Yes I believe it means you can only use the blocks given, the question is misworded.
I also imagine that the blocks are lined up, not added
0,1,10,11,100,101,110,1001,1010,1100 That is 10 numbers.
\((1 + i\sqrt{3})^6\\ =\binom{6}{0}(i\sqrt3)^0+\binom{6}{1}(i\sqrt3)^1+\binom{6}{2}(i\sqrt3)^2 +\binom{6}{3}(i\sqrt3)^3+\binom{6}{4}(i\sqrt3)^4+\binom{6}{5}(i\sqrt3)^5+\binom{6}{6}(i\sqrt3)^6\\ =1+6*(i\sqrt3)^1+15(i\sqrt3)^2+20(i\sqrt3)^3+15(i\sqrt3)^4+6(i\sqrt3)^5+(i\sqrt3)^6\\ =1+(6\sqrt3)i-15(\sqrt3)^2-20(\sqrt3)^3i+15(\sqrt3)^4+6(\sqrt3)^5i-(\sqrt3)^6\\ =1+(6\sqrt3)i-15*3-60(\sqrt3)i+15*9+6*9\sqrt3 i-27\\ =1+6\sqrt3i-45-60\sqrt3i+135+54\sqrt3 i-27\\ =1-45+135-27+(6\sqrt3-60\sqrt3+54\sqrt3) i\\ =1-45+135-27+(6\sqrt3-60\sqrt3+54\sqrt3) i\\ =64\)
You need to check my working. there may also have been a simpler path.
LaTex
(1 + i\sqrt{3})^6\\ =\binom{6}{0}(i\sqrt3)^0+\binom{6}{1}(i\sqrt3)^1+\binom{6}{2}(i\sqrt3)^2 +\binom{6}{3}(i\sqrt3)^3+\binom{6}{4}(i\sqrt3)^4+\binom{6}{5}(i\sqrt3)^5+\binom{6}{6}(i\sqrt3)^6\\ =1+6*(i\sqrt3)^1+15(i\sqrt3)^2+20(i\sqrt3)^3+15(i\sqrt3)^4+6(i\sqrt3)^5+(i\sqrt3)^6\\ =1+(6\sqrt3)i-15(\sqrt3)^2-20(\sqrt3)^3i+15(\sqrt3)^4+6(\sqrt3)^5i-(\sqrt3)^6\\ =1+(6\sqrt3)i-15*3-60(\sqrt3)i+15*9+6*9\sqrt3 i-27\\ =1+6\sqrt3i-45-60\sqrt3i+135+54\sqrt3 i-27\\ =1-45+135-27+(6\sqrt3-60\sqrt3+54\sqrt3) i\\ =1-45+135-27+(6\sqrt3-60\sqrt3+54\sqrt3) i\\ =64
4C3 = 4
4/16 = 1/4
This is the second time you posted this and you got the same response both time.
Most of the answerers here will ignore a string of questions in the same post like this.
There is strictly a one post = one question policy.
I am blocking both your threads.
I do not want people to do all your homework for you.
I've done this one before somewhere ... so has Heureka.
I'll start you off (although there may be an easier way)
3 + x == 2^2 (mod 3^3)
3 + x == 4 + 3^3K
x == 1 + 3^3K
this would also mean that
x == 1 + 3K
5 + x == 3^2 (mod 5^3)
5 + x == 9 + 5^5L
x == 4 + 5^3L
x == 4+ 5L
x == -1 + 5L
7 + x == 4^2 (mod 7^3)
7 + x == 16 + 7^3M
x == 9 + 7M
x == 2 +7M
First I solved 1+3K = -1+5L and got x= -1 mod 15
Then I solved x=1+15a with x = 2+7M and got x=79+105b
So I got an answer of 79 but I could easily have made a careless error.
I use the Euclidean Algorithm to solve the diophantine equations.
Don't be rude Proyaop, it gets you nowhere.
Besides I think you are better than that.
When someone tries to help you, you say thank you! (And you do not take points off them!)
Do you still want help? You didn't state, one way or the other.
Gino is right,
You have copied the question incorrectly.
ABCD is a regular tetrahedron (right pyramid whose faces are all equilateral triangles).
If M is the midpoint of line segment CD , then what is cos
See if you can make sense of my pic and then use cosine rule.
20 consonants, 6 vowels,
3 letter words, at least one vowel
All letters different 6*25*24*3! = 900
2 vowels the same 6*25*3 = 450
2 consonants the same 6*20*3 = 360
all three the same 6 = 6
900+450+360+6 = 1716
Feel free to aks questions.
Do not aks the same question twice like this.
You are just wasting people's time!
+118737 
