Melody

x=88^8888(mod 7)

\(\quad 88^{8888}\quad mod7\\ =4^{8888}\quad mod7\\ =(2^2)^{8888}\quad mod7\\ =(2)^{17776}\quad mod7\\ =(2)^{17776}\quad mod7\\ =(2)^{3*5925+1}\quad mod7\\ =(2)^{3N+1}\quad mod7\\ =2 \)

To work this out I used this

\(2^0=1=1(mod7)\\ 2^1=2=2(mod7)\\ 2^2=4=4(mod7)\\ 2^3=8=1(mod7)\\ ...so \;\;if\;\;N\in Z\\ 2^{3N}=1\\ 2^{3N+1}=2(mod7)\\ 2^{3N+2}=4(mod7)\\\)

Watch this and you will find out :))

I can see you have been doing some digging Max   

Yes I do not mind a little socializing on the forum so long as it does not get too overpowering.

Mind you, something as mindless as "UHHHH" would normally be deleted straight away.

Actually, CPhill and I actually like to see some play here.  It is more fun for us as well.  :)

Put you question in the body of a new post.

Do not put it in the heading.

ضع سؤالك في الجسم وظيفة جديدة.
لا وضعها في البند.

(((5000*1.08^8))(*1.10^10)*1.07^6)

(((5000*1.08^8))(1.10^10)*1.07^6) = 36023.8035764331287797

− 6+( − 3 − 3)2÷ 3

There should be a times sign between the ) and the 2.

= − 6+( − 3 − 3)*2÷ 3

= − 6+( -6)*2÷ 3

= − 6+-12÷ 3

= − 6+-4

= -6-4

= -10

A long dining room table perhaps.

A wooden onament of a multi legged creature.

Maybe a rissole spider   

(x-1)(x+2)=(x-3)(x+5)

\((x-1)(x+2)=(x-3)(x+5)\\ x^2+x-2=x^2+2x-15\\ x-2=2x-15\\ x-2-x=2x-15-x\\ -2=x-15\\ x=13\)

atan(10/14)=

atan(10/14) = 35.537677791974 degrees

Zero IS a number.