Melody

The number of possible paths is  10*3*3 = 90

You need to end on  0, 4, 8 or 12     no other multiples of  4 are possible

Say you choose the number 1   the only multiple of 4 you can get to is 0

    1 -1 +0    or     1+0-1    that is 2 favourable outcomes if you start with the number 1

Now do the same with all the other starting numbers to work out the number of favourable paths.

then the prob will be    number of favourable outcomes / 90

Please let guest finish this on their own

Nicely done guest.  :)

By convention, the real part must go first.     -1+12i

What number must we add to 3-5i  to get 2+7i ? Here, i^2=-1

3+ -1=2

-5+ 12 = 7

so the answer is     -1+12i

For addition and subtraction you can treat the i like it is just an ordinary pronumeral.

Nice try guest.

It took me quite a while to get my head around the question too.

The number is 3p

both 3 and p are prime numbers so  the only factors are  1,3, p and  3p

the sum of the factors is   4+4p

4+4p= 600

p+1=150

p=149    

A different interpretation.

10 to 99    is   90 numbers in total

which ones have BOTH digits either 8 or 9      

only 88, 89,98 and 99

so that is 4/90 

the wording is strange but this is how I have interpreted it.

11 001 010 100 101 011 +    Base 10

    110 100 011 000 100        Base 10

11 111 110 111 101 111           n

11 001 010 100 101 011 +    Base 2

    110 100 011 000 100        Base 2

11 111 110 111 101 111             m

n-m=0

Answer 1 is correct but there is a typo.

It is not c=121

It is the product of all possible c is 121

\(c=-9\pm\sqrt{40}\;i\)

Thanks very much guest for finding that ancient thread  (I assume there is a newer one as well)

It seems this question has already been answered to the best ability of most, if not all, people here.

I have unlocked the linked question if anyone wants to add to it.   

Do not answer directly on this thread