All Answers 356759 Answers

For those who may be skeptical of the "Guest's" calculation, or rather the accuracy of the calculation, can refer to Wolfram/Alpha's calculation here:

https://www.wolframalpha.com/input/?i=birthday+problem+100+people

Ginger couldn’t a soar with the eagle because she’s hanging with a turkey.

LMAO I did. ... I did. I did lay a big, fat Turkey Egg.   LOL

I’ll hatch it for you. Then you’ll have an intelligent companion by Christmas.

Ginger:  Your LaTex is really cool.   

Mine is never as good as I want it to be :((

There is actually a rather complicated-looking  summation formula that calculates the probability and looks like this: 1 - ∑365!100! / {n! (100-2n)! (365-100 +n )! 2^n 365^100}, n=0 to 50

I summed up the first 25 terms and this is what I got:

0.0000003072 4892785157 735709                       
0.0000057176 0222881694 70586
0.0000508909 4268458230 219
0.0002886351 9731554141 541
0.0011725134 2701322632 6
0.0036356601 5220545585 22
0.0089549931 7932524642 57
0.0180040514 1305516981
0.0301304065 5435743848 6
0.0425931051 2915248603 5
0.0514369825 9415105677 2
0.0535378343 2065788517 8
0.0483676643 9980012911 8
0.0381427347 9231609518 2
0.0263757108 7405165721 6
0.0160515040 4620857996 3
0.0086219711 4581710867 67
0.0040969650 1255139206 62
0.0017251648 9044419630 59
0.0006445389 9539204962 057
0.0002138286 3864673084 781
0.0000630164 3196881678 332
0.0000164976 4999120274 6726
0.0000038355 0437597406 85323
0.0000007913 2162082163 957176
________________________________
1 - 0.354135321464259511361245756459
=64.59% - This  is the probability of at least 3 people sharing the same birthday !!!.

The notation is used to denote which function we are referring to on a graph....for instance, we might have the functions f(x), g (x), h(x), etc., on a graph........all are "functions of x".....they all take some x value and return a y value (where they are defined......)

The symbol f(x) represents a function. 

f is the function, x is the input. 

f(1) = 3

f(2) = 6

You can figure out that the function is times 2

f(x) = 2x

Here is a function problem:

f(x) = ax^2 + bx + c

f(0) = 4

f(1) = 15

f(2) = 36

What is product abc?

Anyone is welcome to answer this problem :)

Notes: Function does not have to be f, it could be any letter

p(x) or b(x) both work!

I hope this helped, 

Supermanaccz

Happy Thanksgiving! 

I'm just wondering why you posted this on a math website...

ty 

Same to you, Neptune.....!!!

Well..... if one train runs at  "a" kilometers per hour and travels for an hour, it travels "a" kilometers.....and if the other train travels at "b" kilometers per hour and meets the first train, the distance between the stations is  ( a + b)  kilometers

It is also called a lesser panda.

Fun fact: Master Shifu in Kungfu Panda was based on this cute guy.

GingerAle: It looks like you laid a big, fat "Turkey Egg" !! Happy Thanksgiving !.

The first answer of 333331 is the right answer, even though I don't understand how Guest #1 got the answer, since it satisfies the two modular equations:

333331 mod 3331 =231 and 333331 mod 1361=1247

If Gingerale's answer is 197140, it does not work. I also don't understand the method he/she used.

Thanks Ginger,

I used The Euclidean formula and the extended euclidean formula as well but the working went for ever.  Maybe  I will give it another go.  I thought myabe our guest knew what s/he was talking about and might explain it simply. ://

I'm not so sure I follow what you did either I will have to study it ://

After clicking on this link I heard a gobbling and clucking noise; I thought Mr. BB gave the forum a turkey for Thanksgiving. However on closer inspection, it’s easy to tell it’s just a chicken with a thyroid condition, stuffed with Blarney Dressing.   ^Yum!

\(\begin{array}{rcll} n &\equiv& {\color{red}3331} \pmod {{\color{green}231}} \\ n &\equiv& {\color{red}1361} \pmod {{\color{green}1247}} \\ \text{Set } m &=& 231\cdot 1247 = 288057\\ \\ \end{array}\)

\(\begin{array}{rcll} n &=& {\color{red}3331} \cdot {\color{green}1247} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ {\color{green}1247}^{\varphi({\color{green}231})-1} \pmod {{\color{green}231}} ] }_{=\text{modulo inverse 1247 mod 231} } }_{=1247^{230-1} \mod {231} }}_{=1247^{229} \mod {231}}}_{=113} + {\color{red}1361} \cdot {\color{green}231} \cdot \underbrace{ \underbrace{ \underbrace{ \underbrace{ [ {\color{green}231}^{\varphi({\color{green}1247})-1} \pmod {{\color{green}1247}} ] }_{=\text{modulo inverse 231 mod 1247} } }_{=231^{1246-1} \mod {1247} }}_{=231^{1245} \mod {1247}}}_{=637}\\\\ n &=& {\color{red}3331} \cdot {\color{green}1247} \cdot [ 113] + {\color{red}1361} \cdot {\color{green}231} \cdot [637] \\ n &=& 469374541 + 200267067 \\ n &=& 669641608 \\\\ && n\pmod {m}\\ &=& 669641608 \pmod {288057} \\ &=& 197140 \\\\ n &=& 197140 + k\cdot 288057 \qquad k \in Z\\\\ \mathbf{n_{min}} & \mathbf{=}& \mathbf{197140 } \end{array}\)

.

\(\small \text {Related formulas and principles compliments of Leonhard Euler }\scriptsize \text {(totient function),} \\ \small \text {Euclid of Alexandria }\scriptsize \text {(Extended Euclidean algorithm), and Brilliant Chinese mathematicians “Chinese Remainder Theorem” } \\ \small \text {LaTex layout and coding adapted from Heureka’s mathematical solution and Latex presentation:}\\ \)

You lost me after the word 'simple'.

tan (x)  = - √3

This is true at 60° in quadrant 2  =  2pi/3    and at 60° in quadrant 4  = 5pi/3

So...the general solution is that  x =   2pi/3  ± n* pi     where n is an integer

sec^2(x)  + sec^2(x)  - 1  = 3    subtract 1 from both sides

2sec^2(x)  = 2     divide both sides by 2

sec^2(x)   = 1   subtract 1 from both sides

sec^2(x)  - 1  = 0     factor

( sec (x)  - 1)  (sec (x)  + 1 )   = 0

Set  each factor to 0  and solve

sec(x)  - 1  = 0                              sec(x)  +  1  = 0

sec(x)  = 1                                    sec(x)  = -1

And this is true at                        And this is true at

0  ± n*2pi                                       pi ± n*2pi

Where n is an integer

sec^2 (x)  = 2 tan^2 (x)

Note that   tan^2(x)  + 1  =  sec^2 (x)....so we can write

tan^2(x)  +  1  =  2tan^2(x)      rearrange as

tan^2(x)  = 1   subtract 1 from both sides

tan^2 (x)  - 1  =  0     factor

(tan (x) + 1) ( tan (x)  - 1)  = 0

Set each factor to ) and solve

tan (x)  +  1  =  0                                       tan (x) - 1  = 0

tan (x)  = -1                                               tan (x)  =  1

And this is true at                                     And this is true at 

3pi/4  ± n*pi                                             pi/4  ± n*pi

Where n is an integer

8 tan (x)  + 11  = 19     subtract 11 from  both sides

8 tan (x)  =  8      divide both sides by  8

tan ( x)  = 1

And this is true when x  =  pi/4     and   x  = 5pi/4   in  the requested interval

This graphs a line because for every increase in t by 1, x changes by 2 and y changes by -3

We only need two points to find a line

If t  = 0   we have the point  (3,3)

If t  = 1, we have the point  (5,0)

The slope of this line is       -3/2    [ just as we suspected  !!!! ] 

And using (5,0), the equation of the line is

y  = (-3/2)(x - 5)

y =  (-3/2)x + 15/2

To see that this is true.....when  t  = 2,   x = 7    and y  = -3

So    

y = (-3/2)(7) + 15/2

y = -21/2 + 15/2

y = -6/2   =    -3