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)?

[implosion]

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

[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

[StrangerCoug]

Still waking up, but I believe it's 1/4. Will write out a full proof later if someone hasn't by then.

[D3f3nd3r]

I feel like it’s going to be in the ballpark of 1/3. The two things we need aren’t quite independent (2/3 of the small numbers are odd whereas 1/3 of the big numbers are odd).

[StrangerCoug]

Oh, yeah, missed that.

[implosion]

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

I understand, I came up with a heuristic argument that made me think it would be 1/2 anyway asymptotically. But it certainly wasn't rigorous or anything.

[Scigatt]

Making a few assumptions about the relevant probability distribution, I get P = arctan(sqrt(2))/pi ~= 30.4%. Those assumptions could very easily be wrong though.

[Who]

I get the same result, also making some assumptions, but I feel reasonably confident in them.

How I get it:
Let A be the number of times a 1 is rolled minus the number of times a 4 is rolled, let B be the number of times a 3 is rolled minus the number of times a 6 is rolled, let C be the number of times a 2 is rolled minus the number of times a 5 is rolled. The event is now:
A+B>|C|
Also note that A, B, and C are identically distributed and approximately normal with mean 0 via CLT. Normalizing to make them variance 1 doesn't change the equation, so if we assume they are independent it's now
X>|Y| where X is normal with mean 0 variance 2, Y is normal with mean 0 variance 1. We then do basic integration (Set up the integral, note that it's the standard e^(-r^2/2) with r from 0 to infinity but over a sector with a weird angle) and end up with the answer scigatt gave. A,B, and C are obviously not independent, for example if A=N then that means that every roll was a 1 so B and C must both be 0. But they are [assumption] approximately independent because that is unlikely to happen.

[Scigatt]

I believe that a more rigorous approximation for P is the following:



I think from this the limit value of arctan(sqrt(2))/pi follows easily.

[mith]

How I've been keeping busy during a pandemic...

[2 718281828459]

Here's one I saw somewhere: given that x,y are chosen randomly uniformly in [0,1], what is the probability that floor(x/y) is even?

[Mitillos]

Spoiler: Solution for e trillion's problem

Let's see if this works: Take the unit square [0, 1] X [0, 1], which has area 1. The points corresponding to floor(x/y) being even within this square are as follows:
- Everything to the left (or above) of y = x.
- Everything between x = 2y and x = 3y.
- Everything between x = 4y and x = 5y.
etc., forming triangles between the lines x = 2ky and x = (2k + 1)y, for positive integers k.
These triangles will have an apex at (0, 0) and the corresponding base on the line x = 1. The end-points of the base will be at 1/(2k) and 1/(2k + 1) respectively. In other words, the area of the kth triangle will be (1/2) * [1/(2k) - 1/(2k + 1)]. Note that the first triangle (left of y = x) is not included in this calculation and has area 1/2.
Therefore, the total area of the triangles will be 1/2 + Sum (k = 1 to infinity) (1/2) * [1/(2k) - 1/(2k + 1)] = (1/2) * [2 + Sum (n = 1 to infinity) [(-1)^n]/n]. This sum is the negative of the alternating harmonic series, so the area comes to (1/2) * (2 - ln 2) = 1 - (ln 2)/2, which is the probability requested and approximately 65.34%.

[petapan]

How I've been keeping busy during a pandemic...

great channel, got introduced earlier this year as a way to destress. it's very relaxing to watch the solves. i can only do the easy puzzles though, if it's a little hard i always get completely stumped.

[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.

[Ircher]

Here is a fun integral to do from my probability and statistics course last semester:

The probability density function for the standard normal distribution is given by $f_Y(y)=ke^{-y^2/2}$ where $k=1 / \sqrt{2\pi}$. Show that the variance of the standard normal distribution is equal to 1 by computing the following integral:

$$\int_{-\infty}^{\infty}{y^2f_Y(y)dy}-\left(\int_{-\infty}^{\infty}{yf_Y(y)dy}\right)^2.$$

(You will probably want to copy the equations inside dollar signs into Desmos or something since the forums don't support rendering math really well at all.) You can show your work by just posting the latex code involved.

[Scigatt]

The ye^{(-y^2)} term integrates to zero. Using integration by parts with u = y and dv = ye^{(-y^2)} dy, the y²e^{(-y^2)} integral reduces to the standard normal integral, which is 1.

Here's something simple: Prove that every finite field has a quadratic extension.

[Ircher]

You should prove the standard normal integral is 1 as well if you are going to use that argument. (Yes, it is a probability density function, so it equals 1, but the solution should be understandable by someone who has only taken math through multivariable calculus. This makes the problem more interesting.)

[Who]

If the field has odd characteristic:
For any nonzero x, x and -x are distinct and square to the same value. Thus there are (|F|-1)/2 nonzero squares so there must be (|F|-1)/2 nonzero nonsquares, so let c be a no square then x^2-c is irreducible and thus extending by the root of that is a quadratic extension.

If the field is characteristic 2:
Apply the same argument but instead of x and -x take x and x+1, and instead of both have the same square we have that both have the same x(x+1). Thus there’s some c which is not of the form x(x+1), and we have that x(x+1)+c is irreducible.

[Scigatt]

You should prove the standard normal integral is 1 as well if you are going to use that argument. (Yes, it is a probability density function, so it equals 1, but the solution should be understandable by someone who has only taken math through multivariable calculus. This makes the problem more interesting.)

Square the standard integral, change the variable of one to get the 2D version, recast to polar co-ords and then integrate by substitution to 1.

If the field has odd characteristic:
For any nonzero x, x and -x are distinct and square to the same value. Thus there are (|F|-1)/2 nonzero squares so there must be (|F|-1)/2 nonzero nonsquares, so let c be a no square then x^2-c is irreducible and thus extending by the root of that is a quadratic extension.

If the field is characteristic 2:
Apply the same argument but instead of x and -x take x and x+1, and instead of both have the same square we have that both have the same x(x+1). Thus there’s some c which is not of the form x(x+1), and we have that x(x+1)+c is irreducible.

I had another solution in mind.

The number of reducible monic quadratic polynomials in GF(q) is q + (q*(q-1))/2 = (q*(q+1))/2 , while there are q^2 total monic quadratic polynomials. Since q > 1, then there exists an irreducible quadratic which can be used to generate an extension.

[StrangerCoug]

Here's one to breathe some life in this:

Prove that the cube of any integer ending in 5 is always an odd multiple of 125.

[Farren]

Not sure I can do this formally, but I can at least see that it's true.

Spoiler: Math ...

We'll define

n

as an integer. 10

n

+ 5 would yield the set of all integers with the final digit as a 5.

Part one: showing that the cube is divisible by 125.

(10

n

+ 5) * (10

n

+ 5) * (10

n

+ 5) = ...
(100

n²

+ 100

n

+ 25) * (10

n

+ 5) = ...
(1000

n³

+ 1500

n²

+ 750

n

+ 125)

Each term in that final sum is evenly divisible by 125 (since

n

is an integer); therefore, the entire sum is divisible by 125.

Part two: showing that the cube is odd.

Any integer multiplied by 1000, 1500, or 750 has to be divisible by two. 125 is odd. even + even + even + odd = odd. Therefore, the entire sum is odd.