Cash cows
Two men have equal shares in a herd of N cows. They sell all the cows for $N each. The money they get is in $10 bills plus less than ten $1 bills. One at a time they take a $10 bill. The man who takes the first $10 bill also takes the last one. The second man then takes all the $1 bills. How much does the first man owe the second?
Labels: mathemagic
posted by Chris at
1:56 AM
16 Comments:
(10-x)/2 Dollar, where x is the number of 1$ Bills
Hi t:b:H. I want an actual number of $s. It's somewhere between 0 and 9.
There are any number of solutions.
N = 6 and $2 owed is the smallest non-trivial solution, i.e. first man takes 2 x $10 and the second takes $10 plus $6.
Hi WoO. I'm going to be ultra nitpicking and say you're wrong, just so I can "milk" this one a bit more and because I know you can do better ;)
How much more do you want?
Total takings are N^2. Agreed?
First man takes an odd number of $10 bills and second man takes less than ten $1 bills.
So we have N^2 = 10x + y where x is odd and 0 < y < 10.
The first square that meets this criterion is 16 but this is trivial.
The next is 36, giving my previous answer.
The next is 196, for 14 cows and again $2 owed.
Then comes 256, 16 cows, $2.
It looks like any number of cows ending in 4 or 6 fits the bill, with $2 being owed in each case.
If your reply is an attempt to get more posts, then you've succeeded. But I can't see where my first answer was wrong.
But it's getting late here in Ozland so tell me in the morning.
Hi WoO. The question has the unique solution $2, not many solutions. My response was to let you know that you had correctly found the amount owed, but that you had left open the possibility that there might be another solution. I now think you really meant that there are an infinite number of Ns.
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I'll post my full prepared solution soon.
Tsk Tsk..... Always gotta be right, huh Chris?
LOL. That doesn't even make sense.
Let N = 10x+y where 0≤y<10. The money received=N²=(10x+y)²
=100x²+20xy+y². The first 2 terms are an even multiple of $10, so we can disregard them. As there were an odd number of $10 bills we must have 10≤y²<20 (allowing for case of no $1 bills). The only possibility is that y²=16. So there were 6 $1 bills, hence etc.
Chris do something else. Get a life!!!
If I did, what would you have left to do with all your valuable time? LOL
Anonymous, if you continue to post boring stuff like above, I won't bother responding. I really liked your Marbles 2 posts. There were tears of laughter rolling down my cheeks when I read them. You'd earnt a response then.
Hi, Chris! First of all, if you don't have a google account or an OpenID, you get to type in a name or just click "Anonymous", so the Anonymous above is probably not the one who posted in Marbles 2.
Secondly, there is a mistake in your solution. We can't conclude that 10≤y²<20, all we can conclude is that y²/10 rounded down must be uneven. For y∈{1,..., 9} that only applies for y=4 and y=6, and luckily y²'s last digit is 6 in both cases, so $2 is still the only possible answer.
Hi mo. No mistake. Is just a bound interval on y² based purely on there being one $10 bill and that there must be less than $10 worth of $1 bills. It was a sufficiently tight bound given that the only perfect square number in that range is 16.
I knew that it might be a different Anonymous. I decided not to worry about that as either or both would get the message.
Hi Chris, why must y create only one $10 bill? Have you thought of y=6? y² = 36 > 20. 36 = 3 ten dollar bills (uneven) + 6 (coincidently) one dollar bills. Every other y∈{1,...,9}/{4,6} would create an even number of $10 bills.
Hi mo. You're right. I also failed to correctly interpret your previous post. You said nothing wrong, I just didn't follow your argument - I don't know why not, I might have been too tired. Well spotted (yet again, again :)) and thank you for explaining the error.
Re the Anonymous troll post - there's a story behind the response. Enough said.
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