MAths problem... can u answer it?

[manyu882]

the smallest amount is ((((((5+1)5+1)5+1)5+1)5+1)5+1= 19531

unfortunately, incorrect...
a bit smaller

i know already. the next morning they divide can left one coconut

((((((0+1)5+1)5+1)5+1)5+1)5+1=3906

[h]

uh... dude, that's actually colder than the previous answer.

fails at the 2nd division.

[h]

...care for another one?

common sense, but can you prove these?

a) prove that, of all rectangles having the same given perimeter P, the square encloses the biggest area.

b) prove that, of all rectangles having the same given area S, that of smallest perimeter is the square.

just a note:
you can use the theorems on arithmetic and geometric means for this.
√(a*b) ≤ (a + b)/2
equality holds only when a = b

[Canuck eh]

k, I've had monkeys and coconuts on the brain all day long, starting to bug me

Got as far as x=(4/5)^5*(1/5y+1)
where y=starting number and x the number just prior to the division in the morning...

Missing the key/trick to eliminate one of the variables...

[h]

hahaha, i saw your log and I was wondering where you live... had no idea it was about this.

[h]

...how about this:

In chess, is it possible for the knight to go (by allowable moves) from the lower left-hand corner of the board to the upper right-hand corner and in the process to light exactly once on each square?

note: look at a chessboard, then it's easy.

i dunno if anyone's still working on this, or if anyone ever started...

but here's a hint: a1 and h8 are same color

[h]

k, I've had monkeys and coconuts on the brain all day long, starting to bug me

Got as far as x=(4/5)^5*(1/5y+1)
where y=starting number and x the number just prior to the division in the morning...

Missing the key/trick to eliminate one of the variables...

well, there's 2 (common) ways of going about solving this.
I suppose the easiest is to simply work backwards.

i.e.
N = the initial number of coconuts (before robbery)
n = the number of coconuts each man has received.

the number of coconuts in the last pile (before distribution) is: 5n + 1
the amount stolen by the last man is: (5n+1)/4
the # of coconuts before the last robbery is: 5(5n+1)/4 + 1

...

not so elegant, but if you work your way backwards, you'll eventually find a function of N in terms of n
N = f(n),
and you can work on finding n that satisfies the condition (N must be an integer, and as small as possible)

[Weib]

dude you should sell math products, you'd be rich.
Given that singapore digs such stuff..

[h]

there's already more than enough books out there...

[Weib]

There are so many strength coaches out there, and will be more coaches in future..so whats the deal?
Just have to make it simple for people to understand..

[h]

i think some of you will love this one:

prove that 22225555 + 55552222 is divisible by 7

in case anyone's wondering:

22225555 + 55552222 = (22225555 + 45555) + (55552222 - 42222) - (45555 - 42222)

work from there...

[h]

There are so many strength coaches out there, and will be more coaches in future..so whats the deal?
Just have to make it simple for people to understand..

no offense, but let's face it... it takes a lot less to get published in fitness. i mean, mario lopez got a book right?

and i don't know enough to write a book.

[Canuck eh]

k, I've had monkeys and coconuts on the brain all day long, starting to bug me

Got as far as x=(4/5)^5*(1/5y+1)
where y=starting number and x the number just prior to the division in the morning...

Missing the key/trick to eliminate one of the variables...

well, there's 2 (common) ways of going about solving this.
I suppose the easiest is to simply work backwards.

i.e.
N = the initial number of coconuts (before robbery)
n = the number of coconuts each man has received.

the number of coconuts in the last pile (before distribution) is: 5n + 1
the amount stolen by the last man is: (5n+1)/4
the # of coconuts before the last robbery is: 5(5n+1)/4 + 1

...

not so elegant, but if you work your way backwards, you'll eventually find a function of N in terms of n
N = f(n),
and you can work on finding n that satisfies the condition (N must be an integer, and as small as possible)

Isn't the amount stolen by the last man (5n+1)/5?

My formula is the result of all that working backwards and simplifying. I think you could write it as [(n-1)/n]^n*(1/nx+1) where x is your starting number of nuts and n the number of people you are splitting between. I'd hate to think you have to brute force from there. I'm assuming there is a solution and you have it...

[alex]

Lol

Who remembered using "model blocks" to solve problem sums before?

Simple example to illustrate

Sam[ ][ ]
Paul[ ][12]

[h]

i think this one is well-known:

an island is inhabited by five men and a pet monkey. One afternoon the men gathered a large pile of coconuts, which they proposed o divide equally among themselves the next morning.

during the night one of the men awoke and decided to help himself to his share of the nuts. in dividing them into five equal parts he found that there was one nut left over. This one he gave to the monkey. He then hid his one-fifth share, leaving the rest in a single pile. later during the night another man awoke with the same idea in mind. He went to the pile, divided it into five equal parts, and found that there was one coconut left over. This he gave to the monkey, and then he hid his one-fifth share, restoring the rest to one pile. during the same night each of the other three men arose, one at a time, and in ignorance of what had happened previously, went to the pile, and followed the same procedure. each time one coconut was left over, and it was given to the monkey.

The next morning, all five men went to the diminished nut pile and divided into five equal parts, finding that one nut remained over. what is the least number of coconuts the original pile could have contained?

see the part in bold

n = the amount distributed to each man
the amount in the last pile: 5n + 1

5n + 1 + X = Y
X = amount stolen by the last man
Y = 2nd last pile

X = (1/5) Y (he stole 1/5 of the pile)
5n + 1 = (4/5) Y
X = (5n + 1)/4

[h]

...how about this:

In chess, is it possible for the knight to go (by allowable moves) from the lower left-hand corner of the board to the upper right-hand corner and in the process to light exactly once on each square?

note: look at a chessboard, then it's easy.

i dunno if anyone's still working on this, or if anyone ever started...

but here's a hint: a1 and h8 are same color

I'm guessing this one isn't drawing that much interest...
unfortunately, I guess you won't get this one if you don't play chess.

If you study legal moves of a knight:
if a knight is on a light square, he can only move to a dark square. vice versa.
a1 is dark, h8 is dark.

on odd number of moves, the knight will be on a light square, and on even number of moves, the knight will be on a dark square.
the knight must be on h8 when the 63rd move is completed.

however, 63 is odd, so the knight cannot land on a dark square.

[Canuck eh]

I see how you got there now...

I started slightly differently

Pile prior to distribution
z=y/5+1
pile the 5th guy wakes up to
y=x-x/5+1, which is repeated for the others, so I think we end up in a similar place with two variables...

[h]

ok, I guess I'll just post the answer: it's too much of a pain to type the whole thing on a web browser:

after working backwards, you'll arrive at:

N = 15n + 11 + (265(n+1)/1024)

N must be an integer:
265(n+1) must be divisible by 1024
1024 and 265 are relative prime, thus the least integral value of n which will make 265(n+1) divisible by 1024 is 1023.

N = 15(1023) + 11 265 = 15621

[Canuck eh]

Thanks, at least I won't be preoccupied with coconuts and monkeys tonight. (Funny enough, there are monkeys kind of near to where I stay, but not close enough to keep me awake at night)

The math skills are showing some rust...

[h]

yea, these were easier when you had to do math 1-2/hr a day...

[h]

anyone want more?

[Canuck eh]

bring it

[Twxian]

chiu seem to be very free at work....

[Canuck eh]

chiu seem to be very free at work....

Yeah, trying to do some marketing stuff for work...

[h]

no more monkeys... for now...


two brothers sold a herd of sheep which they owned. for each sheep they received as many $ as the number of sheep they sold (e.g. if they sold 4 sheep, they received $16, 9 sheep, then $81). the money was then divided in the following manner:

First, the older brother took $10, then the younger brother took $10, after which the older brother took another $10, and so on.

At the end of the division the younger brother, whose turn it was, found that there were fewer than $10 left, so he took what remained. To make the division just, the older brother gave the younger his penknife. How much was the penknife worth?

[h]

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.

[Canuck eh]

no more monkeys... for now...


two brothers sold a herd of sheep which they owned. for each sheep they received as many $ as the number of sheep they sold (e.g. if they sold 4 sheep, they received $16, 9 sheep, then $91). the money was then divided in the following manner:

First, the older brother took $10, then the younger brother took $10, after which the older brother took another $10, and so on.

At the end of the division the younger brother, whose turn it was, found that there were fewer than $10 left, so he took what remained. To make the division just, the older brother gave the younger his penknife. How much was the penknife worth?

I think you have a typo, should be 81 instead of 91.

The answer is 6, can't write a proper proof for it though...

[h]

thanks, edited.

no, the penknife is not worth $6... but 6 is a key number.

[Canuck eh]

haha, my bad, 10-6=4=worth of penknife

[h]

uh... still a bit off... if the older brother gave him a penknife worth $4... did they get equal value from the transaction?