The Fruity Diophantine

Let $a$ be the number of apples, $b$ the number of oranges, and $c$ the number of pears he ate. Then the problem asks for solving the following linear system with integer unknowns: $\left\{\begin{array}{l}7x-a=2y\\11x-b=3y\\9x-c=3y\\a+b+c=21\end{array}\right.$ This system has infinitely many solutions given by $x=7-\frac{8}{9}\lambda,\quad y=21-3\lambda,\quad a=7-\frac{2}{9}\lambda,\quad b=14-\frac{7}{9}\lambda,\quad c=\lambda,$ for some non-negative integer $\lambda$ . Since $x=7-\frac{8}{9}\lambda$ must be a positive integer, we deduce that $\lambda=0$ . This implies that $y=21$ , from where the number of fruits that are left is given by $2y+3y+3y=8\cdot 21=\boxed{168}$ .

The solution is unique and we don't need to use diophantine equation techniques.

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 = 3x3x3, 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. a = 21, means there were 168 fruits left (8a)