Easy math puzzles
Poll
 | 25 votes (47.16%) | ||
| | 17 votes (32.07%) | ||
| | 7 votes (13.2%) | ||
| | 4 votes (7.54%) | ||
| | 12 votes (22.64%) | ||
| | 4 votes (7.54%) | ||
| | 6 votes (11.32%) | ||
| | 5 votes (9.43%) | ||
| | 12 votes (22.64%) | ||
| | 10 votes (18.86%) |
53 members have voted
August 19th, 2026 at 6:12:07 PM
permalink
Quote: DogHandQuote: WizardYou have a 100-gallon tub full of pure water. Someone pours two gallons of salt into it, which is immediately absorbed. Pure water flows into the tank at a rate of five gallons per minute. Meanwhile, water leaves a hole in the bottom at the same rate. How long will it take for the water to contain 0.1 gallons of salt?
link to original post
Wiz,
Here is my solution:
Dog Hand
link to original post
I have a slightly difference answer.
At time t=T min, salt=0.1 gal, water=100 gal.
At time t=x, salt=y. We have dy/dx=-5y/100; therefore, the general solution to the differential equation is
𝑦=A* exp(-x/20)=2*exp(-x/20).
Where we have used: at time t=0 min, salt=2 gal, water=100 gal. Therefore
x=-20 ln(y/2)= -20 ln (0.1/2) = 59.915 min.
August 19th, 2026 at 7:18:57 PM
permalink
Quote: DogHandQuote: WizardYou have a 100-gallon tub full of pure water. Someone pours two gallons of salt into it, which is immediately absorbed. Pure water flows into the tank at a rate of five gallons per minute. Meanwhile, water leaves a hole in the bottom at the same rate. How long will it take for the water to contain 0.1 gallons of salt?
link to original post
Wiz,
Here is my solution:
Dog Hand
link to original post
I agree based on your interpretation of the question. I meant to say the initial 2 gallons of walt was immediately absorbed, like a sponge, into the pure water. Thus, the starting volume of saltwater was still 100 gallons. However, I'll take the blame for the ambiguous wording and give you credit. This one was beer worthy.
"No great mind has ever existed without a touch of madness." -- Aristotle
August 19th, 2026 at 8:13:45 PM
permalink
In chemistry, volumes are not additive. The volume of the final solution will be between 100.8 and 101.5 gallons, depending on whether the 2 gallons of salt are measured as a loose powder or a solid block.
Gambling is a math contest where the score is tracked in dollars. Try not to get a negative score.
October 7th, 2026 at 7:17:52 PM
permalink
There is a city with a square grid of three streets running east to west and 3 running north to south, creating 4 blocks. A drunk is at the north west corner of the grid. His home is at the south east corner. Every turn he walks in a random direction, all equally likely, of all available streets to him. How many turns will it take, on average, to get home?
"No great mind has ever existed without a touch of madness." -- Aristotle
October 7th, 2026 at 11:05:05 PM
permalink
13
?
?
It’s all about making that GTA
October 8th, 2026 at 4:22:43 AM
permalink
Quote: Ace213
?
link to original post
I disagree, but could be wrong.
"No great mind has ever existed without a touch of madness." -- Aristotle
October 8th, 2026 at 7:18:16 AM
permalink
44 4/7
Let T(r,c) be the number of turns it will take to get from (r,c) to (3,3); the solution is T(0,0)
Because of symmetry, T(r,c) = T(c,r), which reduces the number of equations involved
T(3,3) = 0
T(3,2) = 1 + 1/3 T(2,2) + 1/3 T(3,1) {+ 1/3 T(3,3)}
T(3,1) = 1 + 1/3 T(3,2) + 1/3 T(2,1) + 1/3 T(3,0)
T(3,0) = 1 + 1/2 T(3,1) + 1/2 T(2,0)
T(2,2) = 1 + 1/2 T(3,2) + 1/2 T(2,1)
T(2,1) = 1 + 1/4 T(3,1) + 1/4 T(2,2) + 1/4 T(1,1) + 1/4 T(2,0)
T(2,0) = 1 + 1/3 T(3,0) + 1/3 T(2,1) + 1/3 T(1,0)
T(1,1) = 1 + 1/2 T(2,1) + 1/2 T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 T(0,0)
T(0,0) = 1 + T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 (1 + T(1,0))
2/3 T(1,0) = 4/3 + 1/3 T(2,0) + 1/3 T(1,1)
T(1,0) = 2 + 1/2 T(2,0) + 1/2 T(1,1)
This is 8 equations in 8 unknowns; I won't bother with the details of solving them here, but T(1,0) = 305/7, so T(0,0) = 23 + 1 = 312/7
Because of symmetry, T(r,c) = T(c,r), which reduces the number of equations involved
T(3,3) = 0
T(3,2) = 1 + 1/3 T(2,2) + 1/3 T(3,1) {+ 1/3 T(3,3)}
T(3,1) = 1 + 1/3 T(3,2) + 1/3 T(2,1) + 1/3 T(3,0)
T(3,0) = 1 + 1/2 T(3,1) + 1/2 T(2,0)
T(2,2) = 1 + 1/2 T(3,2) + 1/2 T(2,1)
T(2,1) = 1 + 1/4 T(3,1) + 1/4 T(2,2) + 1/4 T(1,1) + 1/4 T(2,0)
T(2,0) = 1 + 1/3 T(3,0) + 1/3 T(2,1) + 1/3 T(1,0)
T(1,1) = 1 + 1/2 T(2,1) + 1/2 T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 T(0,0)
T(0,0) = 1 + T(1,0)
T(1,0) = 1 + 1/3 T(2,0) + 1/3 T(1,1) + 1/3 (1 + T(1,0))
2/3 T(1,0) = 4/3 + 1/3 T(2,0) + 1/3 T(1,1)
T(1,0) = 2 + 1/2 T(2,0) + 1/2 T(1,1)
This is 8 equations in 8 unknowns; I won't bother with the details of solving them here, but T(1,0) = 305/7, so T(0,0) = 23 + 1 = 312/7
Edit: I read the problem wrong - I had 4 streets in each direction
Last edited by: ThatDonGuy on Oct 8, 2026
October 8th, 2026 at 7:32:24 AM
permalink
I'm slightly confused by the "three roads" puzzle, especially given the above solution, as my interpretation of the puzzle was different.
If there are three roads across and three roads down, then do I assume the drunk is walking down the middle of these roads, and hence there are only nine junctions/places where they can change direction.
Thus the drunk could, if lucky, go $ABEZ but also $A$CFGZ etc.
This would imply the points are (0,0)(1,0)(2,0)(0,1)...(2,2).
Or is the puzzle using the side-walks so has (0,0) thru (3,3).
If there are three roads across and three roads down, then do I assume the drunk is walking down the middle of these roads, and hence there are only nine junctions/places where they can change direction.
$AB
CDE
FGZ
This would imply the points are (0,0)(1,0)(2,0)(0,1)...(2,2).
Or is the puzzle using the side-walks so has (0,0) thru (3,3).
October 8th, 2026 at 7:45:34 AM
permalink
18
Last edited by: ThatDonGuy on Oct 8, 2026
October 8th, 2026 at 9:53:45 AM
permalink
18
For a 3x3 puzzle there is some symmetry so some of the junctions would have the same values as the ones. Also the middle square has to go to A or C, so has the same value as B. The puzzle therefore simplifies to $-A=B=C-X where
(i) $tart has to go to A. Hence EV(A)=EV($tart)+1.
(ii) A has one chance to go to $tart or two chances to go to B. Thus the expected number of turns from A to exit is 1 + average{ $,B,B }.
(iii) B has two chances to go to A, or two chances to go to C (i.e. has equal chances). (ev{B}=1 + average{A,C} )
(iv) C has one chance to eXit or two chances to return to B. (ev{C}=1 + average{0, B, B} )
(v) X is the exit, so has zero more moves!Using the above assuming the original value EV($)=x. Then EV{A}=x-1.Now EV{A}= average ($ B B) + 1; hence $+B+B=3(x-2) : this means B=x-3.Using similar logic B is average(A C)+1, giving C = x-7
Now we know is the average of (x-3, x-3 -?-)+1. This gives it as x-18.But we know that's the exit. Hence x=18.
For a 3x3 puzzle there is some symmetry so some of the junctions would have the same values as the ones. Also the middle square has to go to A or C, so has the same value as B. The puzzle therefore simplifies to $-A=B=C-X where
(i) $tart has to go to A. Hence EV(A)=EV($tart)+1.
(ii) A has one chance to go to $tart or two chances to go to B. Thus the expected number of turns from A to exit is 1 + average{ $,B,B }.
(iii) B has two chances to go to A, or two chances to go to C (i.e. has equal chances). (ev{B}=1 + average{A,C} )
(iv) C has one chance to eXit or two chances to return to B. (ev{C}=1 + average{0, B, B} )
(v) X is the exit, so has zero more moves!
$ A B
a B C
b c X
x x-1 -?-
... -?- ...
... ... ...
x x-1 x-3
... x-3 ...
... ... ...
x x-1 x-3
... x-3 x-7.
... ... ...
x x-1 x-3
... x-3 x-7
... ...x-18
Last edited by: charliepatrick on Oct 8, 2026
October 8th, 2026 at 1:03:54 PM
permalink
Quote: ThatDonGuy18
link to original post
I agree!
Yes, it is 3x3 roads. It can be simplified as follows:
ABC
BDE
CEH
Where
A = starting location
H = home
B = 17
C = 15
D = 15
E = 11
It's not a coincidence C and D are the same since they both have a 50/50 chance of leading to B or E.
"No great mind has ever existed without a touch of madness." -- Aristotle
October 8th, 2026 at 9:39:48 PM
permalink
Same problem but the drunk starts in the center top and must get to the center bottom.
BAB
CDC
EXE
In other words, he must get from A to X. How many moves does it take on average?
BAB
CDC
EXE
In other words, he must get from A to X. How many moves does it take on average?
"No great mind has ever existed without a touch of madness." -- Aristotle
October 9th, 2026 at 6:44:16 AM
permalink
12
I also get 12 from B, 10 from C, 9 from D, and 6 from E
I also get 12 from B, 10 from C, 9 from D, and 6 from E
October 9th, 2026 at 9:05:45 AM
permalink
Quote: ThatDonGuy12
I also get 12 from B, 10 from C, 9 from D, and 6 from E
link to original post
I agree!
"No great mind has ever existed without a touch of madness." -- Aristotle
October 9th, 2026 at 9:17:28 AM
permalink
Here's a variation on a puzzle that I think I saw here once:
You have a pile of N coins, where N is 11 or higher. 10 of the coins are heads-up, and the rest are tails-up.
You are blindfolded, and have to divide the coins into two piles, each of which has the same number of heads-up coins.
Once you are blindfolded, you cannot tell which coins are heads-up (and they are mixed after you are blindfolded, so you cannot go from memory), but you are allowed to flip over any number of coins, both before and after dividing the coins into piles.
How do you do it?
You have a pile of N coins, where N is 11 or higher. 10 of the coins are heads-up, and the rest are tails-up.
You are blindfolded, and have to divide the coins into two piles, each of which has the same number of heads-up coins.
Once you are blindfolded, you cannot tell which coins are heads-up (and they are mixed after you are blindfolded, so you cannot go from memory), but you are allowed to flip over any number of coins, both before and after dividing the coins into piles.
How do you do it?
Note that N does not have to be an even number
October 9th, 2026 at 5:28:36 PM
permalink
Quote: ThatDonGuyHere's a variation on a puzzle that I think I saw here once...
link to original post
1. Divide the coins into two piles so that pile 1 has 10 coins and pile 2 has the rest.
2. Flip over all the coins in pile 1.
Here is why it works.
Before the flip, here is the distribution of coins by pile, where t is the total:
Pile 1: h heads, 10-h tails
Pile 2: 10-h heads, t-20+h tails
After the flip:
Pile 1: 10-h heads, h tails
Pile 2: 10-h heads, t-20+h tails
This one is an old classic I've seen in many puzzle books.
"No great mind has ever existed without a touch of madness." -- Aristotle
October 9th, 2026 at 8:26:26 PM
permalink
When I was in college I came up with a slightly interesting problem
X! - X = X^X
^ is the exponent operator
This one is pretty easy
X! - X = X^X
^ is the exponent operator
This one is pretty easy
October 10th, 2026 at 5:41:55 AM
permalink
Transform it to this equation:
X! = X^X *+ X
When X=1, 1!=1, 1^1+1=2;
When X=2, 2!=2, 2^2+1=5;
When X=3, 3!=6, 6^6+1=46,657.
When n is large, this trend continues, so there is no solution.
X! = X^X *+ X
When X=1, 1!=1, 1^1+1=2;
When X=2, 2!=2, 2^2+1=5;
When X=3, 3!=6, 6^6+1=46,657.
When n is large, this trend continues, so there is no solution.
October 10th, 2026 at 6:20:05 AM
permalink
that 0^0 = 1.
There is actually a considerable amount of debate over the value of 0^0.
