Blackpill IQ Test Math Problem Inspired by @Ellipsis’s Quiz Threads

[thecel]

Bob’s SMV is an integer that’s equal to the sum of 3 non-negative integers:

• face
• height
• status

If face ≤ 5, height ≤ 8, and money ≤ 3, how many ways are there for Bob to have a total SMV of 11?

In other words, f + h + m = 11. How many unique solutions does this equation have, if f ≤ 5, h ≤ 8, m ≤ 3, and all 3 variables are non-negative integers?

---

Tagging the math cucks:

@Ellipsis @sibience @mrriceguy

 

[Deleted member 23218]

Bob’s SMV is the sum of 3 non-negative integers:

• face
• height
• status

If face ≤ 5, height ≤ 8, and money ≤ 3, how many ways are there for Bob to have a total SMV of 11?

In other words, f + h + m = 11. How many unique solutions does this equation have, if f ≤ 5, h ≤ 8, and m ≤ 3?

---

Tagging the math cucks:

@Ellipsis @sibience @mmriceguy

money is cope md im an incel it just makes me lazy

 

[Shieda_Kayn]

I always knew you were smarter than @Collagen or rope

 

[lasnt]

f(x) * h(x) * m(x) = x^0 + 3x^1 + 6x^2 + 10x^3 + 15x^4 + 21x^5 + 25x^6 + 27x^7 + 27x^8 + 24x^9 + 19x^10 + 13x^11 + 6x^12 + 2x^13

The coefficient of x^11 is 13, so there are 13 ways for Bob to have a total SMV of 11 if face ≤ 5, height ≤ 8, and money ≤ 3.

 

[thecel]

f(x) * h(x) * m(x) = x^0 + 3x^1 + 6x^2 + 10x^3 + 15x^4 + 21x^5 + 25x^6 + 27x^7 + 27x^8 + 24x^9 + 19x^10 + 13x^11 + 6x^12 + 2x^13

The coefficient of x^11 is 13, so there are 13 ways for Bob to have a total SMV of 11 if face ≤ 5, height ≤ 8, and money ≤ 3.

Explain this. Why the huge exponents?

 

[lasnt]

Explain this. Why the huge exponents?

How you can code a game but can't understand this

 

[thecel]

How you can code a game but can't understand this

Devs in 2023 are 90-IQ LeetCode bootcamp cucks who have CoPilotGPT installed into VSCode and have 100 Stack Overflow tabs open.

 

[Sprinkles]

 

[Deleted member 18879]

anyone can do STEM btw if u end up in a min wage job its your fault btw

 

[thecel]

f(x) * h(x) * m(x) = x^0 + 3x^1 + 6x^2 + 10x^3 + 15x^4 + 21x^5 + 25x^6 + 27x^7 + 27x^8 + 24x^9 + 19x^10 + 13x^11 + 6x^12 + 2x^13

The coefficient of x^11 is 13, so there are 13 ways for Bob to have a total SMV of 11 if face ≤ 5, height ≤ 8, and money ≤ 3.

How’d you get the equation? What’s x, and why’re f, h, and m functions?

 

[thecel]

anyone can do STEM btw if u end up in a min wage job its your fault btw

I can do STEM, but I can’t do min wage jobs.

 

[Deleted member 18879]

I can do STEM, but I can’t do min wage jobs.

people on this forum think everyone has the iq to do STEM brutal

so many low empathy psychopathic autists here sadge

 

[lasnt]

How’d you get the equation? What’s x, and why’re f, h, and m functions?

Holy
f= face
h= height
m= money


f(x) * h(x) * m(x) = (x^0 + x^1 + x^2 + x^3 + x^4 + x^5) * (x^0 + x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8) * (x^0 + x^1 + x^2 + x^3) "multiply all possible values"

We get this =
f(x) * h(x) * m(x) = x^0 + 3x^1 + 6x^2 + 10x^3 + 15x^4 + 21x^5 + 25x^6 + 27x^7 + 27x^8 + 24x^9 + 19x^10 + 13x^11 + 6x^12 + 2x^13

so there are 13 ways for Bob to have a total SMV of 11

 

[Deleted member 25938]

I'm assuming that we're leaving out 0 as an argument?

 

[thecel]

I'm assuming that we're leaving out 0 as an argument?

non-negative integers

0 is a non-negative integer.

 

[thecel]

Holy
f= face
h= height
m= money


f(x) * h(x) * m(x) = (x^0 + x^1 + x^2 + x^3 + x^4 + x^5) * (x^0 + x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8) * (x^0 + x^1 + x^2 + x^3) "multiply all possible values"

We get this =
f(x) * h(x) * m(x) = x^0 + 3x^1 + 6x^2 + 10x^3 + 15x^4 + 21x^5 + 25x^6 + 27x^7 + 27x^8 + 24x^9 + 19x^10 + 13x^11 + 6x^12 + 2x^13

so there are 13 ways for Bob to have a total SMV of 11

but why do you raise x (where does this extra variable x come from?) to the power of all possible values? Doesn’t make any sense to me.

f, h, and m are variables, not functions.

 

[thecel]

Bob’s SMV is an integer that’s equal to the sum of 3 non-negative integers:

• face
• height
• status

Sorry about this mistake. Status should be money instead.

 

[thecel]

I'm assuming that we're leaving out 0 as an argument?

I’m pissed off about wrongly typing “status” when it should be money. Or for not making the m variable status to match the bullet.

 

[Deleted member 25938]

0 is a non-negative integer

Fuck you, know I have to redo all my calculations.

Also, 0 shouldn't be applicable, as it wouldn't make sense to have 0 PSL, 0 height, or 0 status/money irl.

 

[thecel]

Fuck you, know I have to redo all my calculations.

Also, 0 shouldn't be applicable, as it wouldn't make sense to have 0 PSL, 0 height, or 0 status/money irl.

I have 0 money.

 

[Deleted member 25938]

I have 0 money.

But you have some status, which still negates 0. 0 is not possible in any of these categories; 0 is a limit.

 

[thecel]

But you have some status, which still negates 0.

not even gonna try to debunk this

 

[Deleted member 25938]

not even gonna try to debunk this

"Muh bluepill"

Think about it mathematically. The only way your status could truly have a value of 0 is if you never existed in the first place, or if you died/killed yourself and a very large interval of time passed such that nobody in the coming generations knows you or about you.

0 is a limit that is approached. No need to bring in blackpill/bluepill dichotomy when it's about the mathematical legitimacy/applicability of the question you presented.

 

[lasnt]

but why do you raise x (where does this extra variable x come from?) to the power of all possible values? Doesn’t make any sense to me.

f, h, and m are variables, not functions.

Extra variables was just extra examples for you to understand after 13x^11...

For example
coefficient of 6x^3

"6" represent all possibilities to obtain 3 SMV after multiplying f,h,m polynomials

 

[thecel]

"Muh bluepill"

Think about it mathematically. The only way your status could truly have a value of 0 is if you never existed in the first place, or if you died/killed yourself and a very large interval of time passed such that nobody in the coming generations knows you or about you.

0 is a limit that is approached. No need to bring in blackpill/bluepill dichotomy when it's about the mathematical legitimacy/applicability of the question you presented.

Some people have negative status.

 

[thecel]

Extra variables was just extra examples for you to understand after 13x^11...

For example
coefficient of 6x^3

"6" represent all possibilities to obtain 3 SMV after multiplying f,h,m polynomials

Why does the variable x exist?

 

[Deleted member 25938]

Some people have negative status.

Negative status is status nonetheless.

Either for the parameters of the question the negative status will have to be taken into account as an absolute value, or your entire question is flawed.

 

[lasnt]

Why does the variable x exist?

Because function needs variable

 

[thecel]

Negative status is status nonetheless.

Either for the parameters of the question the negative status will have to be taken into account as an absolute value, or your entire question is flawed.

The question assumes that these are the upper and lower bounds for Bob’s potential. E.g. the tallest height he can get to is 8.

11 is the SMV threshold to enter a vaginal canal.

 

[thecel]

Because function needs variable

f, h, and m are variables, not functions.

 

[lasnt]

How are they variables if they have set of values are you retard son ?

 

[thecel]

How are they variables if they have set of values are you retard son ?

They’re variables with constraints.

A function needs a set of inputs and a set of outputs.

 

[Deleted member 25938]

Alrighty.

Assuming 0 counts as a non-negative integer, we get the following combinations (I did this using brute power, no formulae or systems, so there might be errors or some I missed):

8 3 0
8 2 1
8 1 2
8 0 3
7 4 0
7 3 1
7 2 2
7 1 3
6 5 0
6 4 1
6 3 2
6 2 3
5 5 1
5 4 2
5 3 3
4 5 2
4 4 3
3 5 3

Leaving us with 18 total combinations Bob can have to attain an SMV of 11.

 

[lasnt]

They’re variables with constraints.

A function needs a set of inputs and a set of outputs.

I already put set of inputs and outputs

the input to the function h(x) is an integer between 0 and 8

the output of the function is the sum of possible values for Bob's height that satisfy the given constraints.

So manlet h(0) would return 1 because there is only one way that Bob can have a height of 0, which is manlet

 

[thecel]

Alrighty.

Assuming 0 counts as a non-negative integer, we get the following combinations (I did this using brute power, no formulae or systems, so there might be errors or some I missed):

8 3 0
8 2 1
8 1 2
8 0 3
7 4 0
7 3 1
7 2 2
7 1 3
6 5 0
6 4 1
6 3 2
6 2 3
5 5 1
5 4 2
5 3 3
4 5 2
4 4 3
3 5 3

Leaving us with 18 total combinations Bob can have to attain an SMV of 11.

Thanks for checking my work for me, manual laborcuck.

My Solution:

Spoiler: My Solution

(5+3-1)!/(5!*(3-1)!)-(2+2-1)!/(2!*(2-1)!) = 18

 

[lasnt]

Thanks for checking my work for me, manual laborcuck.

My Solution:

Spoiler: My Solution

(5+3-1)!/(5!*(3-1)!)-(2+2-1)!/(2!*(2-1)!) = 18

Brutal got cucked by two Mongolianoids

 

[Deleted member 21661]

I have 0 money.

No you don't. You're wearing clothes, using a computer, live in a house, and have parents with money. So 0 shouldn't really be an option here which would change the answer.

 

[qsabathi]

Pretty sure it’s just C(16 , 11) which is 4368 ways.

 

[qsabathi]

Pretty sure it’s just C(16 , 11) which is 4368 ways.

Whoops nvm, I’m over counting.

 

[Deleted member 29261]

Thx for the looksmaxxing advice

 

[thecel]

Alrighty.

Assuming 0 counts as a non-negative integer, we get the following combinations (I did this using brute power, no formulae or systems, so there might be errors or some I missed):

8 3 0
8 2 1
8 1 2
8 0 3
7 4 0
7 3 1
7 2 2
7 1 3
6 5 0
6 4 1
6 3 2
6 2 3
5 5 1
5 4 2
5 3 3
4 5 2
4 4 3
3 5 3

Leaving us with 18 total combinations Bob can have to attain an SMV of 11.

High-IQ, Midwit, and Low-IQ Solutions:

High-IQ Solution:

I messed up the solution in the previous reply, but I found a way to use math gymnastics to make it correct, so I don’t have to admit I was wrong (yay, ego boost!).

Spoiler: solution

Using the formula (k+n-1)!/(k!*(n-1)!) = number of combinations with replacement

5 + 8 + 3 = 16. The target is 11, so we need to subtract five ones from the variables. We have 5 objects to distribute amongst the 3 variables.

(5+3-1)!/(5!*(3-1)!) = 21

But that includes combinations where m is negative. The minimum m in the above combos is -2. We need to subtract the number of combinations of f + h + m = 11 where -2 ≤ m < 0, f ≥ 5, and h ≥ 8.

Let’s introduce a variable called w to non-negatify m. w = m + 2. m is negative when w < 2. Thus the maximum integer value of w (for m to be negative) is 1.

Then we calculate the number of unique solutions for f + h + w = 13 where f ≤ 5, h ≤ 8, and w ≤ 1. The target sum is 13, not 11, because w = m + 2. The sum of the 3 variables’ maximum allowed values is 14, which means we have one excess one to subtract in order to make the sum 13.

There’re 3 ways to do that.

(5+3-1)!/(5!*(3-1)!) - 3 = 18

My old solution is (5+3-1)!/(5!*(3-1)!)-(2+2-1)!/(2!*(2-1)!) = 18. Is it wrong? Not if you do mental gymnastics:

How many numbers are ≥ 1 and ≤ 3? Obviously 3 numbers: 1, 2, and 3. Here’s how to convert that to a combinatorics problem:

Imagine an a lineup of 3 Looksmax members. At least 1 is autistic. How many possible rates of autism are there in this sample of 3?

We assume 1 member is autistic because 1 is the minimum. Then the next 2 members can either be autistic or not autistic. The order doesn’t matter, and the labels “autistic” and “not autistic” can be repeated. So we use the same formula:

(k+n-1)!/(k!*(n-1)!) = number of combinations with replacement

2 members; 2 labels. The labels are the n; the members are the k.

(2+2-1)!/(2!*(2-1)!) = 3

We have now cheated my old solution into being correct.

Midwit Solution:

Spoiler: Solution

5 + 8 = 13.

When m = 0, f + h = 11.
13 - 11 = 2.
There’re 3 ways to distribute the 2 excess ones between f and h.

When m = 1, f + h = 10.
13 - 10 = 3.
There’re 4 ways to distribute the 3 excess ones between f and h.

When m = 2, f + h = 9.
13 - 9 = 4.
There’re 5 ways to distribute the 4 excess ones between f and h.

When m = 3, f + h = 8.
13 - 8 = 5.
There’re 6 ways to distribute the 5 excess ones between f and h.

3+4+5+6 = 18.

Low-IQ Solution:

Spoiler: Solution

your solution and my previous solution

 

[Deleted member 25938]

High-IQ, Midwit, and Low-IQ Solutions:

High-IQ Solution:

I messed up the solution in the previous reply, but I found a way to use math gymnastics to make it correct, so I don’t have to admit I was wrong (yay, ego boost!).

Spoiler: solution

Using the formula (k+n-1)!/(k!*(n-1)!) = number of combinations with replacement

5 + 8 + 3 = 16. The target is 11, so we need to subtract five ones from the variables. We have 5 objects to distribute amongst the 3 variables.

(5+3-1)!/(5!*(3-1)!) = 21

But that includes combinations where m is negative. The minimum m in the above combos is -2. We need to subtract the number of combinations of f + h + m = 11 where -2 ≤ m < 0, f ≥ 5, and h ≥ 8.

Let’s introduce a variable called w to non-negatify m. w = m + 2. m is negative when w < 2. Thus the maximum integer value of w (for m to be negative) is 1.

Then we calculate the number of unique solutions for f + h + w = 13 where f ≤ 5, h ≤ 8, and w ≤ 1. The target sum is 13, not 11, because w = m + 2. The sum of the 3 variables’ maximum allowed values is 14, which means we have one excess one to subtract in order to make the sum 13.

There’re 3 ways to do that.

(5+3-1)!/(5!*(3-1)!) - 3 = 18

My old solution is (5+3-1)!/(5!*(3-1)!)-(2+2-1)!/(2!*(2-1)!) = 18. Is it wrong? Not if you do mental gymnastics:

How many numbers are ≥ 1 and ≤ 3? Obviously 3 numbers: 1, 2, and 3. Here’s how to convert that to a combinatorics problem:

Imagine an a lineup of 3 Looksmax members. At least 1 is autistic. How many possible rates of autism are there in this sample of 3?

We assume 1 member is autistic because 1 is the minimum. Then the next 2 members can either be autistic or not autistic. The order doesn’t matter, and the labels “autistic” and “not autistic” can be repeated. So we use the same formula:

(k+n-1)!/(k!*(n-1)!) = number of combinations with replacement

2 members; 2 labels. The labels are the n; the members are the k.

(2+2-1)!/(2!*(2-1)!) = 3

We have now cheated my old solution into being correct.

Midwit Solution:

Spoiler: Solution

5 + 8 = 13.

When m = 0, f + h = 11.
13 - 11 = 2.
There’re 3 ways to distribute the 2 excess ones between f and h.

When m = 1, f + h = 10.
13 - 10 = 3.
There’re 4 ways to distribute the 3 excess ones between f and h.

When m = 2, f + h = 9.
13 - 9 = 4.
There’re 5 ways to distribute the 4 excess ones between f and h.

When m = 3, f + h = 8.
13 - 8 = 5.
There’re 6 ways to distribute the 5 excess ones between f and h.

3+4+5+6 = 18.

Low-IQ Solution:

Spoiler: Solution

your solution and my previous solution

See @BongMog I told you I'm low IQ.

 

[Deleted member 25938]

Thanks for checking my work for me, manual laborcuck.

If this is considered manual labor then you are definitely a NEETcel.

 

[BongMog]

See @BongMog I told you I'm low IQ.

You IQ mog me to death

 

[Deleted member 25938]

High-IQ, Midwit, and Low-IQ Solutions:

High-IQ Solution:

I messed up the solution in the previous reply, but I found a way to use math gymnastics to make it correct, so I don’t have to admit I was wrong (yay, ego boost!).

Spoiler: solution

Using the formula (k+n-1)!/(k!*(n-1)!) = number of combinations with replacement

5 + 8 + 3 = 16. The target is 11, so we need to subtract five ones from the variables. We have 5 objects to distribute amongst the 3 variables.

(5+3-1)!/(5!*(3-1)!) = 21

But that includes combinations where m is negative. The minimum m in the above combos is -2. We need to subtract the number of combinations of f + h + m = 11 where -2 ≤ m < 0, f ≥ 5, and h ≥ 8.

Let’s introduce a variable called w to non-negatify m. w = m + 2. m is negative when w < 2. Thus the maximum integer value of w (for m to be negative) is 1.

Then we calculate the number of unique solutions for f + h + w = 13 where f ≤ 5, h ≤ 8, and w ≤ 1. The target sum is 13, not 11, because w = m + 2. The sum of the 3 variables’ maximum allowed values is 14, which means we have one excess one to subtract in order to make the sum 13.

There’re 3 ways to do that.

(5+3-1)!/(5!*(3-1)!) - 3 = 18

My old solution is (5+3-1)!/(5!*(3-1)!)-(2+2-1)!/(2!*(2-1)!) = 18. Is it wrong? Not if you do mental gymnastics:

How many numbers are ≥ 1 and ≤ 3? Obviously 3 numbers: 1, 2, and 3. Here’s how to convert that to a combinatorics problem:

Imagine an a lineup of 3 Looksmax members. At least 1 is autistic. How many possible rates of autism are there in this sample of 3?

We assume 1 member is autistic because 1 is the minimum. Then the next 2 members can either be autistic or not autistic. The order doesn’t matter, and the labels “autistic” and “not autistic” can be repeated. So we use the same formula:

(k+n-1)!/(k!*(n-1)!) = number of combinations with replacement

2 members; 2 labels. The labels are the n; the members are the k.

(2+2-1)!/(2!*(2-1)!) = 3

We have now cheated my old solution into being correct.

Midwit Solution:

Spoiler: Solution

5 + 8 = 13.

When m = 0, f + h = 11.
13 - 11 = 2.
There’re 3 ways to distribute the 2 excess ones between f and h.

When m = 1, f + h = 10.
13 - 10 = 3.
There’re 4 ways to distribute the 3 excess ones between f and h.

When m = 2, f + h = 9.
13 - 9 = 4.
There’re 5 ways to distribute the 4 excess ones between f and h.

When m = 3, f + h = 8.
13 - 8 = 5.
There’re 6 ways to distribute the 5 excess ones between f and h.

3+4+5+6 = 18.

Low-IQ Solution:

Spoiler: Solution

your solution and my previous solution

Over. But ig I still got it correct, imagine if I got the answer incorrect

Spoiler

Also I mog you

 

[Deleted member 25938]

You IQ mog me to death
View attachment 2200896

Well, in all fairness, my mathematical skills are not developed enough. I probably mog almost everyone here in musical IQ though.

 

[DefinitelyNT]

Gee I wonder what the Asians of .org are up to today

Oh

 

[Struggler03]

OP is homeworkmaxxing

 

[thecel]

OP is homeworkmaxxing

I finished a final exam 3 days late and couldn’t turn it in

 

[thecel]

Well, in all fairness, my mathematical skills are not developed enough. I probably mog almost everyone here in musical IQ though.

My musical IQ is 101, and I scored 87 on a subtest. Never began for earcels.