Math and Logic Puzzles: Redux

[Sirius9121]

6. What is the last digit of 7^49?

It should be 7.

Proof:
=mod(7^1,10) = 7
=mod(7^2,10) = 9
3
1
repeat (7,9,3,1)

[Sirius9121]

4. Prove that a number is a multiple of 2 if the last digit is a multiple of 2, and that a number is multiple of 5 if the last digit is 0 or 5.

Let (number) = 10x+y where y is smallest possible interger value when x is an integer or 0.

10x = 2 * 5 * x
If y = 2m, 10x+y = 2(5x+m)

If y = 5m, 10x+y = 5(2x+m)

[Sirius9121]

Prove that a number is multiple of 3 if the sum of its digits is a multiple of 3

Let x = a * 10^1 + b * 10^2 + c * 10^3 ... ? * 10^(digits of x)

x = a * (10^1-1) + b * (10^2-1) ... ? * 10^(digitsofx) + (a + b + c...)
Since (10^integer-1) must be 9 repeated (integer) times, (10^integer-1) must be a multiple of 3

x = 3(something) + (a + b + c...)

if a + b + c... = 3n, x = 3(something+n)

[Sirius9121]

Let us assume we roll a fair dice a very large number of times. What is the probability that more (123s) than (456s) are rolled and ALSO have more (135s) than (246s)?

[Sirius9121]

I suspect it's 1/2 but am not confident and am intrigued to try to show it formally and will think about it more later

please note that both requirements need to be fulfilled... its lower than 1/2

[Sirius9121]

Mitillos's solution seems to be correct.

Spoiler: why

This is me, running this simulation 500000 times by randomizing x and y then dividing them. The probablity by this method was around 65.3574%. Since this is a completely random simulation, it makes sense to have the result differ a bit, therefore I agree with Mitillos's solution.