Jugs Riddle

If $\frac{1}{12}$ liters replace 1, $\frac{2}{3}$ liter is 8, $\frac{5}{4}$ liter is 15, and 1 liter is 12.

$x=4,y=6$

Since these jugs need full transfer and the numbers of fillings or wastings are integers n & m, we can write up equation:

$\frac{2n}{3} + \frac{5m}{4} =1$

Thus, $8n + 15m =12$ .

Since m is an integer, $15m = 12-8n = 4(3-2n)$ , then $4|m$ . Otherwise, n won't be an integer.

Similarly, we'll obtain $3|n$ .

Then we can rewrite the equation for some integers s,t:

$24s + 60t = 12$

$2s+5t =1$

The least possible absolute values are $s=-2$ and $t=1$ .

That means $n =y= -6$ or 6 wastings and $m=x=4$ or 4 fillings.

Therefore, $x+y=\boxed{10}$ .

Checking answers, $(\frac{2}{3})(-6)+(\frac{5}{4})(4) = 5-4 = 1$ .

2/3 x + 5/4 y = 1 where each letter represents the number of fills or wastes for each jar.

Looking for an integer set answer of a point (x,y) so that the sum abs(x) + abs(y) = minimum.

The smallest integer pair on a coordinate grid is x = -6 and y = 4 (graph the equation from line 1).

Thus our minimum is 10.

In the first step we fill the big jug from the faucet $(\frac54l=\frac{15}{12}l)$ and pour the water into the small jug until is filled up $(\frac23l=\frac8{12}l)$ , and emptied it, and pour the rest of the water from the big jug into the small jug $(\frac7{12}l)$ ; so, $x=1, y=1$ . In the second step we fill again the big jug from the faucet $(\frac{15}{12}l)$ and fill up the small jug $(\frac1{12}l)$ , then it is emptied. In the big jug there remains $(\frac{14}{12}l)$ . A Part of this amount $(\frac8{12}l)$ is poured into the small jug, then it is emptied, and pour the water from the big jug into the small one $(\frac6{12}l)$ . Now $x=2, y=3$ . In the third step we fill again the big jug from the faucet and fill up the small jug $(\frac2{12}l)$ , then it is emptied. In the big jug there remains $(\frac{13}{12}l)$ . A Part of this amount $(\frac8{12}l)$ is poured into the small jug, then it is emptied, and pour the water from the big jug into the small one $(\frac5{12}l)$ . Now $x=3, y=5$ . In the fourth step we fill again the big jug from the faucet and fill up the small jug $(\frac3{12}l)$ , then it is emptied. In the big jug there remains $(\frac{12}{12}l=1l)$ . So $x=4, y=6$ , i.e. $x+y=10$ .

Hello, I ran a simulation in python to find this out, I think the solution is 9, if we don't bother about emptying the last full jug with 2/3 and just use the 1 from the 5/4 jug. Otherwise it's 10