Monday, January 8, 2007

Word Problem

Two brothers sold a flock of sheep. For each sheep, they received as many dollars as the total number of sheep in the original flock. The money was divided as follows: The older brother took 10 dollars, after which the younger brother took 10 dollars. This process repeated until it was the younger brothers turn and there wasn't a full 10 dollars left. He took what remained, and in order to even things up, the old brother gave him a penknife. What was the penknife worth?


Kimota94 aka Matt said...

Let x be the # of sheep in the original flock.
Therefore $x is the amount received for each sheep sold.
Therefore $x*x = the total money the brothers received for the sheep when they sold them.

For the older brother to get his full $10 in a round in which the younger brother gets less than $10, that means x*x can't have an even 2nd last digit, since that would mean that the older brother got the extra leftover. With an odd 2nd last digit, the divvying up of money, $10 at a time, will end with a ful $10 going to the older brother and less than $10 going to the younger (eg. $3_, $5_, $7_, etc.)

When squaring any positive integer, the last 2 digits of the result will be determined by the last digit of the starting integer for the following reason. The starting integer can be represented as n+m, where n is a multiple of 10, and m is the final digit of the integer. Thus n*n, as the square of a multiple of 10, will have at least 2 zeroes at the end. Therefore the resulting square ends with the 2 digits resulting from m*m.

Looking at m*m values:
0*0 = 00
1*1 = 01
2*2 = 04
3*3 = 09
4*4 = 16
5*5 = 25
6*6 = 36
7*7 = 49
8*8 = 64
9*9 = 81

m can't be 0, 1, 2, 3, 5, 7, 8 or 9, as those produce even digits in the 2nd last position.

m must therefore be 4 or 6, both of which produce an odd second last digit.

Since the last digit, in each case, is 6, this means the penknife is representing 4 dollars of inequity between what the brothers received. We can't narrow down the original number of sheep as being anything other than from the set {4, 6, 14, 16, 24, 26, ...} but we can safely say the penknife is "worth" 4 dollars.

Jimmy said...

Excellent solution, Kimota!

You got the 'remainder' right...the younger brother got $6. That would make the penknife worht $2, not $4 though.

Kimota94 aka Matt said...

Are you saying $2 because the older brother "lost" it to his younger brother (in other words, he's down $2, and the younger is up $2)? For all we know, the older brother had it given to him, making it worth nothing to him, but it's worth $4 to the younger, in order to make him feel like he 'got' the same amount of money out of the deal as his older sibling.

You and your damn trick questions!

cjg said...

It isn't a trick question, just that the question said "to make it fair". That means that both brothers had to have 10*x "value" at the end. Which means the remainder plus the penknife add to 10.

I spent a long time before I looked at this posting, trying to solve it fully. That was a mistake, but I did find also that n=14 works, with a remainder of "6" so the knife is worth $4.

I have two follow on questions, neither of which I have an answer for:

1) Give an expression for all the values that satisfy the problem.

2) Is the knife always worth $4?

Given simple testing, (2) may be affirmative. And I hypothesize, without proof (due to tiredness) that numbers ending in 4 or 6 will be valid answers.

Jimmy said...

Let x be the number of sheep in the flock.
Let k be the number of 'rounds' where both brothers get $10.
Let r be the number of dollars the younger brother gets in the last round.

x^2 = 20k + 10 + r

Take that mod 20 we see:

x^2 = 10 + r

Now, as Kimota94 points out x^2 mod 20 can be one of:

0, 1, 4, 5, 9, 16

16 is the only one greater than 10. therefore r must be 6.

The inequity is $4, so the penknife must be worth $2.

Anonymous said...

cell formula
a4 num of sheep
b4 total $ to divvy up
c4 older takes
d4 whats left
e4 younger takes
f4 remaining $ to divvy up

a5 14
b5 =a5*a5
c5 10
d5 (blank)
e5 10
f5 =b5-(c5+e5)

b7 new total pot

b8 =f5
c8 10
d8 =b8-c8
e8 10
f8 =b8-(c8+e8)

b9 =f8

take c8 thru f8 and paste into c9 thru f9

take b9 thru f9 and paste into b10 thru f21

d16 should have the first positive number between 9 and 1 (in this case 6

to make sure your formulas are working correctly here's the 1st 3 lines of numbers

line 5
14 196 10 blank 10 176

line 8
176 10 166 10 156
line 9
156 10 146 10 136

I would have to agree w/ Jimmy though because take a $2 knife from older and give it to younger....younger is missing $4 so take $2 knife from older (now younger is down only $2) and give to younger - now younger is down $0 so it sort of depends on how you look at it - then again maybe not - yes - (after re-reading what I just wrote...) i would totally agree w/ Jimmy...$2

"and that's my final answer"