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The number begins with 5258103566533240.............
and ends with........6502764323392192512.
Note: The method used to calculate these numbers is relatively involved in that you have to have an arbitrary precision calculator or be able to use a program(code) to calculate them. I used the latter in C++.

A more algebraic way would be:\(3(10x+y)+6=10y+x\) . Now, we have \(30x+3y+6=10y+x\) . Solving, we get \(29x+6=7y. \) Solve it, and you should get 15.

The number is = 15

The two digits are = 1 + 5 = 6

Reverse 15 and you get 51

3 x 15 + 6 = 51

Numbers = 33  and   36

LCM[33, 36] = 396

396 / 33         = 12

GCD[33, 36] = 3

SUM =33 + 36 =69

A = 30

B = 40

LCM[30, 40] = 120

GCD[30, 40] = 10

The interest rate can be found as :

(12240 - 12000] / 12000  =  .02   =  2%

So.....the interest rate after this is  2% / 2  = 1% = .01

So.....the amount after 3 years is

12240  ( 1 + r) ^2   =   

12240 ( 1 + .01)^2  =

12240 (1.01)^2  = 

£12486.02

Thanks Rom!

I'm not the question asker, but I'm confused as to how you did question 8?

\(\theta_1 = 5 \theta_2 - 13\\ \theta_1 + \theta_2 = 140\\ \text{now substitute the first equation into the second and solve for }\theta_2\\ \text{and finally plug that value into the first equation to find }\theta_1\\ \text{You can do eet!}\)

I don't know what you mean by "the plan" but what the problem seems to be getting after

is that as all points on a circle are equidistant from the center we have the following equation

\(\forall (x,y) \ni (x,y) \text{ is on a circle with radius }R \text{ and center }(x_c, y_c)\\ d = \sqrt{(x-x_c)^2 + (y-y_c)^2} = R\\ \text{and squaring both sides we get}\\ (x-x_c)^2 + (y-y_c)^2 =R^2\)

By the change-of-base theorem.....

log₃ 7  =  log 7 / log 3

So

a^(log 7 /log 3) = 27

So

a^(og₃ 7)^2 =

a^ [ (log 7/log3) * (log7 /log3 ) ] = 27^(log7/log3)

Similarly

b^[ log 11/ log 7]^2    =

b^[ (log 11/log 7) *(log11/log 7) ] = 49^(log11/log 7)

And

c^[(log25/log11) * (log25/log11)]  =   sqrt(11)^(log25/log11)

So

27^(log7/log3) + 49^(log11/log7) + sqrt(11)^(log 25/log11)   =

27^(log₃ 7) + 49^(log₇ 11)  +  sqrt (11)^(log₁₁ 25)  =

(3^3)^(log ₃ 7)  +  (7^2)^(log ₇ 11 ) +  (11^[1/2] )^(log₁₁ 25)  =

(3 ^log₃ 7)^3  + (7 ^ log ₇ 11)^2  +   (  11^log₁₁ 25 )^(1/2)

By a log property.....a^(log _a b )  = b

So we have

(7)^3    + 11^2   +  25^(1/2)  =

343   +  121 +  5  =

469

18,859,680^2 - 10,080^2 =17!

3 1/4 * 2 * 4 1/2

= 13/4  * 2  *  9/2  =  234/8 = 29 1/4

17! =   355687428096000

So....one possibiiity is

355687428096002 -  2

We can do this with a graph

See here : https://www.desmos.com/calculator/2hkwxitnsh

The region formed by  0 ≤ x ≤ 8 and 0 ≤ y ≤ 4 is an  8 x 4 rectangle with an area of 32

The  inequality  x + y   ≤ 4     forms a triangular region within this rectangle ......the base and height of this  triangle  = 4....so its area =  (4 * 4)  / 2      =   16 / 2 = 8   

So.....the probability that a point falls within this triangle is 8 / 32   = 1 / 4

what

Thanks Rom,

I do not know how to do that but I did end up making the diagram.

I only know how to use Desmos and GeoGebra   

One trouble is, I actually do not understand what the question is asking. 

Are a,b and c the side lengths of the triangle ?

Anyway, here are my graphs of the two most extreme r values.    that is r=1 and r=0.8

radius = 1

radius 0.8

I can give you the parametric equations for that plot.

What graphing tool are you using?

The two integers are   4  and 36

LCM =  36

GCD =  4

So

36  : 4      =     9

And their sum = 40

By the Pythagorean Theorem, BD = √ [ AB^2 - AD^2]  =√  [ 13^2 - 5^2] = √[ 169 - 25] = √144 = 12

And, again, by the Pythagorean Theorem, 

BC = √ [ DC^2 - BD^2 ] = √ [ 20^2 - 12^2 ] = √ [ 400 - 144 ] = √256  =  16

Call the intersection of BD and AC , E

Angle AED  = Angle CEB   [ vertical angles ]

Angle ADE = Angle CBE   [ right angles ]

Therefore ....triangle AED is similar to triangle CEB

And 

AD/BC = DE /BE

5/16 =  DE/BE

Therefore....there are 21 equal parts of BD...and BE is 16 of these...and DE is 5 of these....

So BE =  (16/21) *BD = (16/21)* 12 =  64/7

And DE = (5/21)* BD = (5/21)  *12 = 60/21 = 20/ 7

And by the Pythagorean Theorem

EC = √ [BE^2 + BC^2]  = √ [ (64/7)^2 + 16^2 ]

And

AE = √[ AD^2 + DE^2 ] = √ [ 5^2 + (20/7)^2 ]

So

AC  = AE  + EC  =  √ [ (64/7)^2 + 16^2 ] + √ [ 5^2 + (20/7)^2 ]   ≈  24.2 cm

This was a good post to make Coolstuff.   

I also wish everyone a very happy and productive 2019   !

thank you very much!

8) verify that 13! = 78912^2 -288^2 without actually calculating the value of 13!

78912^2 - 288^2  =

(78912 + 288) (78912 - 288)  =

(79200) (78624)

(2^5 * 3^2  * 5^2* 11 )  ( 2^5 * 3^3 * 7 * 13)

( 2  * 3 *  2^2 * 5 * ( 2  * 3)  *  2 * 5 * 11)  ( 2 * 2 * 2^3 * 3^2 * 3  * 7 * 13)  = 

( 2 * 3 * 4 * 5 * 6 )  ( 2 *5* 11) ( 7 *  2^3 * 3^2 ) ( 2 * 2 * 3 * 13 )  =

(2 * 3 * 4 * 5 * 6 * 7 * 8 * 9)  (  (2 * 5) * 11 * ( 2 * 2 * 3) * 13 )  =

1 * (2 * 3 * 4 * 5 * 6 * 7 *8 * 9) ( 10 * 11 * 12 * 13 )   =   13!  

11.    x^3  -  y^3

x^3 - x^2 y + x^2y - xy^2 + xy^2 - y^3   =  

x^2 ( x - y)  + xy (x - y) + y^2 (x - y) =

(x - y) ( x^2 + xy + y^2)