Given cups that are $\frac12$, $\frac13$, $\frac14$, $\frac15$, $\frac18$, $\frac19$, $\frac1{10}$ full, can we pour to get a cup $\frac16$ full?

Question

There are seven cups, $C_1$, $C_2$, $\ldots$, $C_7$ and they have the same capacity $V$.

Initial:

• Water of $C_1$ occupies $\frac{1}{2}V$
• Water of $C_2$ occupies $\frac{1}{3}V$
• Water of $C_3$ occupies $\frac{1}{4}V$
• Water of $C_4$ occupies $\frac{1}{5}V$
• Water of $C_5$ occupies $\frac{1}{8}V$
• Water of $C_6$ occupies $\frac{1}{9}V$
• Water of $C_7$ occupies $\frac{1}{10}V$

Allow pouring all water from one cup to another if the water does not overflow or pour water from one cup to another until it is full. Can we, after a number of times pouring water, have cup that occupies $\frac{1}{6}V$?

This is my attempt:

• We consider cup A and cup B have the amount of water, respectively, a and b where $0\le a,b\le 1$. If you pour water from cup A to cup B, the following will happen:
• If $a+b<1$. Then after pouring, cup A will be empty and cup B takes up $a+b$ cup.
• If $a+b<1$. Then after pouring, cup A will takes up $a-b$ cup and cup B is full.

I just think something here and I can't solve this problem ! Can you help me this stuck!

Answer 1

Whenever you pour one cup $a$ into cup $b$, you are either left with an empty cup and a cup with $a+b$, or a full cup and a cup with $a + b -1$. Either way, the action leaves you with a cup with $a+b \mod 1$ water in it. That means if we want to have $1/6$ left, we need to find a combination of cups that adds up to $1/6\mod 1$.

A straightforward way is to put them all on a common denominator, namely $360$. Then we have $180$, $120$, $90$, $72$, $45$, $40$, and $36$, and we need these to add up to $60$ or $420$. Clearly this can't be done, so it's not possible.

Answer 2

No, it's not possible. While it is a simple observation, I'd like to just point out that a full cup is useless. Pouring it into another cup will swap the full cup's contents to another cup (regardless of how much water was in the other cup). An empty cup is similarly useless.

We can weed out some other useless cups by their denominator. Suppose we have $n > 1$ cups, filled with $\frac{a_i}{b_i} V$ volume of water inside, where $\gcd(a_i, b_i) = 1$ and $0 < p_i < q_i$ (i.e. we may have done some operations, and we are ignore empty/full cups). Further, suppose $p^m \mid b_n$ where $p$ is a prime, $m \ge 1$, but $p^m \not\mid b_i$ for $i < n$. I claim that any operation we perform involving the $n$th cup will produce a cup $\frac{a}{b}V$ in lowest terms (and an empty/full cup) where $p^m \mid b$.

This is relatively straightforward to verify. If we combine with $\frac{a_i}{b_i} V$, then either the sum is less than $V$ or greater than $V$. Note it can never be equal to $V$, otherwise $b_i = b_n$, against assumption.

If the sum is less than $V$, then the non-empty cup comes to $\frac{a_i b_n + a_n b_i}{b_i b_n} V$. Note that $p^m \mid b_i b_n$. If $p^m \mid a_i b_n + a_n b_i$, then $p^m \mid a_n b_i$ since $p^m \mid b_n \mid a_i b_n$. We obviously can't have $p^m \mid a_n$, otherwise $\gcd(a_n b_n) \ge p^m > 1$, so $p^m \mid b_i$. But this contradicts our assumption that $p^m \not\mid b_i$.

If the sum is greater than $V$, the argument is very much the same. The improper fraction's least denominator is divisible by $p^m$, and hence the same is true when subtracting $1$.

By induction, we can see that, any sequence of operations involving the $n$th cup will continue to make one cup, but not the others, have a denominator divisible by $p^m$.

Using this fact, we can see that in our original problem, cups $C_3$ to $C_7$ are useless, containing either factors of $5$, or too high factors of $2$ or $3$. We can also quickly verify that nothing can be done with just $C_1$ and $C_2$, so it is impossible, as claimed.

Answer 3

I believe the answer is no. Note that the $\frac15$ and $\frac1{10}$ serve no use here; you should show that any combination of them (positive or negative) with the other cups will always yield amounts with a denominator that is some multiple of 5. Similarly, you should be able to show that using the $\frac19$ or $\frac18$ blocks either positively or negatively will yield denominators divisible by $9$ or $8$, respectively. This means it's down to $\frac12$, $\frac13$ and $\frac14$, and no combination of those gives either $\frac16$ or $\frac76$.