorial of a positive integer $x$ is the product of all positive integers less than or equal to $z$ that are congruent to $z$ modulo $n$. For example, for the number 16, its 2-factorial is $16 \times 14 \times 12 \times 10 \times 8 \times 6 \times 4 \times 2$, its 3-factorial is $16 \times 13 \times 10 \times 7 \times 4 \times 1$ and its 18-factorial is 16. A positive integer is called [i]olympic[/i] if it has $n$ digits, all different than zero, and if it is equal to the sum of the $n$-factorials of its digits. Find all positive olympic integers.

Find all natural numbers $m$ and $n$ such that $3^m+n!-1$ is the square of a natural number.

For positive integers $a$ and $b$, define \[a!_b=\prod_{1\le i\le a\atop i \equiv a \mod b} i\] Let $p$ be a prime and $n>3$ a positive integer. Show that there exist at least 2 different positive integers $t$ such that $1<t<p^n$ and $t!_p\equiv 1\pmod {p^n}$.

For a positive integer $n>1$, let $g(n)$ denote the largest positive proper divisor of $n$ and $f(n)=n-g(n)$. For example, $g(10)=5, f(10)=5$ and $g(13)=1,f(13)=12$. Let $N$ be the smallest positive integer such that $f(f(f(N)))=97$. Find the largest integer not exceeding $\sqrt{N}$

A [i]Beaver-number[/i] is a positive 5 digit integer whose digit sum is divisible by 17. Call a pair of [i]Beaver-numbers[/i] differing by exactly $1$ a [i]Beaver-pair[/i]. The smaller number in a [i]Beaver-pair[/i] is called an [i]MIT Beaver[/i], while the larger number is called a [i]CIT Beaver[/i]. Find the positive difference between the largest and smallest [i]CIT Beavers[/i] (over all [i]Beaver-pairs[/i]).

Let $m$ be the smallest positive integer such that $m^2+(m+1)^2+\cdots+(m+10)^2$ is the square of a positive integer $n$. Find $m+n$

Find the number of ordered pairs $(a,b)$ such that $a,b \in \{10,11,\cdots,29,30\}$ and \\ $\hspace{1cm}$ $GCD(a,b)+LCM(a,b)=a+b$.

Let $p$ be a prime and suppose $2^{2p} \equiv 1 (\text{mod}$ $ 2p+1)$ is prime. Prove that $2p+1$ is prime$^{1}$ [size=75]$^{1}$This is a special case of Pocklington's theorem. A proof of this special case is required.[/size]

A four-digit number $n$ is said to be [i]literally 1434[/i] if, when every digit is replaced by its remainder when divided by $5$, the result is $1434$. For example, $1984$ is [i]literally 1434[/i] because $1$ mod $5$ is $1$, $9$ mod $5$ is $4$, $8$ mod $5$ is $3$, and $4$ mod $5$ is $4$. Find the sum of all four-digit positive integers that are [i]literally 1434[/i]. [i]Proposed by Evin Liang[/i] [hide=Solution] [i]Solution.[/i] $\boxed{67384}$ The possible numbers are $\overline{abcd}$ where $a$ is $1$ or $6$, $b$ is $4$ or $9$, $c$ is $3$ or $8$, and $d$ is $4$ or $9$. There are $16$ such numbers and the average is $\dfrac{8423}{2}$, so the total in this case is $\boxed{67384}$. [/hide]

How many multiples of $17$ are there between $23$ and $227$?

Find all triples of positive integers $(a,b,c)$ such that: $a^2bc-2ab^2c-2abc^2+b^3c+bc^3+2b^2c^2=11$

A four-digit number $n$ is said to be [i]literally 1434[/i] if, when every digit is replaced by its remainder when divided by $5$, the result is $1434$. For example, $1984$ is [i]literally 1434[/i] because $1$ mod $5$ is $1$, $9$ mod $5$ is $4$, $8$ mod $5$ is $3$, and $4$ mod $5$ is $4$. Find the sum of all four-digit positive integers that are [i]literally 1434[/i]. [i]Proposed by Evin Liang[/i] [hide=Solution] [i]Solution.[/i] $\boxed{67384}$ The possible numbers are $\overline{abcd}$ where $a$ is $1$ or $6$, $b$ is $4$ or $9$, $c$ is $3$ or $8$, and $d$ is $4$ or $9$. There are $16$ such numbers and the average is $\dfrac{8423}{2}$, so the total in this case is $\boxed{67384}$. [/hide]

Let $S$ be the sum of all positive integers $n$ such that $\frac{3}{5}$ of the positive divisors of $n$ are multiples of $6$ and $n$ has no prime divisors greater than $3$. Compute $\frac{S}{36}$.

Suppose $a,b,c$ are positive integers such that \[\gcd(a,b)+\gcd(a,c)+\gcd(b,c)=b+c+2023\] Prove that $\gcd(b,c)=2023$. [i]Remark.[/i] For positive integers $x$ and $y$, $\gcd(x,y)$ denotes their greatest common divisor. [i]Ivan Novak[/i]

For a positive integer $m$, let $\varphi(m)$ be the number of positive integers $k \le m$ such that $k$ and $m$ are relatively prime, and let $\sigma(m)$ be the sum of the positive divisors of $m$. Find the sum of all even positive integers $n$ such that \[ \frac{n^5\sigma(n) - 2}{\varphi(n)} \] is an integer.

A triangle is called a parabolic triangle if its vertices lie on a parabola $y = x^2$. Prove that for every nonnegative integer $n$, there is an odd number $m$ and a parabolic triangle with vertices at three distinct points with integer coordinates with area $(2^nm)^2$.