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Aha, integer factorization.

So 1676962425600 is 2^8 * 3^2 * 5^2 * 7 * 11^2 * 37 * 929

>If they were all prime  and you were told how many there were then you could

so I have 2,3,5,7,11,37,929. How can I determine correct exponents?

Thanks.

EP: Harmonic mean of 2 numbers is = 2 /{1/a + 1/b}

The answer should be: a≈37.7641  and  b≈5362.24

Geometric mean = Sqrt{37.7641 x 5362.24} = 450

Solve x(y + z) = 39, y(x + z) = 60, z(x + y) = 63.

69years and 420 years

two numbers   x  and y

harmonic mean:                ( 2/(1/x+1/y) )=75   re-arranging

                                              2 (x * y)/(x+y) = 75

arithmetic mean :

(x+y)/2 = 2700   re-arranging

    y = 5400- x       sub this into red equation

2x(5400-x)/5400 = 75   re - arrange to

   2 x^2 -10800x + 405000 = 0    Quadratic Formula shows x =  5362.24     and y = 37.76

Geometric mean:    sqrt (x*y) =~   450

(Thanx Guest for noticing my error....I have corrected this post)

A frog is at the bottom of the well which is 20 meters deep. Everyday the frog jumps 5 meters upwards and fall 4 meters down.On which day the frog will reach the top?

I think on the 15th day the frog will only be at 15 m   after he has fallen back 4m...then on THE SIXTEENTH day the frog's 5 m jump will clear the top of the well.....

going off your answer, if the frog DID reach the top the moment it goes over the well top, regardless how much it falls:

the well is 20m deep, the frog jumps 5m. So the frog needs to reach 20 - 5 = 15m.

jumps 5m and falls 4m, so it jumps 5 - 4 = 1m

15m/1m=15

the frog reaches the top on the 15th day

Is reaching to top considerd to be when the frog falls back 4 meters and his height is 20 m from the bottom?

  if so, then each day he gains 1 meter in height    .....   it will take 20 days.  Not that for several days the frog's 5 meter jump will raise it higher than the lip of the well....but the frog falls back down.....

arctan(1) + arctan(2) + arctan(3) = 3.141592653589      from the Web2.0 calculator ...as guest found.

I'm not sure if this is the smallest prime, but here are 3 different ways of summing it up. 
43 = 3 + 17 + 23
43 = 7 + 5 + 31
43 =11 + 13 + 19

Hi Melody

Consider it from the point of view of the equation

\(\displaystyle x^{n}-1=(x-1)k(x). \)

All that you are looking for is that \(\displaystyle (x-1) \text{ is a factor of }x^{n}-1.\)  

Hi Guest,

using your logic

f(n)=15^n-1      would always have a facter of  14 . is this true?

if n =1 it is true

if n=2 then f(2)=224   yes, that is divisable by 14

if n=3 then f(3)=3374  yes that is divisable by 14

I am surprised, I did not expect that to work.

I am still not completely convinced that this will always be the case but maybe it is.

I am intrigued.

Can anyone prove that this will always be the case?

Do all 3 sides have to be different in length? 

What makes you think it is an isosceles triangle? How about sides a = 3, side b= 6 and side c = 4

Yoy still have: a x c =2b =3 x 4 = 2 x 6 ??

Hi Melody: I'm not sure if I know how to prove it formally, but notice the following:

1 - Your result of 18^n - 1 = 17 when n = 1

2 - The result of 18^2 - 1 = 17 x 19

3 - The result of 18^3 - 1 = 17 x 7^3

.

.

.

10 - The result of 18^10 - 1 =17 x  11 x 19........etc.

Because you are subtracting 1 from 18 =17, no matter what value n takes, the result will ALWAYS have a 17 as a prime factor. I don't know if this makes sense mathematically, but it seems clear to me that you will always have 17 as factor for any value of n.

It 's PI!! 

3.14........

Thanks guest, I'll take another  look.

2^n*3^(2n) - 1

\(2^n*3^{2n} - 1=18^n-1 \)

ok, your answer could well be correct, it certainly works for the first 3 terms,

This is not a proof that it will always be so.  Do you know how to prove it?

Let the side lengths of the triangle ABC  be:  a=1,  c=1,  and  b= ac/2 = 0.5

cos(A) /a + cos(C)/c=(cos75.522°/1)+(cos75.522°/1) = 0.25 + 0.25 = 0.5 

Let   a be the longer side of the  original rectangle and  b be  the shorter side

So

 a*b  =12

b  =  12/a   (1)

When folded in half, the  longer side  =  b   and the shorter side  is  a/2

a/ b  =  b / (a/2)

a/b  = 2b / a

a^2  = 2b^2

a = √2b

So

a * b   = 12

√2 b* b  = 12

b^2  =  12/√2

b^2 =  6√2

b^2 = √72

b = ⁴√72    

And a  = √2 ⁴√72  

So

a / b =  √2 ⁴√72  / [  ⁴√72  ] =  √2

And   b/ [a/2]  =   2b / a  =  2  ⁴√72  / [√2 ⁴√72 ] =  2/√2  = √2

So

a  = √2 ⁴√72 

b =  ⁴√72 

And

a₁  = b =   ⁴√72 

b₁  =  a/2  =   [√2 ⁴√72] / 2

BA =  (1, 0, -1)

CA  = ( 1, -1, 2)

Find the  cross-product   BA x CA

i     j      k        i         j

1    0     -1      1       0

1   -1      2      1       -1

[ 0i  - 1j  - 1k ]  - [ 0k + 1i + 2j ]  =   -1i - 3j - 1k   =   ( -1, -3 , -1)  = the  normal vector

Note that if we multiply this  vector by -1  we get

( 1, 3 , 1)

And adding these components gives us  5

Nice, Melody.....!!!

\(k, rk, r^2k   \qquad    \qquad      k, Rk, R^2k      \)

\(r^2k -  R^2k = 3(rk-Rk)\\ r^2 -  R^2 = 3(r-R)\\ (r-R)(r+R) = 3(r-R)\\ r+R=3 \)

The sum of the ratios is 3

Hi Melody: The way you understand it, 2^n*3^(2n) - 1, it gives the following 10 terms:

GCD = (17, 323, 5831, 104975, 1889567, 34012223, 612220031, 11019960575, 198359290367, 3570467226623) = 17