Math Quiz

[Deleted member 21661]

@Ellipsis
Testing these as alternate sample questions.

1. The primorial of a number is the product of all primes less than or equal to that natural number. For example, the primorial of 3 is 2 * 3 = 6. Compute the number of natural numbers whose primorial is 210


2. Compute the greatest value of n that satisfies n + 20[√ n] - 20[∛n] = 1000


3. For every subset T of U = {1, 2, 3..., 18}, let s(T) be the sum of the elements of T, with s(θ) defined to be 0. If T is chosen at random among all subsets of U, the probability that s(T) is divisible by 3 is m/n, where m and n are relatively prime positive integers. Find m.


Three questions of varying difficulty, which one would be better? Maybe a mix? Maybe all too easy?
@Psychophilly

 

[Deleted member 25938]

I'm nowhere near a level of solving these. At least not while making another demo quiz.

 

[Deleted member 16960]

1.The prime factorization of 210 is $2\cdot3\cdot5\cdot7$. Therefore, the primorial of 210 is $2\cdot3\cdot5\cdot7 = 210$. In general, the primorial of a number that has at least four distinct prime factors is greater than the number itself. Therefore, the only natural number whose primorial is 210 is 210 itself. Hence, the number of natural numbers whose primorial is 210 is 1.

2.We note that $\sqrt{n}$ and $\sqrt[3]{n}$ are both integers if and only if $n$ is a perfect square and a perfect cube. Letting $k = \sqrt{n}$ and $m = \sqrt[3]{n}$, we can rewrite the given equation as $k^2 + 20k m - 20m^3 = 1000$. Since $k$ and $m$ are integers, we can apply Simon's Favorite Factoring Trick by adding 400 to both sides and factoring to get $(k+10m)^2 - 20^2m^2 = 1400$. We can simplify this equation by dividing both sides by $20^2$ to get $(\frac{k}{20}+ \frac{m}{2})^2 - m^2 = \frac{7}{25}$. Letting $x = \frac{k}{20}+ \frac{m}{2}$, we have $x^2 - m^2 = \frac{7}{25}$, which we can rewrite as $(x-m)(x+m) = \frac{7}{25}$. Since $x$ and $m$ are integers, we must have $x-m$ and $x+m$ both of the same parity. The only pairs of factors of $\frac{7}{25}$ that have the same parity are $\frac{1}{25}\cdot7$ and $-\frac{1}{25}\cdot(-7)$. Therefore, we must have $x-m = \frac{1}{25}$ and $x+m = -\frac{7}{25}$, or $x-m = -\frac{1}{25}$ and $x+m = \frac{7}{25}$. Solving these systems of equations gives us $(x,m) = (\frac{3}{25}, \frac{2}{5})$ or $(-\frac{1}{25}, -\frac{2}{5})$. Therefore, we have $\frac{k}{20}+ \frac{m}{2} = \frac{3}{25}$ or $-\frac{1}{25}$, which gives us $k+m = \frac{6}{5}$ or $\frac{2}{5}$. Since $k$ and $m$ are positive integers, the only solution is $k = 4$ and $m = 1$. Therefore, the greatest value of $n$ that satisfies the equation is $n = k^2m^3 = 4^2\cdot1^3 = 16$.

3. We note that $18\equiv0\pmod3$, so the sum of the elements in any subset of $U$ is congruent to $0$, $1$, or $2$ modulo $3$. Let $S_i$ be the set of all subsets of $U$ whose elements sum to $i$ modulo $3$, for $i=0,1,2$. We want to find the probability that a subset chosen uniformly at random from all subsets of $U$ belongs to $S_0$


DECISIVE KINH VICTORY

 

[Deleted member 21661]

1.The prime factorization of 210 is $2\cdot3\cdot5\cdot7$. Therefore, the primorial of 210 is $2\cdot3\cdot5\cdot7 = 210$. In general, the primorial of a number that has at least four distinct prime factors is greater than the number itself. Therefore, the only natural number whose primorial is 210 is 210 itself. Hence, the number of natural numbers whose primorial is 210 is 1.

2.We note that $\sqrt{n}$ and $\sqrt[3]{n}$ are both integers if and only if $n$ is a perfect square and a perfect cube. Letting $k = \sqrt{n}$ and $m = \sqrt[3]{n}$, we can rewrite the given equation as $k^2 + 20k m - 20m^3 = 1000$. Since $k$ and $m$ are integers, we can apply Simon's Favorite Factoring Trick by adding 400 to both sides and factoring to get $(k+10m)^2 - 20^2m^2 = 1400$. We can simplify this equation by dividing both sides by $20^2$ to get $(\frac{k}{20}+ \frac{m}{2})^2 - m^2 = \frac{7}{25}$. Letting $x = \frac{k}{20}+ \frac{m}{2}$, we have $x^2 - m^2 = \frac{7}{25}$, which we can rewrite as $(x-m)(x+m) = \frac{7}{25}$. Since $x$ and $m$ are integers, we must have $x-m$ and $x+m$ both of the same parity. The only pairs of factors of $\frac{7}{25}$ that have the same parity are $\frac{1}{25}\cdot7$ and $-\frac{1}{25}\cdot(-7)$. Therefore, we must have $x-m = \frac{1}{25}$ and $x+m = -\frac{7}{25}$, or $x-m = -\frac{1}{25}$ and $x+m = \frac{7}{25}$. Solving these systems of equations gives us $(x,m) = (\frac{3}{25}, \frac{2}{5})$ or $(-\frac{1}{25}, -\frac{2}{5})$. Therefore, we have $\frac{k}{20}+ \frac{m}{2} = \frac{3}{25}$ or $-\frac{1}{25}$, which gives us $k+m = \frac{6}{5}$ or $\frac{2}{5}$. Since $k$ and $m$ are positive integers, the only solution is $k = 4$ and $m = 1$. Therefore, the greatest value of $n$ that satisfies the equation is $n = k^2m^3 = 4^2\cdot1^3 = 16$.

3. We note that $18\equiv0\pmod3$, so the sum of the elements in any subset of $U$ is congruent to $0$, $1$, or $2$ modulo $3$. Let $S_i$ be the set of all subsets of $U$ whose elements sum to $i$ modulo $3$, for $i=0,1,2$. We want to find the probability that a subset chosen uniformly at random from all subsets of $U$ belongs to $S_0$


DECISIVE KINH VICTORY

mirin how chat gpt got them wrong

 

[saiya_online]

1.The prime factorization of 210 is $2\cdot3\cdot5\cdot7$. Therefore, the primorial of 210 is $2\cdot3\cdot5\cdot7 = 210$. In general, the primorial of a number that has at least four distinct prime factors is greater than the number itself. Therefore, the only natural number whose primorial is 210 is 210 itself. Hence, the number of natural numbers whose primorial is 210 is 1.

2.We note that $\sqrt{n}$ and $\sqrt[3]{n}$ are both integers if and only if $n$ is a perfect square and a perfect cube. Letting $k = \sqrt{n}$ and $m = \sqrt[3]{n}$, we can rewrite the given equation as $k^2 + 20k m - 20m^3 = 1000$. Since $k$ and $m$ are integers, we can apply Simon's Favorite Factoring Trick by adding 400 to both sides and factoring to get $(k+10m)^2 - 20^2m^2 = 1400$. We can simplify this equation by dividing both sides by $20^2$ to get $(\frac{k}{20}+ \frac{m}{2})^2 - m^2 = \frac{7}{25}$. Letting $x = \frac{k}{20}+ \frac{m}{2}$, we have $x^2 - m^2 = \frac{7}{25}$, which we can rewrite as $(x-m)(x+m) = \frac{7}{25}$. Since $x$ and $m$ are integers, we must have $x-m$ and $x+m$ both of the same parity. The only pairs of factors of $\frac{7}{25}$ that have the same parity are $\frac{1}{25}\cdot7$ and $-\frac{1}{25}\cdot(-7)$. Therefore, we must have $x-m = \frac{1}{25}$ and $x+m = -\frac{7}{25}$, or $x-m = -\frac{1}{25}$ and $x+m = \frac{7}{25}$. Solving these systems of equations gives us $(x,m) = (\frac{3}{25}, \frac{2}{5})$ or $(-\frac{1}{25}, -\frac{2}{5})$. Therefore, we have $\frac{k}{20}+ \frac{m}{2} = \frac{3}{25}$ or $-\frac{1}{25}$, which gives us $k+m = \frac{6}{5}$ or $\frac{2}{5}$. Since $k$ and $m$ are positive integers, the only solution is $k = 4$ and $m = 1$. Therefore, the greatest value of $n$ that satisfies the equation is $n = k^2m^3 = 4^2\cdot1^3 = 16$.

3. We note that $18\equiv0\pmod3$, so the sum of the elements in any subset of $U$ is congruent to $0$, $1$, or $2$ modulo $3$. Let $S_i$ be the set of all subsets of $U$ whose elements sum to $i$ modulo $3$, for $i=0,1,2$. We want to find the probability that a subset chosen uniformly at random from all subsets of $U$ belongs to $S_0$


DECISIVE KINH VICTORY

Go chinaman!

 

[Deleted member 21661]

I'm nowhere near a level of solving these. At least not while making another demo quiz.

The first one is fairly easy, don't let it intimidate you

Think of the first few prime numbers and how it relates to 210 and its factors, because remember that these are just the products of various prime numbers

 

[Deleted member 16960]

Go chinaman!

I'm a Kinh, not a chinaman, retarded red ape (so-called white man)

Kinh (so-called master race)

Chinkite (so-called chinese, japanese, korean, filipino, thai etc)

 

[saiya_online]

I'm a Kinh, not a chinaman, retarded red ape (so-called white man)

Kinh (so-called master race)

Chinkite (so-called chinese, japanese, korean, filipino, thai etc)

Wouldn't japanese be at the top of the mongoloid chart? They contain more aesthetic features no? Japanese on average tend to have a more defined nosebridge

Korean plastic surgery ltbs worship her just because of this feature, that any average white woman has jfl

 

[Deleted member 16960]

Wouldn't japanese be at the top of the mongoloid chart? They contain more aesthetic features no? Japanese on average tend to have a more defined nosebridge
View attachment 2184262
Korean plastic surgery ltbs worship her just because of this feature, that any average white woman has jfl

Jap monkeys are disgusting

I would never waste my Kinh genes by reproducing with oriental subhumans

Only Kinh women and Nordic Scadinavian women are worth passing on my genes

 

[Deleted member 21661]

Jap monkeys are disgusting

I would never waste my Kinh genes by reproducing with oriental subhumans

Only Kinh women and Nordic Scadinavian women are worth passing on my genes

Dw we will create a hapa master race

 

[Deleted member 25938]

The first one is fairly easy, don't let it intimidate you

Think of the first few prime numbers and how it relates to 210 and its factors, because remember that these are just the products of various prime numbers

Ight then...

1) If we attain the product of each ascending prime number consecutively, we get 2 * 3 * 5 * 7, which will give us 210. Would that mean the answer to this question is 4 numbers (?)

3)

If T is chosen at random among all subsets of U, the probability that s(T) is divisible by 3 is m/n

I'm a little confused with this portion of the question. Is this stating that P( s(T) / 3 = m/n ) is true after performing division of s(T) by the chosen element T? Or am I overstepping the problem?

 

[Deleted member 21661]

Ight then...

1) If we attain the product of each ascending prime number consecutively, we get 2 * 3 * 5 * 7, which will give us 210. Would that mean the answer to this question is 4 numbers (?)

Correct

I'm a little confused with this portion of the question. Is this stating that P( s(T) / 3 = m/n ) is true after performing division of s(T) by the chosen element T? Or am I overstepping the problem?

You're overstepping

 

[Deleted member 25938]

You're overstepping

Fak. Okay.

I'm working out rn. I'll attempt to solve it tomorrow. I feel I need to up my math skills, so this'll help a lot.

 

[Deleted member 21661]

Fak. Okay.

I'm working out rn. I'll attempt to solve it tomorrow. I feel I need to up my math skills, so this'll help a lot.

There are pretty good resources. I think that a straight up math quiz isn't as good (as in, just taking random straightforward calculus, geometry, etc questions), as questions that are sort of a puzzle that assume you have a certain baseline of knowledge.
Not exactly what you'd see on tests but they're more fun and probably make you smarter

 

[Deleted member 21661]

@thecel