geno3141

The sum of the digits used to write the whole numbers 1 through 100 (inclusive):

First, the units digits:   1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9  =  45

These units digits will be repeated for each of the sets: 10 ... 19, 20 ... 29, 30 ... 39, ..., 90 ... 99

So the total of the units digits is 10 x 45 = 450 

Now, the tens digits:

--  in writing the numbers from 10 through 19 the "1" will be written 10 times:  10 x 1  =  10

--  in writing the numbers from 20 through 29 the "2" will be written 10 times:  10 x 2  =  20

--  in writing the numbers from 30 through 39 the "3" will be written 10 times:  10 x 3  =  30

So, we also have 10 x 4 = 40, 10 x 5 = 50, 10 x 6 = 60, 10 x 7 = 70, 10 x 8 = 80, and 10 x 9 = 90

Summing the tens digits:  450.

Total for both the ones digits and the tens digits = 900

Plus the "1" in 100:  901.

To make the reading simpler, let  x  =  ( sqrt(3) + sqrt(2) )   --->   a  =  x^-3

and                                         let  y  =  ( sqrt(3) - sqrt(2) )   --->   b  =  y^-3

This means that  x · y  =  ( sqrt(3) + sqrt(2) ) · ( sqrt(3) - sqrt(2) )  =  3 - 2  =  1

      and  x³y³  =  ( xy )³  =  ( 1 ) =  1

1 / (a + 1)  =  1 / ( x^-3 + 1 )  =  x³ / ( 1 + x³ )                (multiplying both the numerator and denominator by x³)

1 / (b + 1)  =  1 / ( y^-3 + 1 )  =  y³ / ( 1 + y³ )                (multiplying both the numerator and denominator by y³)

1 / (a + 1)  +  1 / (b + 1)  =  x³ / ( 1 + x³ )   +   y³ / ( 1 + y³ ) 

using the common denominator of  ( 1 + x³ ) · ( 1 + y³ ) :

                                       =  { [ x³( 1 + y³ ) ] + [ y³( 1 + x³ ) ] }  /  [ ( 1 + x³ )( 1 + y³ ) ]

multiplying out both the numberator and denominator:

                                             [ x³ + x³y³ + y³ + x³y³ ] / [ 1 + x³ + y³ + x³y³ ]

since x³y³  =  1:                   [ x³ + y³ + 2 ] / [ x³ + y³ + 2 ]  =  1.

SAMBHRANT

The "word" must begin and end with the letter "A"

So, it could be just AA -- 1 "word"

But, it could be three letters long: 

     AxA where x represents any of the other 7 letters  -- 7 "words"

Or, it could be four letters long:

     AxxA with 7 choices for the first x and 6 choices for the second x  -- 7·6 = 42 "words"

Or, it could be five letters long:

     AxxxA with 7 choices for the first x and 6 choices for the second x and 5 choices for the third x

              --  7·6·5 =  210 "words"

and, you can take it from here.

You'll need to add up all the individual possibilites to get the total number of possibilities.

A very interesting question ...

The integers:   ..., -3, -2, -1, 0, 1, 2, 3, ...

If you arranged them as:  0, 1, -1, 2, -2, 3, -3, 4, -4, ...

and added them this way:  (0) + (1 + -1) + (2 + -2) + (3 + -3) + ...

       you would get:             0 + 0 + 0 + 0 + ...  =  0

but, if you added them this way:  (0 + 1) + (-1 + 2) + (-2 + 3) + (-3 + 4) + ...

       you would get:                       1 + 1 + 1 + 1 + 1 + ...  which tends to positive infinity

and, if you added them this way:  (0 + -1) + (1 + -2) + (2 + -3) + (3 + -4) + ...

       you would get:                        -1 + -1 + -1 + -1 + ...  which tends to negative infinity

So, there is no particular sum  --  no sum ...

You might want to try to rearrange the integers to get other "answers" ...

( x² + y² - z² )² - 4x²y²  =  ( x² + y² - z² )² - ( 2xy )²  

  difference of squares  =  [ ( x² + y² - z² ) + ( 2xy) ] · [ ( x² + y² - z² ) - ( 2xy) ]

  rewrite                         =  [ x² + y² - z² +  2xy ] · [ x² + y² - z² -  2xy ]

  rewrite                         =  [ x² + 2xy + y² - z² ] · [ x² - 2xy + y² - z² ]

  rewrite                         =  [ ( x² + 2xy + y² )  -  ( z )² ] · [ ( x² - 2xy + y² )  -  ( z )² ]

  rewrite                         =  [ ( x + y )²  - ( z )² ] · [ ( x - y )² -  ( z )² ]

  difference of squares  =  [ x + y + z ] · [ x + y - z ] · [ x - y + z ] · [ x - y - z ]  

a)  V     (x·x)⁴  =  (x²)⁴  =  x⁸          x⁴·x⁴  =  x⁸

b)  F

c)  F

d)  V     (x + y)⁰  =  1

Opposite rays occur on a line. From point X, Y will be in one direction and Z will be in the opposite direction.

If you are standing at any vertex of a convex polygon, there are three vertices to which you cannot draw a diagonal.

You cannot draw a diagonal to the vertex on your left (it would be a side, not a diagonal); you cannot draw a

diagonal to the vertex that you are standing on; and you cannot draw a diagonal to the vertex to your right

(it also would be a side). 

You can draw diagonals to each of the other vertices. So, if there are n-vertices, you can draw n - 3 diagonals.

This will be true for each of the vertices:  n·(n - 3).

However, when you draw a diagonal from vertex A to vertex G, this will be redrawn as a diagonal from 

vertex G to vertex A; so, we'll have to divide the expression above by 2.

The final formula for the number of diagonals in a convex n-gon is:  n·(n - 3) / 2

For an octagon, it will be:  8·(8 - 3) / 2

Using the change-of-base formula, writing each of these logs as logs with base 10:

log₂(3)  =  log(3) / log(2)

log₃(4)  =  log(4) / log(3)

log₄(5)  =  log(5) / log(4)

...

log₁₂₆(127)  =  log(127) / log(126)

log₁₂₇(128)  =  log(128) / log(127)

When you multiply all of these together, notice that only  log(3)  from the first factor and  log(128)

from the last factor remain, all the rest cancel out.

log(128) / log(2)  =  log( 2⁷ ) / log( 2 ) =  7 · log( 2 ) / log( 2 )  =  7

The total number of degrees in all the interior angles of an n-gon is  (n - 2)180^o.

10 sides would be too few because they would create (10 - 2)180^o  =  1440^o; but this is close.

11 sides create (11 - 2)180^o  =  1620^o. Subtracting 1485^o from this number gives 135^o;

so if 10 of the angles give a total of 1485^o and the 11^th angle has 135^o, we would get an 11-sided polygon.