Goi's Fitness Forum

6 times

Canuck eh wrote:6 times

???
sorry, there was a typo in the question... but I don't think you read the question carefully?

Ah yes, my bad...

a) I'm assuming this is a how to... split the 80 into 3 piles and put the two evenly numbered piles onto the scales. If they balance, then counterfeit in the unweighed pile. Rinse and repeat and done in four steps.

b) on a related note then the number of steps required for n coins is the 3rd root of n where n is a integer.

What you guys doing answering Math questions in the wee hours of the morning?

cuz they're there

Canuck eh wrote:Ah yes, my bad...

a) I'm assuming this is a how to... split the 80 into 3 piles and put the two evenly numbered piles onto the scales. If they balance, then counterfeit in the unweighed pile. Rinse and repeat and done in four steps.

b) on a related note then the number of steps required for n coins is the 3rd root of n where n is a integer.

a) corect.
note: if the counterfeit is in the pile with 26 coins, you can add another coin (from another pile) to make three piles of 9 coins. (does it make a difference? no, but this seems to be the preferred answer for some reason)


b) yea, kinda:
the number of steps, k, satisfies inequalities:
3^(k)≥ n
3^(k - 1) < n

galapogos wrote:What you guys doing answering Math questions in the wee hours of the morning?

i was waiting for cards vs eagles line to move...
yea, i need life.

...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.

With the new forum upgrade, you can do stuff like this!
3k≥ n
3k - 1 < n

Instead of the old way like this:
3^(k)≥ n
3^(k - 1) < n

Cool, no?

a)
let "x" be length of sq, "a" be the length subtracted from x
area of sq =x*x=x2
area of rect= (x-a)(x+a)=x2-a2, thus x2 > x2-a2 (proven)
thus x2 > x2-a2 (proven)

galapogos wrote:With the new forum upgrade, you can do stuff like this!
3k≥ n
3k - 1 < n

Instead of the old way like this:
3^(k)≥ n
3^(k - 1) < n

Cool, no?

eh why my ans kena cut off...

Instead of chess boxing, we can train and solve math problems...

Yeah, all you Charlie Eppes wannabes...

a)
let "x" be length of sq, "a" be the change in length
area of sq =x*x=x2
area of rect= (x-a)(x+a)=x2-a2
thus, x2 > x2-a2 (proven) ( a>0)

b)
as notice, length of rect is always 2 times bigger than the length of sq
and width is 2 times smaller than the length of sq if both has the same area

given both the same area,
x^2=x/2+2x

perimeter of sq= 4x
perimeter of rect = x+4x

thus 4x < x+4x (proven) (x=/=0)

uh... ok, i get what you're saying... but only partial credit, because:

you didn't use P and S in your answers...

if you got the time...

c) prove that, of all triangles with the same given perimeter, the greatest area is enclosed by the equilateral triangle. (use Heron's formula)

h wrote:uh... ok, i get what you're saying... but only partial credit, because:

you didn't use P and S in your answers...

a)
let "p" be perimeter, "a" be the change in length
area of sq =(p/4)*(p/4)=(p/4)2
area of rect= ((p/4)-a)((p/4)+a)=(p/4)2-(p/4)2
thus, (p/4)2 > (p/4)2-a2 (proven) ( a>0)

b)
as notice, length of rect is always 2 times bigger than the length of sq
and width is 2 times smaller than the length of sq if both has the same area

given both the same area, s
s=(sqrt s)^2=(sqrt s)/2+2(sqrt s)

perimeter of sq= 4(sqrt s)
perimeter of rect (sqrt s) x+4(sqrt s)

thus 4(sqrt s) < (sqrt s)+4(sqrt s) (proven) (s is not less than 0)

manyu882 wrote: b)
as notice, length of rect is always 2 times bigger than the length of sq
and width is 2 times smaller than the length of sq if both has the same area

hmm... how about
area: 16
length: 16
width: 1
?

h wrote:

manyu882 wrote: b)
as notice, length of rect is always 2 times bigger than the length of sq
and width is 2 times smaller than the length of sq if both has the same area

hmm... how about
area: 16
length: 16
width: 1
?

you make a point. i should say that if the length of rect is always z times bigger than the length of sq
and width is z times smaller than the length of sq if both has the same area.

where z is the factor

manyu882 wrote:

h wrote:

manyu882 wrote: b)
as notice, length of rect is always 2 times bigger than the length of sq
and width is 2 times smaller than the length of sq if both has the same area

hmm... how about
area: 16
length: 16
width: 1
?

you make a point. i should say that if the length of rect is always z times bigger than the length of sq
and width is z times smaller than the length of sq if both has the same area.

where z is the factor

as notice, length of rect is always z times bigger than the length of sq
and width is z times smaller than the length of sq if both has the same area

given both the same area, s
s=(sqrt s)^2=((sqrt s)/z)*z(sqrt s)

perimeter of sq= 4(sqrt s)
perimeter of rect (sqrt s) x+4(sqrt s)

thus 4(sqrt s) < (sqrt s)+4(sqrt s) (proven) (s is not less than 0)

whoa hyo i didnt know u were good at math...

crazy bunch. ahhH!

flamekid wrote:whoa hyo i didnt know u were good at math...

...you can't get into banking industries without decent math.
well, you could, but you'll just be a pencil pusher.

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?

h wrote:

flamekid wrote:whoa hyo i didnt know u were good at math...

...you can't get into banking industries without decent math.
well, you could, but you'll just be a pencil pusher.

Only the modellers need to do some hardcore math. For the rest, you have to be able to look past the numbers and analyze the problems qualitatively...

well, i guess it really depends on what you do...

i think some of you will love this one:

prove that 22225555 + 55552222 is divisible by 7

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

h wrote:i think some of you will love this one:

prove that 22225555 + 55552222 is divisible by 7

cham laio.. i addicted to math qn.. my cad also put one side

manyu882 wrote:the smallest amount is ((((((5+1)5+1)5+1)5+1)5+1)5+1= 19531

unfortunately, incorrect...
a bit smaller