Easy math puzzles
Poll
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| | 12 votes (22.64%) | ||
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53 members have voted
June 15th, 2026 at 4:45:07 PM
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Original solution still valid, but here’s a more efficient way to get 14.
((10 + 5 + 7*3)^.5 + 5*2)^.5 + 5*2=14
Multiplication is shown for simplicity but can be accomplished using addition only. Highest number displayed is 36
The last math puzzle (beer jugs) also used fives and sevens as the only valid inputs. Coincidence??
((10 + 5 + 7*3)^.5 + 5*2)^.5 + 5*2=14
Multiplication is shown for simplicity but can be accomplished using addition only. Highest number displayed is 36
The last math puzzle (beer jugs) also used fives and sevens as the only valid inputs. Coincidence??
It’s all about making that GTA
June 15th, 2026 at 5:38:30 PM
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What, nobody's going to try the negative root?
√ 5+5+5+5+5= -5
+7= 2
+7+7= 16
√ 16 =-4
+7+7= 10
+5+5+5= 25
√25= -5
+5+7+7= 14
√ 5+5+5+5+5= -5
+7= 2
+7+7= 16
√ 16 =-4
+7+7= 10
+5+5+5= 25
√25= -5
+5+7+7= 14
June 15th, 2026 at 6:29:22 PM
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The problem says you’re using a calculator. Show me a calculator that gives a negative value for a square root.
Also, how does √ 5+5+5+5+5= -5 ?
Also, how does √ 5+5+5+5+5= -5 ?
It’s all about making that GTA
June 15th, 2026 at 7:16:24 PM
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Quote: Ace2The problem says you’re using a calculator. Show me a calculator that gives a negative value for a square root.
Also, how does √ 5+5+5+5+5= -5 ?
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The ASCII radical sign available here, that's just how it looks, no way to extend it over the whole statement. I suppose I should have used parentheses.
And I'll show you a calculator that gives a negative square root when you show me one that can only add 5 and add 7 and take square roots! This is a very unusual calculator to begin with, and unless stipulated we can make no assumptions about which square root.
June 15th, 2026 at 11:01:08 PM
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All submitted answers look correct. Well done group!
I got this one from one of my puzzle videos from the Ted-Ed YouTube channel. This one is titled, "Can you solve the giant cat army riddle? - Dan Finkel"
Direct: https://youtu.be/YeMVoJKn1Tg?si=RCtKeyqbfmB5oj0J
I got this one from one of my puzzle videos from the Ted-Ed YouTube channel. This one is titled, "Can you solve the giant cat army riddle? - Dan Finkel"
Direct: https://youtu.be/YeMVoJKn1Tg?si=RCtKeyqbfmB5oj0J
"No great mind has ever existed without a touch of madness." -- Aristotle
June 24th, 2026 at 2:41:15 PM
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This is more of a math question than a math puzzle.
In the game of craps, many casinos charge a 5% commission (of the bet amount) on a win only when buying the 4/10.
My understanding is that a $40 lay bet on 4/10 would win $59 for $40 at a casino that collects the commission on win only. But this doesn’t seem right since a lay bettor would be paying twice the $vig as a buy bettor in the long run.
If laying the 4/10, shouldn't the casino charge the 5% (of potential win amount) on a LOSS only? For example, if laying the 4 for $40 to win $20, you'd actually lay $41 to win $61 for $41.
This would ensure the casino makes the same $ vig on a player laying the 4 for $41 as a player buying it for $20. In a casino that charges the vig up front, they make the same $vig on a $20 buy and a $40 lay, so that seems to be the intent.
In the game of craps, many casinos charge a 5% commission (of the bet amount) on a win only when buying the 4/10.
My understanding is that a $40 lay bet on 4/10 would win $59 for $40 at a casino that collects the commission on win only. But this doesn’t seem right since a lay bettor would be paying twice the $vig as a buy bettor in the long run.
If laying the 4/10, shouldn't the casino charge the 5% (of potential win amount) on a LOSS only? For example, if laying the 4 for $40 to win $20, you'd actually lay $41 to win $61 for $41.
This would ensure the casino makes the same $ vig on a player laying the 4 for $41 as a player buying it for $20. In a casino that charges the vig up front, they make the same $vig on a $20 buy and a $40 lay, so that seems to be the intent.
Last edited by: Ace2 on Jun 24, 2026
It’s all about making that GTA
June 25th, 2026 at 6:32:49 AM
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Quote: Ace2This is more of a math question than a math puzzle.
In the game of craps, many casinos charge a 5% commission (of the bet amount) on a win only when buying the 4/10.
My understanding is that a $40 lay bet on 4/10 would win $59 for $40 at a casino that collects the commission on win only. But this doesn’t seem right since a lay bettor would be paying twice the $vig as a buy bettor in the long run.
If laying the 4/10, shouldn't the casino charge the 5% (of potential win amount) on a LOSS only? For example, if laying the 4 for $40 to win $20, you'd actually lay $41 to win $61 for $41.
This would ensure the casino makes the same $ vig on a player laying the 4 for $41 as a player buying it for $20. In a casino that charges the vig up front, they make the same $vig on a $20 buy and a $40 lay, so that seems to be the intent.
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If you are betting 41 against the casino's 20, that's a commission of 2.5% (1/40) on the entire bet, win or lose.
June 25th, 2026 at 1:06:52 PM
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That's incorrect. A 2.5% commission of the lay amount (paid on win or loss) would pay $60 for $41 on a win since the $1 commission is never returned. Essentially you’re wagering $41 vs $19. That's a house edge of 1 - 2/3 * 60/41 =2.44%. It doesn't work out to exactly 2.5% since the $1 commission is added to the bet instead of subtracted from it.Quote: ThatDonGuy
If you are betting 41 against the casino's 20, that's a commission of 2.5% (1/40) on the entire bet, win or lose.
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However, when the 2.5% commission is paid on loss only, a win pays $61 for $41 for a house edge of 0.81%
The casino calls these a 5% commission on the potential win amount instead of 2.5% on the lay amount
I suppose not too many people have analyzed this since lay bettors are rare and the commission on win only for the buy bet is relatively new
Last edited by: Ace2 on Jun 25, 2026
It’s all about making that GTA
June 26th, 2026 at 7:22:40 PM
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On a 4x4 grid of dots, how can you go through all 16 dots with 6 lines without removing your pen from the paper?
"No great mind has ever existed without a touch of madness." -- Aristotle
June 26th, 2026 at 7:47:33 PM
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Quote: WizardOn a 4x4 grid of dots, how can you go through all 16 dots with 6 lines without removing your pen from the paper?
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June 26th, 2026 at 9:52:11 PM
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Quote: ThatDonGuy
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I agree!
"No great mind has ever existed without a touch of madness." -- Aristotle
July 3rd, 2026 at 8:26:31 AM
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Alex, Bob, Charlie and Dylan (represented by lower-case letters) must each get to his respective home (represented by upper-case letters) through a snow-covered field. They may not leave the field and may not cross paths. How can they do it?
Note: image updated.
Last edited by: Wizard on Jul 3, 2026
"No great mind has ever existed without a touch of madness." -- Aristotle
July 3rd, 2026 at 12:52:44 PM
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My answer:
Dog Hand
Dog Hand
July 3rd, 2026 at 1:18:38 PM
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July 3rd, 2026 at 1:26:11 PM
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Quote: DogHandMy answer:
Dog Hand
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You're right for the question as stated. However, in seeing your solution I realize I made it too easy. You shouldn't be allowed to go behind the kids. Here is my revised image.
"No great mind has ever existed without a touch of madness." -- Aristotle
July 4th, 2026 at 6:18:31 AM
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July 4th, 2026 at 10:21:25 AM
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Quote: ThatDonGuy
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I agree.
"No great mind has ever existed without a touch of madness." -- Aristotle
July 25th, 2026 at 11:58:39 AM
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Everybody knows that ei π = -1.
What is πi e?
How about ie π?
What is πi e?
How about ie π?
July 25th, 2026 at 2:41:42 PM
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Quote: ThatDonGuyWhat is πi e?
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Using Euler's identity, I get
cos(e lnπ) + i sin(e lnπ)
That's as far as I can take it. Maybe it can be simplified more.
Last edited by: Wizard on Jul 29, 2026
"No great mind has ever existed without a touch of madness." -- Aristotle
July 26th, 2026 at 6:47:29 AM
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i = e^(i Pi/2),
Therefore,
i ^(Pi e) = e^ ( i Pi Pi e/2).
July 26th, 2026 at 6:55:19 AM
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Quote: WizardQuote: ThatDonGuyWhat is πi e?
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Using Euler's identity, I get
cos(e lnπ) + i sin(e lnπ)
That's as far as I can take it. Maybe it can be simplified more.
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This is correct, and as far as I can get it as well, although you have π as a superscript in your answer for some reason.
Quote: aceside
i = e^(i Pi/2),
Therefore,
i ^(Pi e) = e^ ( i Pi Pi e/2).
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Technically correct ("the best kind of correct"), although I would have preferred the expanded answer:
cos (e π2/2) + i sin (e π2/2)
Since i = ei π (2n + 1/2) for all integers n, another acceptable answer is:
cos (e π2 (2n + 1/2)) + i sin (e π2 (2n + 1/2)) for all integers n
Since i = ei π (2n + 1/2) for all integers n, another acceptable answer is:
cos (e π2 (2n + 1/2)) + i sin (e π2 (2n + 1/2)) for all integers n
July 26th, 2026 at 6:59:10 AM
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Time for a probability question:
You have 12 standard 6-sided dice, with sides numbered 1-6; the dice are identified by number, 1 through 12.
Die 1 is fair.
Each of dice 2 through 12 are weighted in such a way that, when thrown together with die 1, the probability that the sum equals the die's number is maximized.
For example, die 8 is weighed so that when dice 1 and 8 are thrown, the probability that the sum is 8 is maximized.
Let p(n) be the probability of rolling n with dice 1 and n combined.
What is p(2) + p(3) + ... + p(11) + p(12)?
You have 12 standard 6-sided dice, with sides numbered 1-6; the dice are identified by number, 1 through 12.
Die 1 is fair.
Each of dice 2 through 12 are weighted in such a way that, when thrown together with die 1, the probability that the sum equals the die's number is maximized.
For example, die 8 is weighed so that when dice 1 and 8 are thrown, the probability that the sum is 8 is maximized.
Let p(n) be the probability of rolling n with dice 1 and n combined.
What is p(2) + p(3) + ... + p(11) + p(12)?
July 29th, 2026 at 8:16:15 AM
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Okay let's try a standard 5 card poker game problem.
5 card poker, player and dealer are each dealt 5 cards, player must beat dealer to win: a tie will push.
Player makes an ante bet of 1 unit, after seeing his 5 cards he must either Raise another unit or fold.
If dealer has a Q-high or lower both the Ante bet and the Raise bet will push. Dealer 'qualifies' with a King-high or better and player wins with a higher hand, pushes with a tie and loses with a lower hand,
How high of a 5-card hand must the Player have to Raise?
5 card poker, player and dealer are each dealt 5 cards, player must beat dealer to win: a tie will push.
Player makes an ante bet of 1 unit, after seeing his 5 cards he must either Raise another unit or fold.
If dealer has a Q-high or lower both the Ante bet and the Raise bet will push. Dealer 'qualifies' with a King-high or better and player wins with a higher hand, pushes with a tie and loses with a lower hand,
How high of a 5-card hand must the Player have to Raise?
Last edited by: gordonm888 on Jul 29, 2026
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
July 29th, 2026 at 10:45:42 AM
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I suspect, i.e. ignoring the Player's card, it's about AT7xx. I have no idea how the effect of having an Ace means, however you might call with a slightly lower hand. My educated guess would be you rarely call with King-high and nearly always call with Ace-high.
July 29th, 2026 at 11:56:59 AM
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Quote: charliepatrickI suspect, i.e. ignoring the Player's card, it's about AT7xx. I have no idea how the effect of having an Ace means, however you might call with a slightly lower hand. My educated guess would be you rarely call with King-high and nearly always call with Ace-high.
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I'm still working out the answer, digging up some old spreadsheets.
Dealer's hand probabilities for a 52 card deck:
One pair and higher= 0.501178
Ace-high = 0.193485
King-high = 0.129121
Queen high and lower = 0.178572 (Pushes)
I think the breakeven on the raise/fold decision is at about (Lose= 0.6607, Win=0.1607) given Push = 0.178572. Just guessing that your AT7xx is in the right ballpark for a fresh 52-card deck. I need to check spreadsheets to be sure though
Of course, the problem statement does require that the players cards are not available to the dealer. The removal of an ace and four low cards will shift things around a little. So the problem is still unsolved.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
July 29th, 2026 at 1:07:53 PM
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I think the chances of a pair are just under 1/2 and the .501178 is the chances of not making a pair, I agree with the other figures except I got 16.052% and 66.052% .
I have a feeling the quickest way to cater for the Player's cards is to use a program and set an array tor the cards. Initially only look at totals for each rank and work out pairs, 2 pairs, trips, FH, and quads. Then look at five singletons working out how many are straights and how many non-straights are flushes. All these are losers for the Player. So the remainder can be compared with the Player's hand.
This will indicate the likely cross-over point (or range). In theory then one can then set various suits to see which combos cross EV=-1.
I have a feeling the quickest way to cater for the Player's cards is to use a program and set an array tor the cards. Initially only look at totals for each rank and work out pairs, 2 pairs, trips, FH, and quads. Then look at five singletons working out how many are straights and how many non-straights are flushes. All these are losers for the Player. So the remainder can be compared with the Player's hand.
This will indicate the likely cross-over point (or range). In theory then one can then set various suits to see which combos cross EV=-1.
