Diophantine Equation - Ratio of Fruits

According to the question, 27x-21=8y, where both x & y are positive integers. So, y=(3/8)(9x-7) For y to be integer, (9x-7) must be divisible by 8. When x=7, (9x-7) is divisible by 8, So y=21 Before eating, A=49, O=77, P=63 After eating, A=42, O=63, P=63 So 7 apples, 14 oranges & 0 pears have been eaten

Since we won't be dealing in fractions here, the total number of fruits to begin with (B) must be a multiple of 7+11+9 = 27 and after Timmy pigs out, a multiple of 8 (A). That is because 7, 11, and 9 have no common factors, nor do 2 and 3. Let B = 27 b and A = 8 a. 27b - 21 = 8a We need to solve this Diophantine Equation in integers. b = (8a + 21)/27 27 = 3 × 3 × 3, so 8a + 21 must be divisible by 3, 21 is a multiple of 3, so 8a must also be a multiple of 3. Therefore a must be a multiple of 3. Trying various multiples of 3, we find that the lowest value of a that makes this come out in integers is 21. (There are better ways of doing this, but in this case the slow method works because the numbers are small enough.) a = 21, means there were 168 fruits left (8a): 42 apples, 63 oranges, 63 pears (2a, 3a, 3a) adding the 21 fruits Timmy ate, makes 189 fruits 189/27 = 7, so b = 7 49 apples, 77 oranges, 63 pears. (7b, 11b, 9b) He ate 7 apples, 14 oranges, and 0 pears.

I noticed that all ratios were in their simplest form so all the fruit at that moment has to be a multiple of those ratios. (eg. There could've been 7 apples, 11 oranges and 9 pears; 14 apples, 22 oranges and 18 pears; ...) We can let these ratios be represented as vectors $p$ and $q$ where : $p = \left(\begin{array}{c} 7 \\ 11 \\ 9 \end{array}\right),\: q = \left(\begin{array}{c} 2 \\ 3 \\ 3 \end{array}\right)$ They represent the ratio of fruit before and after the eating respectively. We will also say that vector $r$ is how much fruit was ate. The amount of fruit altogether can be given as $|ap|$ and $|bq|$ where $a$ and $b$ are integer constants. As $|p| = 27$ , $|q| = 8$ and the number of fruit ate was $|r|$ , this gives us the Diophantine equation $27a - 8b = 21$

Solving this Diophantine can be simple by taking both sides to the mod 27 giving $-8b \equiv 21 \mod 27 \\ \implies b \equiv 21 \mod 27$

Note that $-8b \equiv b \mod 27$ is entirely coincidental.

Let $b=21$ . This will give us $a = 7$ . A quick check will show that $7\times 27 - 21 \times 8 = 21$ . Hence $7.\left(\begin{array}{c} 7 \\ 11 \\ 9 \end{array}\right) - 21.\left(\begin{array}{c} 2 \\ 3 \\ 3 \end{array}\right) \\ = \left( \begin{array}{c} 7 \\ 14 \\ 0\end{array}\right)$

Giving us our final answer of 7 apples, 14 oranges and 0 pears.