real math problems

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Soupboy0

222 posts

Soupboy0

#1 h Mar 25, 2025, 11:40 PM

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Ill be posting questions once in a while. Here's the first question:

What fraction of numbers from $1$ to $1000$ have the digit $7$ and are divisible by $3$ ?

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AbhayAttarde01

1467 posts

AbhayAttarde01

#2 h Mar 26, 2025, 1:12 AM

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i was typing up a casework solution and then realized I could do it a different way lol
kind of casework?
this is my first attempt at this if I'm wrong I'll see where I got it wrong

This post has been edited 1 time. Last edited by AbhayAttarde01, Mar 26, 2025, 1:12 AM

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Soupboy0

222 posts

Soupboy0

#4 h Mar 26, 2025, 11:06 PM

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Next problem: Find the exact value of $\lfloor(\frac{8}{3})^{10}\rfloor$ without a calculator.

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maromex

130 posts

maromex

#5 h Mar 26, 2025, 11:38 PM

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Click to reveal hidden text
hide tag no work we do a little trolling

This post has been edited 1 time. Last edited by maromex, Mar 26, 2025, 11:38 PM

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iwastedmyusername

50 posts

iwastedmyusername

#6 h Mar 27, 2025, 2:03 AM

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maromex wrote:

Click to reveal hidden text
hide tag no work we do a little trolling

yea i was thinking the same thing
i wonder if therers a non bash way to do it

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Soupboy0

222 posts

Soupboy0

#7 h Mar 27, 2025, 3:41 AM

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official approved solution

This post has been edited 2 times. Last edited by Soupboy0, Mar 27, 2025, 7:18 PM

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Soupboy0

222 posts

Soupboy0

#8 h Mar 27, 2025, 7:15 PM

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3) A $m \times n$ grid is made. $mn$ unit squares are created by separating the side with length $m$ in $m$ portions and separating the side with length $n$ in $n$ portions. Find, with proof, a formula for how many rectangles with sides parallel to the grid can be created in terms of $m$ and $n$ . For example, when $m = 2$ and $n = 2$ , it can be found by casework that $9$ rectangles can be created.

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cheltstudent

549 posts

cheltstudent

#9 h Mar 27, 2025, 7:25 PM

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sol

This post has been edited 2 times. Last edited by cheltstudent, Mar 27, 2025, 7:29 PM
Reason: gg

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elizhang101412

1191 posts

elizhang101412

#10 h Mar 27, 2025, 8:01 PM

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cheltstudent wrote:

sol

bro you are not slick with that ai usage :skull:

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cheltstudent

549 posts

cheltstudent

#11 h Mar 27, 2025, 8:02 PM

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wut... I have a parental controlled computer

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Soupboy0

222 posts

Soupboy0

#12 h Mar 28, 2025, 2:18 PM

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4) If positive integers $(a, b, c, d)$ satisfy $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=\frac{13}{40}+\frac{13}{42}$ , find the ordered pair $(a, b, c, d)$

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Mathdreams

1443 posts

Mathdreams

#13 h Mar 28, 2025, 2:19 PM

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Answer

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fruitmonster97

2415 posts

fruitmonster97

#14 h Mar 28, 2025, 2:53 PM

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that's the trivial sol @above, but i'd be suprised if this didn't have multiple sols:
greedy algorithm:
a=2: then 1/b+1/c+1/d=133/840. b=8 yields 1/c+1/d=1/105. The sols here can be found using sfft, some trivial ones for (c,d) are (106,105*106) and (210,210). b=9 and c=9 yields d=45, for example.
a=3: (3,4,21,840)
a=4: (4,4,8,105), (4,5,6,56)

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elizhang101412

1191 posts

elizhang101412

#15 h Mar 28, 2025, 10:49 PM

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cheltstudent wrote:

wut... I have a parental controlled computer

can you stop trying this bruh literally all the text is formatted like an ai

This post has been edited 1 time. Last edited by elizhang101412, Mar 28, 2025, 10:50 PM

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Soupboy0

222 posts

Soupboy0

#16 h Mar 28, 2025, 11:19 PM

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5) If $p$ and $q$ are positive integers, what is the probability that $5^p+7^q$ is divisible by $6$ ?

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corgi61

156 posts

corgi61

#17 h Mar 28, 2025, 11:50 PM

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Solution

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iwastedmyusername

50 posts

iwastedmyusername

#18 h Mar 29, 2025, 12:42 AM

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cheltstudent wrote:

sol

idk man seems kinda sus

Attachments:

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cheltstudent

549 posts

cheltstudent

#19 h Mar 29, 2025, 12:43 AM

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ya, it's same method but completely different wording

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RandomMathGuy500

53 posts

RandomMathGuy500

#20 h Mar 29, 2025, 1:15 AM

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cheltstudent wrote:

sol

yeah it's def AI

Attachments:

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cheltstudent

549 posts

cheltstudent

#21 h Saturday at 10:57 AM

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broooooooooooooooooo, i literally didn't

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maxamc

481 posts

maxamc

#22 h Saturday at 12:39 PM

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cheltstudent wrote:

broooooooooooooooooo, i literally didn't

lying is banned

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Soupboy0

222 posts

Soupboy0

#23 h Saturday at 10:04 PM

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6) Leeoz posts a daily problem every $19, 20$ , or $21$ hours, with equal probability. Soupboy0 posts a daily problem every $18, 21$ , or $22$ hours, with equal probability. If both Leeoz and Soupboy0 post at the same time, what is the expected time that will elapse in hours before Leeoz and Soupboy0 post again?

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elizhang101412

1191 posts

elizhang101412

#24 h Yesterday at 1:03 AM

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is it Click to reveal hidden text

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Soupboy0

222 posts

Soupboy0

#25 h Yesterday at 1:52 PM

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i have no idea lol i just thought that looked like a cool problem

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Bummer12345

116 posts

Bummer12345

#26 h Yesterday at 2:47 PM

Y by

probably ai

In a previous post, in bogus proof marathon, he posted

cheltstudent wrote:

When I tried to post this, the thread was deleted :noo:

S1950

which is probably AI, as nobody writing a solution would write a "Final Analysis" portion, let alone using this wording; not to mention usage of words such as "scrutinize" and "erroneous" in a mathematical proof. This demonstrates that he has used ai before

cheltstudent wrote:

Initialization:
In an $m \times n$ grid, each unit square is created by dividing the sides of lengths $m$ and $n$ . Consequently, we have $m+1$ horizontal lines and $n+1$ vertical lines:

"initialization" :/

cheltstudent wrote:

Total Number of Rectangles:
To find the total number of rectangles $R$ that can be formed, we multiply the number of ways to choose the horizontal lines and the ways to choose the vertical lines:
$R = \binom{m+1}{2} \times \binom{n+1}{2}/4$
Substituting the expressions we derived:
$R = \left(\frac{(m+1)m}{2}\right) \times \left(\frac{(n+1)n}{2}\right) = \frac{(m+1)m(n+1)n}{4}$
Final Conclusion:
Thus, the total number of rectangles that can be formed in an $m \times n$ grid is given by the formula:
$\boxed{\frac{(m+1)m(n+1)n}{4}}$

where does the " $/4$ " come from in the 6th line? It completely disappeared in the boxed answer. If it was kept, there would be a 16 in the denominator, which is incorrect. a person writing a proof for a problem like this would likely notice this error, or more likely

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cheltstudent

549 posts

cheltstudent

#27 h Yesterday at 2:51 PM

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Bogus one was, but not this.
This one came from my cousin who might have used AI idk.
Anyway, BACK ON TOPIC

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cheltstudent

549 posts

cheltstudent

#28 h Yesterday at 2:55 PM

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Reason I used AI for Bogus Proof: It was never an MSM problem, and people would have probably taken ages to solve

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iwastedmyusername

50 posts

iwastedmyusername

#29 h Yesterday at 3:00 PM

Y by

ai didnt even solve it...
it just said 2p choose p is congruent to 2 mod p^3 but 7 isnt divisible by 3 without explaining why

This post has been edited 1 time. Last edited by iwastedmyusername, Yesterday at 3:03 PM
Reason: skibidi toilet

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cheltstudent

549 posts

cheltstudent

#30 h Yesterday at 3:07 PM

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Soupboy0 wrote:

6) Leeoz posts a daily problem every $19, 20$ , or $21$ hours, with equal probability. Soupboy0 posts a daily problem every $18, 21$ , or $22$ hours, with equal probability. If both Leeoz and Soupboy0 post at the same time, what is the expected time that will elapse in hours before Leeoz and Soupboy0 post again?