2. Tine was collecting stamps. For his birthday, he received a new album in which he could store many stamps. He took 2002 tolars from his savings and decided to spend all the money on buying stamps. A friend offered him smaller stamps for 10 tolars and larger ones for 28 tolars. Tine decided to buy as many stamps as possible. How many stamps can he buy?
I/2. Denimo, da bo Tine kupil $x$ znamk po 10 tolarjev in $y$ znamk po 28 tolarjev. Tedaj velja $10 x+28 y=2002$ oziroma $5 x+14 y=1001$, od tod pa $5 x+5 y=1001-9 y$ oziroma $x+y=\frac{1001-9 y}{5}$. Vrednost vsote bo tem večja, čim manjši bo $y$. Ker je $y$ naravno število, lahko poskušamo z vrednostmi $y=1, y=2, y=3 \ldots$, dokler ne dobimo cele vrednosti za vsoto $x+y$. Sicer pa vemo, da bo vsota $x+y$ celo število, ko bodo enice v razliki $1001-9 y$ enake 0 ali 5, to je, ko bodo enice v zmnožku $9 y$ enake 1 ali 6. Najmanjši $y$, ko to velja, je $y=4$. Tedaj je $x+y=\frac{1001-36}{5}=\frac{965}{5}=193$. Tine bo kupil 193 znamk.

Let's assume Tine will buy $x$ stamps at 10 tolars each and $y$ stamps at 28 tolars each. Then we have $10 x + 28 y = 2002$ or $5 x + 14 y = 1001$, from which we get $5 x + 5 y = 1001 - 9 y$ or $x + y = \frac{1001 - 9 y}{5}$. The value of the sum will be greater the smaller $y$ is. Since $y$ is a natural number, we can try values $y=1, y=2, y=3 \ldots$, until we get an integer value for the sum $x + y$. However, we know that the sum $x + y$ will be an integer when the units in the difference $1001 - 9 y$ are 0 or 5, that is, when the units in the product $9 y$ are 1 or 6. The smallest $y$ for which this is true is $y=4$. Then $x + y = \frac{1001 - 36}{5} = \frac{965}{5} = 193$. Tine will buy 193 stamps.