4. Jure wanted to prepare gifts for nine friends. He will give each friend two chocolates. In the store, he saw that a bar of dark chocolate costs 6 tolars more than a bar of milk chocolate, and that each costs an integer number of tolars. He decided to spend 822 tolars in such a way that as many friends as possible would receive two different chocolates. What was the price of the milk chocolate?

MATHEMATICS COMPETITION

April 7, 2001

## PROBLEMS FOR THE SECOND YEAR
I/4. Recimo, da je Jure kupil $k$ mlečnih čokolad in da je bila cena mlečne čokolade $c$ tolarjev. Tedaj je $k c+(18-k)(c+6)=822$, od koder dobimo $3 c=119+k$. Ker bi Jure rad dal čim več prijateljem po dve različni čokoladi, moramo poiskati tako rešitev, da bo $k$ čim bliže številu 9, hkrati pa mora biti $119+k$ deljivo s 3. Tako pridemo do $k=10$ in $c=43$. Mlečna čokolada je stala 43 tolarjev.

Vpeljava dveh neznank za ceno in količino (lešnikovih ali mlečnih) čokolad: 1 točka. Enačba $k c+(18-k)(c+6)=822$: $\mathbf{2}$ točki. Poenostavitev $3 c=119+k$: 1 točka. Sklep: “Iščemo $k$, ki bo čim bližje 9”: $\mathbf{2}$ točki. Rešitev $k=10$ in $c=43$: $\mathbf{1}$ točka.

Let's say Jure bought $k$ milk chocolates and the price of a milk chocolate was $c$ tolars. Then $k c + (18 - k)(c + 6) = 822$, from which we get $3 c = 119 + k$. Since Jure wants to give as many friends as possible two different chocolates, we need to find a solution where $k$ is as close to 9 as possible, while $119 + k$ must be divisible by 3. This leads us to $k = 10$ and $c = 43$. The milk chocolate cost 43 tolars.

Introducing two unknowns for the price and quantity (of dark or milk) chocolates: 1 point. Equation $k c + (18 - k)(c + 6) = 822$: $\mathbf{2}$ points. Simplification $3 c = 119 + k$: 1 point. Conclusion: “We are looking for $k$ that is as close to 9 as possible”: $\mathbf{2}$ points. Solution $k = 10$ and $c = 43$: $\mathbf{1}$ point.