# How to sell at the buying price and still have something in hand? [closed]

A farmer walked into the cattle market and found the price is fixed at "5 for 2 coins", either buy or sell. He thought a little and then bought cattle of 250 cows for 100 coins (250/5 * 2).

Then sold a part of cattle at the "5 for 2 coins" rule for 100 coins and takes the remainder of cattle as his profit.

How did he achieve this?

P.S: There is a chance another version of this question already asked.

Hint:

The cows are not the same size.

Edit: The farmer sold the cattle at the same day.

• In my opinion this puzzle needs more focus. Many possible answers come to my mind, but there's no way of telling, which one is the intended one. Commented Dec 4, 2023 at 10:39
• Can the cow only be sold in multiples of 5, if no, then how's the rounding done? Commented Dec 4, 2023 at 11:08
• @Christoph, I remembered this puzzle from a book I read 50 years ago from a polish author (translated) and actually add some more description. It is simpler than it seems. Commented Dec 4, 2023 at 14:42
• @VivekKumar, yes. The cows are sold in batches of whole numbers that can be interpreted as 5 for 2. Commented Dec 4, 2023 at 14:46

I believe that

At least one of the cows was (very) pregnant

and therefore by the end of the day

the farmer finds himself with at least one additional calf

he can then sell back all the cows

and keep the newborn(s)

• This is a valid and intelligent answer and I voted up because it is correct. But it does not fulfill the mathematical approach of the original designer of the puzzle. Commented Dec 4, 2023 at 14:52
• I think you should accept this answer and post the intended answer yourself. It looks like this thread is about to be closed for inviting speculative answers. Commented Dec 4, 2023 at 14:57

He probably just

Waited a few years for the cows to reproduce, and sold the ones he initially bought. Which would explain why there is a difference in the size of the cows!

• It might be a solution if the cost of feeding and nutrition and a place to keep the cattle is zero. I will edit the question to clarify this. Commented Dec 4, 2023 at 9:37
• @ShahramAlemzadeh While it's fine to edit questions to clarify the puzzle overall, adding an edit to invalidate a specific answer (just because it's not the one you were thinking of) is bad form. Commented Dec 4, 2023 at 10:42
• @Sneftel, the edit was not to invalidate this answer. The answer is logical if it was not a math puzzle, but a business issue. Commented Dec 4, 2023 at 15:06

"5 for 2 coins" is a woefully underspecified bargain - it doesn't say what you're getting 5 of, nor does it state what the value of the 2 coins is.

The farmer needs to simply sell 5 of something that does not amount to a whole cow. The farmer bought "5 heads for 2 coins", but may instead sell "5 hooves for 2 coins", selling only 1.25 cattle for the same price he bought 5 cattle. The farmer buys 250 cattle for 100 coins, but may recoup his investment by selling only 63 cows. Maybe the farmer markets them as emotional support cows and sells therapy sessions as "5 minutes for 2 coins" until he breaks even, and then takes all 250 cows home as profit.

Alternatively, the farmer may sell 5 heads of cattle for 2 coins of greater value than he bought them for. Maybe he bought the cows for half-dollar coins and sells them for dollar coins. Maybe he is a numismatist and only sells cows for coins with collectible value exceeding their nominal value.

Another solution respecting the size of the cows hint, the farmer simply sells the smallest cows and goes home with the largest ones. He is paying some average price-per-pound when buying the 250 cows, and by selling the smallest cows, for the same fixed price, he is getting a higher price-per-pound. Although the price is set per-cow, it isn't actually true that all cows have the same value. The farmer may buy 500,000lbs of cow for 100 coins, sell more than half the herd for more than half the price on a per-cow basis, yet still have more than half of the total weight remaining. If these cows are bred for meat, that extra weight is profit.

A lateral thinking solution could be achieved by

slaughtering a few of the cows and then selling 5 (kg of cow meat) for 2 coins.

Depending on the weight of the cows

slaughtering 1 or 2 should be enough to get 100 coins.

• The cows sold untouched and in their initial condition: Alive and in one piece. Nothing physically was done to the cattle. The lateral-thinking applies to interpret "5 for 2 coins", basically a math problem. Commented Dec 4, 2023 at 14:28

He bought cattle for 100 silver coins, but sold the cattle for 100 gold coins. Then he traded the 100 gold coins he'd received for a larger number of silver coins, and bought the rest of the cattle with some of the silver coins.

• This is a valid answer and I voted it up because there was no condition on the kind of coins in the answer. But it does not cover the mathematical view of the puzzle. Please also consider there are other cattle in the field that are priced as 5 for 2 silver coins, so there should be a reasonable cause the buyer pays gold when he can pay silver. Commented Dec 4, 2023 at 15:01

The farmer already had some cattle with him when he entered the market.

He bought big adult cows and sold young/small cows.

His profit was the difference in worth between his original small cows and the new adult cows (which would take years of work and food to grow up).

• You were really close in considering the size of cows as intended by the puzzle designer. Commented Dec 4, 2023 at 15:30

I think a possible solution is:

Interpret the "remainder of cattle" as the byproduct of cows like: milk, fertilizer, work, etc. This also justifies the different size (a milked cow should have less mass than a cow full of milk).
When the farmer sold them again, the farmer sold them all, and keep the byproduct.

• This is also a valid and reasonable answer, voted up. Commented Dec 4, 2023 at 15:26

The answer as given by the puzzle designer:

The farmer chose 125 calves and 125 cows. Then he tagged each 3 calves for 1 coin, and each 2 cows also for 1 coin. So, he did not break the rule because there are still 5 for 2 coins. He sold 120 cows for 60 coins, and 120 calves for 40 coins for a total of 100 coins. And takes the remining 5 claves and 5 cows as his profit.

• After the farmer has sold 120 calves and 80 cows (40 units of "5 for 2"), his only option is to sell 40 cows for 20 coins. I struggle to see how that can be viewed as "5 for 2 coins". At some point, he must stop selling entire units of "3 calves + 2 cows for 2 coins". This strikes me as rather weaselly - he can sell "5 cows for 2 coins" by charging 2 coins for a single cow and giving the others away for free, but only choosing to actually give you the ones you pay money for. I don't think an "offer" that is impossible to take someone up on is much of an offer. Commented Dec 4, 2023 at 15:39
• This answer is just playing with the rule of "5 for 2 coins". If he sold according to the rule there was no profit. The accepted answer is the most logical one, but the puzzle designer intended a mathematical approach. Commented Dec 4, 2023 at 15:54
• I don’t see how this answer makes sense. How does “sell 120 cows for 60 coins” match the rule “5 for 2 coins”? Commented Dec 4, 2023 at 18:41