Let there be a positive integer n. Show that for any set of n consecutive integers there is one divisible by n.

Let c, c+1, c+2, c+3, ... c+n-1 be the integers in this sequence. If r = (c) mod n, then either r = 0 or c < c+n-r ≤ c+n-1, and therefore in the sequence. Since (n-r) mod n = n-r, (c+n-r) mod n =0. Therefore, there must exist one integer in the sequence that is divisible by n.

Bob and Alice are playing a game of dice throwing. Bob found a eight-sided dice, where in Alice only found two four-sided dice. What are the odds that the sum of Alice's results are greater than Bob's?

Each of Bob's twelve possible out comes; {1, 2, 3, 4, 5, 6, 7, 8} are equally likely to occur, at a 1/8th probability. Alice's outcomes; {2, 3, 4, 5, 6, 7, 8} have a {1/16th, 2/16th, 3/16th, 4/16th, 3/16th, 2/16th, 1/16th} probability, respectively. To determine the odds of Alice rolling greater the Bob, we multiply each of Bob's results by the chance Alice rolls greater than that result. There is a 1/8th chance of Bob rolling a 1, and a 16/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 2, and a 15/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 3, and a 13/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 4, and a 10/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 5, and a 6/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 6, and a 3/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 7, and a 1/16th chance Alice rolls greater. There is a 1/8th chance of Bob rolling a 8, and a 0/16th chance Alice rolls greater. Multiplying those chances together, the total probability p that Alice rolls greater then Bob is as follows; p=(1/8)*(16/16)+(1/8)*(15/16)+(1/8)*(13/16)+(1/8)*(10/16)+(1/8)*(6/16)+(1/8)*(3/16)+(1/8)*(1/16)+(1/8)*(0/16) p=(1/8)*(1/16)*(16+15+13+10+6+3+1+0) p=64/128=0.5 There is a 50% chance of Alice's two four-sided dice roll a higher total than Bob's one eight-sided dice

Given that: x^2 +y^2 = 25 and 2x = y. What are are all possible values for x and y.

Since we know that y = 2x, from to the second equation, we can replace the y in the first equation with 2x. Then x^2+(2x)^2=25 can be simplified to x^2+4(x^2)=25 thanks to the commutative property of multiplication. It can be further simplified to 5(x^2)=25 by the distributive property. Dividing both sides by 5 will leave us with x^2=5, and if we take the square root of both sides, we have x=±√5. So, given that 2x=y, the possible values for x and y are √5 and 2√5, and -√5 and -2√5.