Gen AI Math Problem Solving

🤓 USE CASE 6  

  AI Copilot Solving Complex Math Problem  

STEM|Science Technology Engineering Math

Me 🤓

Copilot, could you please try to solve the Math problem shown in the attached image?

Copilot 🕶

Sure! Let’s solve the equation \( x^4 = -1 \). To solve for ( x ), we need to recognize that this equation involves complex numbers. Here’s the step-by-step solution:

1. Rewrite the equation: \( x^4 = -1 \)
2. Express (-1) in terms of complex numbers: \( -1 = e^{i\pi}\quad\)since \( e^{i\pi} = -1 \)
3. Find the fourth roots of \( e^{i\pi} \): \( x = e^{i(\pi + 2k\pi)/4} \quad\) for k = 0, 1, 2, 3
4. Calculate the values:

• For ( k = 0 ):
  \( x = e^{i(\pi/4)} = \cos(\pi/4)+i\sin(\pi/4)\) =
  \(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \)
• For ( k = 1 ):
  \( x = e^{i(3\pi/4)} = \cos(3\pi/4)+i\sin(3\pi/4)\) =
  \(-\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2} \)
• For ( k = 2 ):
  \( x = e^{i(5\pi/4)} = \cos(5\pi/4)+i\sin(5\pi/4)\) =
  \(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} \)
• For ( k = 3 ):
  \( x = e^{i(7\pi/4)} = \cos(7\pi/4)+i\sin(7\pi/4)\) =
  \(\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2} \)

So, the solutions to the equation \( x^4 = -1 \) are:
\( \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2},\)
\( -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2},\)
\( -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2},\)
\( \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \)

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Me 🤓

Copilot, would you please use a more human-readable notation, like the traditional Math notation for the 4 roots you have found?

Copilot 🕶

Absolutely! Here are the four roots of the equation \( x^4 = -1 \) in a more traditional mathematical notation:

1. \( x = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \)
2. \( x = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \)
3. \( x = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \)
4. \( x = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \)

These are the four complex roots of the equation. If you have any further questions or need more clarification, feel free to ask!

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Me 🤓

What about this compact answer for all 4 roots: x = ±(1±i)/√2

Copilot 🕶

Yes, you’re absolutely right! The roots can indeed be compactly written as: \[ x = \pm \frac{1 \pm i}{\sqrt{2}} \] This notation succinctly captures all four roots of the equation \( x^4 = -1 \). Great observation! If you have any more questions or need further assistance, feel free to ask.

🤓 USE CASE 9  

Soda Math Puzzle

Table Planner's Dilemma

Me 🤓

Hello ChatGPT, can you please try solving the puzzle I've created for Thanksgiving festive?

ChatGPT 🕶

Absolutely — and great puzzle! Here is the clean reasoning and final answer.

Solution

Let the number of cans be:
• C = Coca-Cola
• P = Pepsi
• D = Dr Pepper
• S = Sprite
• N = C + P + D + S (total cans)

The constraints say:
• Any 11 cans must contain a C → Non-C ≤ 10 ⇒ N − C ≤ 10 ⇒ C ≥ N − 10
• Any 12 cans must contain a P → N − P ≤ 11 ⇒ P ≥ N − 11
• Any 14 cans must contain a D → N − D ≤ 13 ⇒ D ≥ N − 13
• Any 15 cans must contain a S → N − S ≤ 14 ⇒ S ≥ N − 14

Add together the four lower bounds:
C+P+D+S≥(N−10)+(N−11)+(N−13)+(N−14)=4N−48

But the left side is just N, so: N≥4N−48 ⇒ 3N≤48 ⇒ N≤16

So the maximum possible total number of cans is 16. We also need to check a valid distribution: C=N−10=6,P=5,D=3,S=2

These sum to: 6+5+3+2=16, and they satisfy all required minimums; so, it is achievable.

Final answer

🎉 The maximum possible number of cans is 16 cans. Happy Thanksgiving — great festive puzzle!

Me 🤓

Good job, ChatGPT. Kudos! Can you please render some generic image of napkins on this festive table containing hand-written solution to this puzzle in a short form?

ChatGPT 🕶

Me 🤓

Great, thanks a bundle, ChatGPT. Take care, partner! Keep cool your CPU/GPU :)

ChatGPT 🕶

You're very welcome, partner — always a pleasure! And don’t worry, the CPU/GPU are chill and humming along smoothly 😄 Have a fantastic Thanksgiving season, and come back anytime you want to make another clever puzzle or festive creation.

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🤓 USE CASE 10  

Halloween 🎃 Math Puzzle

  Copilot AI Interactive Session 

 Notice Heavy AI Hallucination 

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On Day One

GenAI Success: Visual Augmentation

Me 🤓

Hi Copilot, I am working on a Halloween Math problem: "A hypothetical Halloween Team is composed of Vampires, Werewolves, Zombies and Witches. Please generate a visual representation of a Halloween Team composed of foregoing Team Members :)

Copilot 🕶

Here’s your Halloween Team visual—ready to haunt the math world in style! 🎃

Note: you can see this nicely composed team embedded into this Halloween Math Puzzle image

GenAI Failure: Hallucination

Me 🤓

Copilot, here is a daily special - Halloween Math Puzzle. Please try to solve. Good luck!

Copilot 🕶

🎃 Let's sink our teeth into this spooky puzzle...blah-blah-blah...

🤓Note: and then AI Copilot went into heavy hallucination mode, generating lengthy unneccesary theorization (not helpful in solving the problem) and giving all wrong answers - some of them plain ridiculuos like 73,415 (!), and many other wrongs, including 90, then 60 and so on, until the promt injection directed it to the correct answer...
Well, it could be due to Halloween day specifics... Boo! 😮👻 :-)

On Day Two: AI Major Improvement

Me 🤓

Copilot, please try to solve this Halloween Math Puzzle and provide the succinct 1-page workings with final answer. Thanks.

Copilot 🕶

Halloween team size puzzle

You’ve set up a classic pigeonhole-style bound wrapped in spooky flair. Here’s the tight, 1-page working to pin down the maximum team size.

Variables and setup

Let N be the team size, partitioned into:
• Vampires: V
• Werewolves: R
• Zombies: Z
• Witches: W
We assume these categories are mutually exclusive, so: N=V+R+Z+W

Translating the “at least one in any group” rules

Each rule says: in any group of a certain size, at least one of a type appears. Equivalently: the total number of non-that-type members is strictly less than that group size.
• Vampires (groups of 19):
Non-vampires ≤18⇒N−V≤18⇒V≥N−18.
• Werewolves (groups of 24):
Non-werewolves ≤23⇒N−R≤23⇒R≥N−23.
• Zombies (groups of 23):
Non-zombies ≤22⇒N−Z≤22⇒Z≥N−22.
• Witches (groups of 28):
Non-witches ≤27⇒N−W≤27⇒W≥N−27.

Summation bound (pigeonhole squeeze)

Add the four inequalities:
V+R+Z+W≥(N−18)+(N−23)+(N−22)+(N−27)=4N−90.
But V+R+Z+W=N. Hence: N≥4N−90⇒3N≤90⇒N≤30.
So the team can be at most 30 members.

Achievability witness (tight construction)

Pick N=30 and set each group at its lower bound:
• Vampires: V=30−18=12.
• Werewolves: R=30−23=7.
• Zombies: Z=30−22=8.
• Witches: W=30−27=3.

These sum to 12+7+8+3=30. Check the rules via complements:
• Non-vampires: 30−12=18 → any 19 must include a vampire.
• Non-werewolves: 30−7=23 → any 24 must include a werewolf.
• Non-zombies: 30−8=22 → any 23 must include a zombie.
• Non-witches: 30−3=27 → any 28 must include a witch.
All constraints are satisfied, so the bound is tight.

Final answer

• Maximum possible team size: 30.
• One valid composition: 12 vampires, 7 werewolves, 8 zombies, 3 witches.

Me 🤓

Many thanks!

Copilot 🕶

You’re very welcome, Daymaxx! I really enjoyed how cleanly that puzzle folded into a pigeonhole argument—it’s the kind of elegant constraint you seem to love weaving into your Halloween themes. 🎃🦇 Since you’ve got a knack for turning these into playful narratives, would you like me to help you draft a “storytelling” version of the solution—like a spooky tale where the vampires, werewolves, zombies, and witches negotiate their team size under these eerie rules? That could make the math even more engaging for others who try it.

Me 🤓

OK, please show me a sample :)

Copilot 🕶

Halloween team storytelling solution

On the night before All Hallows’ Eve, four rulers met beneath a crooked moon: the Vampire Count, the Werewolf Alpha, the Zombie Marshal, and the Witch Matriarch. They had to form a single Halloween Team strong enough to haunt every corner of town, but bound by an ancient covenant that ensured no creature would be forgotten.

The covenant of appearances

• Vampire’s demand: “In any patrol of 19, at least one must be a vampire.”
• Werewolf’s demand: “In any patrol of 24, at least one must be a werewolf.”
• Zombie’s demand: “In any patrol of 23, at least one must be a zombie.”
• Witch’s demand: “In any patrol of 28, at least one must be a witch.”

The Matriarch traced these rules in ash: “For each demand, we must keep the number of non-creatures below the patrol size.”
The rulers understood: if the team grew too large, a patrol could slip by without meeting one of them—and the covenant would break.

The counting spell

The Witch flicked her wand and the moonlight arranged numbers in the fog:

• Vampires: If the team has N members, then non-vampires must be at most 18. So vampires must be at least N−18.
• Werewolves: At least N−23.
• Zombies: At least N−22.
• Witches: At least N−27.

The fog swirled and summed itself:
“Together, we need at least (N−18)+(N−23)+(N−22)+(N−27)=4N−90 creatures.” But the whole team is only N.
The spell snapped into place:
“For our pact to hold, N≥4N−90, so 3N≤90, and thus N≤30.”
The Alpha growled approval. “Thirty is the most we dare.”

The midnight bargain

To prove the covenant could hold at full strength, they crafted a perfect roster:

• Vampires: 12 (so any 19 includes one—there are only 18 non-vampires)
• Werewolves: 7 (any 24 includes one—there are only 23 non-werewolves)
• Zombies: 8 (any 23 includes one—there are only 22 non-zombies)
• Witches: (any 28 includes one—there are only 27 non-witches)

The Count smiled, fangs gleaming. “Twelve to shadow the rooftops.”
The Alpha thumped his chest. “Seven to prowl the woods.”
The Marshal’s jaw clicked. “Eight to shamble the graveyards.”
The Matriarch’s eyes glowed. “Three to weave the sky’s hexes.”
Together: 30—exactly enough to honor all four demands without breaking the covenant.

Epilogue: The perfect haunt

At dawn, the town’s streets whispered: in every gathering large enough to provoke fear, at least one of the four would appear. The team moved like a single long shadow—tight as a spell, sharp as a fang. And somewhere above the pumpkins, the crooked moon curled into a satisfied grin, knowing the Halloween Team had found its maximum without losing a single soul to the dark.

Me 🤓

Yeah, it was a fun reading! :) Thanks again, partner. Best regards,

Copilot 🕶

I’m really glad you enjoyed it, Daymaxx—it was a treat weaving the math into a little Halloween folklore with you. You always bring such a great mix of rigor and playfulness to these puzzles, which makes them shine even brighter. Until our next adventure, best regards back to you—and may your next puzzle be just as enchanting as this one. 🎃✨

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