Track / Stage 3 / 224 of 260 #224 of 1964

Problem 224

AMC 10/12, early questions

Number theory Difficulty 3.8 Find the answer South African Mathematics Olympiad · South Africa

How many pairs of non-negative integers $x$ and $y$ are solutions of $\frac{x}{20} + \frac{y}{15} = 1$ ?

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Official solution

We can re-write $\frac{x}{20} + \frac{y}{15} = 1$ in the form $3x + 4y = 60$ . Then we can see that $3x$ must be divisible by $4$ , so $x$ must be, and trying successive possible values we see that only the following combinations of $x$ - and $y$ -values are acceptable: $(0; 15)$ , $(4; 12)$ , $(8; 9)$ , $(12; 6)$ , $(16; 3)$ , $(20; 0)$ .

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