Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it Romania

The non-negative integers $a$ , $b$ , $c$ are such that the numbers
$m = \frac{5a + 6b + 7c + 6}{4a + 3b + 2c + 3} \quad \text{and} \quad n = \frac{a + 2b + 3c + 5}{3a + b + 2c + 5}$
are both integers.

a. Prove that $m \ge 2$ .
b. Find $m$ and $n$ .

Solution

a. If $m \le 1$ , then $5a + 6b + 7c + 6 \le 4a + 3b + 2c + 3$ , that is $a + 3b + 5c + 3 \le 0$ – impossible. So $m \ge 2$ .

b. If $n \ge 2$ , then $a + 2b + 3c + 5 \ge 6a + 2b + 4c + 10$ , whence $0 \ge 5a + c + 5$ , false. Since $n > 0$ , $n$ must be equal to 1.
Now $n = 1$ yields $a + 2b + 3c + 5 = 3a + b + 2c + 5$ , hence $b + c = 2a$ .
Suppose $m \ge 3$ . Then $5a+6b+7c+6 \ge 12a+9b+6c+9$ , therefore $c \ge 7a+3b+3$ . This gives $b+c \ge 7a+4b+3$ , that is $2a \ge 7a+4b+3$ , hence $0 \ge 5a+4b+3$ , false. So $m < 3$ , and a) implies $m = 2$ .
The above show that the only possibility is $m = 2$ and $n = 1$ . These values are indeed achieved for $a = b = c$ .

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