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Proove that if you take 5 integer numbers at random, at least two of them differ by a multiple of 4.

Let’s call the #s a, b, c, d, and e. Now let’s look at a and b. There’s four possibilities we look at: either they differ by a multiple of 4, or their difference is a multiple of 4 plus 1, 2, or 3. Now, if they differ by a multiple of 4, we are done. If they don’t, we move on to the next number, c. If c differs from a or b by a multiple of 4, we are done. However, if it is not, it is not differing by a multiple of 4 for both a and b. This means that the it has two different remainders for both a and b when the difference is divided by 4 (since if they were the same, that would mean a and b were differing by a multiple of 4). Now let’s look at d. If it differs by a multiple of 4 for a,b, or c, we are done. However, if it doesn’t, the “remainders” must all be three different numbers (1, 2, and 3), lest a, b, or c differ by a multiple of 4. Now let’s look at e. If it differs by a multiple of 4 for a, b, c, or d, we are done. If it doesn’t, there must be four different remainders for the differences between each other number. However, there are only three possible differences. This means e must have a difference that is a multiple of 4 with the first four numbers, and that therefore among the five numbers, at least two differ by a multiple of four. 

Q. E. D.

How did I do?

It does the trick! 

A shorter way to do it is to establish that there are only 4 possible reminders when they are divided by 4, so there must be two of these 5 numbers with the same remainder. When you substract them you also substract the remainders, so the difference is a multiple of 4. (4n+r)-(4m+r) = 4n-4m=4(n-m).

What is the last digit of 2^2019+3^2019?

keep us on the first  page!

Math is not a matter of having an answer ! It is how you get it. But no it is not 7.

Prove that the square root of 2 cannot be expressed as a whole-number ratio.

We now have 3 pending problems:

1. solve this with blueemu's method

x+y=2

x*y=1

2. My last one: What is the last digit of 2^2019+3^2019?

and blueemu about the square root.

3. Prove that the square root of 2 cannot be expressed as a whole-number ratio.

So ? hitthepin?

Problem #2:

2 ^ 2019 must end in an 8, because successive powers of 2 end in (2, 4, 8 or 6) and 2019 modulo 4 is 3 so we would use the third member of that set (an 8).

Similarly, 3 ^ 2019 must end in a 7 because successive powers of 3 end in (3, 9, 7 or 1) and again 2019 modulo 4 is 3 so we would use the third member of that set (a 7).

So 2 ^ 2019 + 3 ^ 2019 must end in a 5 (because any number ending in 8 added to a number ending in 7 gives a sum ending in 5).

For problem #1 I just get a single answer (or rather, two answers that are identical):

x+y=2

x*y=1

(n+d) + (n-d) = 2
2n = 2
n=1

(n+d) * (n-d) = 1
(1+d) * (1-d) = 1
1 - d^2 = 1
d^2 = 0
d = 0

x = 1
y = 1

nice.

Hypothesis: suppose √2 =a/b with a and b integers relatively prime (proper fraction).

then a=b√2 and a^2 = 2b^2 so a must be even. Let a=2c, then (2c)^2 = 2 b^2.

Dividing by 2, we get  2c^2 = b^2. This means that b must be even. So a and b must be both even numbers so a/b was not a proper fraction, which contradict the hypothesis.

So this fraction does not exist.

I still haven't got to actually solving some calculus questions...

Integrate erf(x) dx. erf(x)=integral of (2*e^x^2)dx/sqrt(pi)

Yup. That's the method.

Using exactly two 2s andany mathematical symbols, make 5.

You always take definite integration for this function because there is no function equal to it that does not use an integral. http://mathworld.wolfram.com/Erf.html this is the cumulative of a normal distribution (with proper coefficient) https://en.wikipedia.org/wiki/Cumulative_distribution_function

@Endgame. This looks more like a Martin Gardner puzzle than a real math question. So let's be creative! You can write 2 as you wish like roman style II and another II.

Now you can join them bars to make two V's, which are V and V or 5 and 5 in roman digits.

Now add a times symbol and a square root symbol. √(V x V) = V or five.

Prove that the set of prime numbers is not finite.