Goi's Fitness Forum

argh... stupid me... $2. That is my final answer Regis

yea... it's $2.

anyway, the answer goes something along the lines of:

N = the $ received
n = # of sheep sold

N = n2

since the older brother got the last $10, N must have odd number of tens.

let:
a = the 2nd digit (10) in n
b = the first didit (1) in n
note: it's irrelevant what digit 100s or 1000's are.

N = (10a + b)2 = 100a2 + 20ab + b2

(100a2 + 20ab) is divisible by 20, meaning it contains even # of tens.

b2 must contain odd # of tens.
among the possible values of b2 (0, 1, 4, 9, 16, 25, 36, 49, 64, 81)
only 16, and 36 have odd # of tens.

the younger brother received $6 at the end.

...then you know the rest.

h wrote:one more...

N rings having different outer diameters are slipped onto an upright peg, the largest rin on the bottom, to form a pyramid.

We wish to transfer all the rings, one at a time, to a second peg, but we have a third peg at our disposal.

During the transfers, it is not permitted to place a larger ring on a smaller one.

what is the smallest number of moves necessary to complete the transfer to peg number 2.

2n-1, where n=number of rings and n>2

yea, that's correct... but why 2 > n?

k(n) = 2n-1

n =2
k(2) = 4 -1 = 3 (correct)

n = 1
k(1) = 2 - 1 = 1 (correct)

n = 0
k(0) = 1 -1 = 0 (correct, but I suppose n = 0 can be ignored)

so it seems to work for all positive integers?

There's no good reason...

I only started with the case of 3 rings. 1 seemed obvious and didn't look at 2

so yes, where n=positive integer

...for the bored:

a) on which of the two days of the wee, Saturday or Sunday, does New Year's Day (not CNY) fall more often?
b) on which day of the week does thirtieth of the month most often fall?