Sam Wang decides to evaluate an expression of the form $x +2 \cdot 2+ y$. However, he unfortunately reads each ’plus’ as a ’times’ and reads each ’times’ as a ’plus’. Surprisingly, he still gets the problem correct. Find $x + y$. [i]Proposed by Edwin Zhao[/i] [hide=Solution] [i]Solution.[/i] $\boxed{4}$ We have $x+2*2+y=x \cdot 2+2 \cdot y$. When simplifying, we have $x+y+4=2x+2y$, and $x+y=4$. [/hide]

Let [i]Revolution[/i]$(x) = x^3 +Ux^2 +Sx + A$, where $U$, $S$, and $A$ are all integers and $U +S + A +1 = 1773$. Given that [i]Revolution[/i] has exactly two distinct nonzero integer roots $G$ and $B$, find the minimum value of $|GB|$. [i]Proposed by Jacob Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{392}$ Notice that $U + S + A + 1$ is just [i]Revolution[/i]$(1)$ so [i]Revolution[/i]$(1) = 1773$. Since $G$ and $B$ are integer roots we write [i]Revolution[/i]$(X) = (X-G)^2(X-B)$ without loss of generality. So Revolution$(1) = (1-G)^2(1-B) = 1773$. $1773$ can be factored as $32 \cdot 197$, so to minimize $|GB|$ we set $1-G = 3$ and $1-B = 197$. We get that $G = -2$ and $B = -196$ so $|GB| = \boxed{392}$. [/hide]

Eddie has a study block that lasts $1$ hour. It takes Eddie $25$ minutes to do his homework and $5$ minutes to play a game of Clash Royale. He can’t do both at the same time. How many games can he play in this study block while still completing his homework? [i]Proposed by Edwin Zhao[/i] [hide=Solution] [i]Solution.[/i] $\boxed{7}$ Study block lasts 60 minutes, thus he has 35 minutes to play Clash Royale, during which he can play $\frac{35}{5}=\boxed{7}$ games. [/hide]

Sam dumps tea for $6$ hours at a constant rate of $60$ tea crates per hour. Eddie takes $4$ hours to dump the same amount of tea at a different constant rate. How many tea crates does Eddie dump per hour? [i]Proposed by Samuel Tsui[/i] [hide=Solution] [i]Solution.[/i] $\boxed{90}$ Sam dumps a total of $6 \cdot 60 = 360$ tea crates and since it takes Eddie $4$ hours to dump that many he dumps at a rate of $\dfrac{360}{4}= \boxed{90}$ tea crates per hour. [/hide]

Sam dumps tea for $6$ hours at a constant rate of $60$ tea crates per hour. Eddie takes $4$ hours to dump the same amount of tea at a different constant rate. How many tea crates does Eddie dump per hour? [i]Proposed by Samuel Tsui[/i] [hide=Solution] [i]Solution.[/i] $\boxed{90}$ Sam dumps a total of $6 \cdot 60 = 360$ tea crates and since it takes Eddie $4$ hours to dump that many he dumps at a rate of $\dfrac{360}{4}= \boxed{90}$ tea crates per hour. [/hide]

If $a \diamondsuit b = \vert a - b \vert \cdot \vert b - a \vert$ then find the value of $1 \diamondsuit (2 \diamondsuit (3 \diamondsuit (4 \diamondsuit 5)))$. [i]Proposed by Muztaba Syed[/i] [hide=Solution] [i]Solution.[/i] $\boxed{9}$ $a\diamondsuit b = (a-b)^2$. This gives us an answer of $\boxed{9}$. [/hide]

Sam Wang decides to evaluate an expression of the form $x +2 \cdot 2+ y$. However, he unfortunately reads each ’plus’ as a ’times’ and reads each ’times’ as a ’plus’. Surprisingly, he still gets the problem correct. Find $x + y$. [i]Proposed by Edwin Zhao[/i] [hide=Solution] [i]Solution.[/i] $\boxed{4}$ We have $x+2*2+y=x \cdot 2+2 \cdot y$. When simplifying, we have $x+y+4=2x+2y$, and $x+y=4$. [/hide]

If $a \diamondsuit b = \vert a - b \vert \cdot \vert b - a \vert$ then find the value of $1 \diamondsuit (2 \diamondsuit (3 \diamondsuit (4 \diamondsuit 5)))$. [i]Proposed by Muztaba Syed[/i] [hide=Solution] [i]Solution.[/i] $\boxed{9}$ $a\diamondsuit b = (a-b)^2$. This gives us an answer of $\boxed{9}$. [/hide]

On day $1$ of the new year, John Adams and Samuel Adams each drink one gallon of tea. For each positive integer $n$, on the $n$th day of the year, John drinks $n$ gallons of tea and Samuel drinks $n^2$ gallons of tea. After how many days does the combined tea intake of John and Samuel that year first exceed $900$ gallons? [i]Proposed by Aidan Duncan[/i] [hide=Solution] [i]Solution. [/i] $\boxed{13}$ The total amount that John and Samuel have drank by day $n$ is $$\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}=\dfrac{n(n+1)(n+2)}{3}.$$ Now, note that ourdesired number of days should be a bit below $\sqrt[3]{2700}$. Testing a few values gives $\boxed{13}$ as our answer. [/hide]

Let [i]Revolution[/i]$(x) = x^3 +Ux^2 +Sx + A$, where $U$, $S$, and $A$ are all integers and $U +S + A +1 = 1773$. Given that [i]Revolution[/i] has exactly two distinct nonzero integer roots $G$ and $B$, find the minimum value of $|GB|$. [i]Proposed by Jacob Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{392}$ Notice that $U + S + A + 1$ is just [i]Revolution[/i]$(1)$ so [i]Revolution[/i]$(1) = 1773$. Since $G$ and $B$ are integer roots we write [i]Revolution[/i]$(X) = (X-G)^2(X-B)$ without loss of generality. So Revolution$(1) = (1-G)^2(1-B) = 1773$. $1773$ can be factored as $32 \cdot 197$, so to minimize $|GB|$ we set $1-G = 3$ and $1-B = 197$. We get that $G = -2$ and $B = -196$ so $|GB| = \boxed{392}$. [/hide]

Eddie has a study block that lasts $1$ hour. It takes Eddie $25$ minutes to do his homework and $5$ minutes to play a game of Clash Royale. He can’t do both at the same time. How many games can he play in this study block while still completing his homework? [i]Proposed by Edwin Zhao[/i] [hide=Solution] [i]Solution.[/i] $\boxed{7}$ Study block lasts 60 minutes, thus he has 35 minutes to play Clash Royale, during which he can play $\frac{35}{5}=\boxed{7}$ games. [/hide]

On day $1$ of the new year, John Adams and Samuel Adams each drink one gallon of tea. For each positive integer $n$, on the $n$th day of the year, John drinks $n$ gallons of tea and Samuel drinks $n^2$ gallons of tea. After how many days does the combined tea intake of John and Samuel that year first exceed $900$ gallons? [i]Proposed by Aidan Duncan[/i] [hide=Solution] [i]Solution. [/i] $\boxed{13}$ The total amount that John and Samuel have drank by day $n$ is $$\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}=\dfrac{n(n+1)(n+2)}{3}.$$ Now, note that ourdesired number of days should be a bit below $\sqrt[3]{2700}$. Testing a few values gives $\boxed{13}$ as our answer. [/hide]