Easy math puzzles
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Quote: Ace2Agree. The exact answer, which I posted on July-29, is:
[(6e^(1/6) - 7)/(e^(1/6) - 1)*(1 - 1/e^(1/6)) + 1/e^(1/6)]/(1 - 5/6*1/e^(1/6))
P.S. other people might care but aren’t capable of solving it
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Thank you! I will write up my solution in a PDF document shortly to share.
It takes Alice and Bill 2 days to paint a house.
It takes Bill and Cindy 3 days to paint a house.
It takes Alice and Cindy 4 days to paint a house.
How long does it take if they all paint?
Quote: WizardI may have asked this one before, but it's a classic.
It takes Alice and Bill 2 days to paint a house.
It takes Bill and Cindy 3 days to paint a house.
It takes Alice and Cindy 4 days to paint a house.
How long does it take if they all paint?
link to original post
Solve this by putting the problem in terms of painting rates.
Let a, b, and c be the rates for Alice, Bill, and Cindy, respectively.
a + b = 1/2, which is a rate of 0.5 house / day.
b + c = 1/3
a + c = 1/4
Sum the three equations to get: 2a + 2b + 2c = 13/12
Then: a + b + c = 13/24 houses / day
1 / (13/24 houses/day) = 24/13 days/house = 1 11/13 days, or about 1 day, 20 hours, and 18 minutes.
Quote: ChesterDogquestions-and-answers/math/34502-easy-math-puzzles/104/#post962824]link to original post
Solve this by putting the problem in terms of painting rates.
Let a, b, and c be the rates for Alice, Bill, and Cindy, respectively.
a + b = 1/2, which is a rate of 0.5 house / day.
b + c = 1/3
a + c = 1/4
Sum the three equations to get: 2a + 2b + 2c = 13/12
Then: a + b + c = 13/24 houses / day
1 / (13/24 houses/day) = 24/13 days/house = 1 11/13 days, or about 1 day, 20 hours, and 18 minutes.
link to original post
I agree! Much simpler than how I did it.
For extra credit, how long would it take each individual person to paint the house?
Quote: WizardQuote: ChesterDogquestions-and-answers/math/34502-easy-math-puzzles/104/#post962824]link to original post
Solve this by putting the problem in terms of painting rates.
Let a, b, and c be the rates for Alice, Bill, and Cindy, respectively.
a + b = 1/2, which is a rate of 0.5 house / day.
b + c = 1/3
a + c = 1/4
Sum the three equations to get: 2a + 2b + 2c = 13/12
Then: a + b + c = 13/24 houses / day
1 / (13/24 houses/day) = 24/13 days/house = 1 11/13 days, or about 1 day, 20 hours, and 18 minutes.
link to original post
I agree! Much simpler than how I did it.
For extra credit, how long would it take each individual person to paint the house?
link to original post
1) a + b = 1/2
2) b + c = 1/3
3) a + c = 1/4
Subtracting 2) from 1) yields:
4) a - c = 1/6
Adding 3) and 4) yields:
2a = 5/12
a = 5/24 houses/day
1 / (5/24 houses/day) = 24/5 days/house = 4.8 days for Alice to paint a house, which is 4 days, 19 hours, and 12 minutes.
b = 1/2 - a = 1/2 - 5/24 = 7/24 houses/day
1 / (7/24 houses/day) = 24/7 days/house = 3 3/7 days for Bill to paint a house, which is about 3 days, 10 hours, and 17 minutes.
c = 1/4 - a = 1/4 - 5/24 = 1/24 houses/day
1 / (1/24 houses/day) = 24 days for Cindy to paint a house
Quote: WizardI may have asked this one before
Yes, you did, sort of; you used 3, 4, 5 instead of 2, 3, 4
1 + 2 + 3 = 6
1 x 2 x 3 = 6 as well.
1. Given two positive numbers a and b, it is always possible to find a positive number c such that a + b + c = abc, except for what pairs (a, b)?
2. Are there any sets of three integers a, b, c besides 1, 2, 3 where a <= b <= c and a + b + c = abc?
Quote: ThatDonGuyHere's one (well, two) from me:
1 + 2 + 3 = 6
1 x 2 x 3 = 6 as well.
1. Given two positive numbers a and b, it is always possible to find a positive number c such that a + b + c = abc, except for what pairs (a, b)?
2. Are there any sets of three integers a, b, c besides 1, 2, 3 where a <= b <= c and a + b + c = abc?
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The answer for 1 is...
Proof:
a+b+c = abc
a+b = (ab-1)c
c = (a+b)/(ab-1)
This exists unless ab = 1 (you can't divide by 0). So clearly you can find c when ab is not 1.
If ab = 1 then we get
abc = a+b+c
c = a+b+c
a+b = 0
which is a contradiction (since a and b must be positive). So c never exists when ab=1
Quote: gerback123Find an approximation for the probability that W is less than 1845.
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I figured out how to solve this problem, but doing the calculations by hand takes too long, so I wrote code in Python to do it.
Die:
(x¹ + x² + x³ + x⁴)⁶⁰⁰
Coin:
(x⁰ + x¹)⁶⁰⁰
The obtained powers and their coefficients give the sum and the occurrence respectively. For example, for 600 tosses of the coin we get:
1·x⁶⁰⁰ + 600·x⁵⁹⁹ + 179700·x⁵⁹⁸ + 35820200·x⁵⁹⁷ ...
That means that there exist:
- 1 combination for the sum of 600,
- 600 combinations for the sum of 599,
- 179700 combinations for the sum of 598, and so on.
Then, we can just combine the above results to check their sums and calculate occurrences.
The final code:
import fractions as frac
die = [1, 2, 3, 4]
coin = [0, 1]
def CombCounter(obj):
counter = {s: 1 for s in obj} # {sum: number of combinations}
for _ in range(599):
prev = counter
counter = {}
for val in obj:
for s, n in prev.items():
Sum = s + val
counter[Sum] = counter.get(Sum, 0) + n
# for k, v in counter.items(): print(k, v)
return counter
die_combs = CombCounter(die)
coin_combs = CombCounter(coin)
n_combs = 0
n_all_combs = 0
for die_sum, die_n_combs in die_combs.items():
for coin_sum, coin_n_combs in coin_combs.items():
n_all_combs += die_n_combs * coin_n_combs
if die_sum + coin_sum < 1845:
n_combs += die_n_combs * coin_n_combs
p_exact = frac.Fraction(n_combs, n_all_combs)
p_approx = float(p_exact)
print('Exact probability:', p_exact, sep='\n')
print()
print('Approximate probability:', p_approx, sep='\n')
The result is:
Exact probability:
16629696692942443221311600805732571727882056503186445797071356619220143871973750380670420495722626540277141546853704453502562909306791685045031373467661131681245447148833330909962703142185937903212616044186346871542074272828072146756783959026590407852531117642742436997304638345486431845969028253743772232874927266087537986207158866084048753547876182678807964235979895578405803847502461039492476417081293332575410523848089271026552194744106232466060091481194107062426685104974467724051884406154898804316213028728752858074007202926320266624549/17862087144182552090100651130756164665976827862122256167423329247233898321830719666540870487544055009398369586776484355671625497973604947199022105534491506565630899537593179994034477830621111528653321226095994410794048389454474256755765855031213008497406701691127284482478696801343353279019313560663355819346556656789780042055905915034991279242393102960416490645747179216448162518823663813131314335790891510706123877326968206993303684221095374172364198108537519867527915452940344040267013916003084424614072759887985824821063642644988081209344
Approximate probability:
0.9310052380053759
Quote: ChesterDog
1) a + b = 1/2
2) b + c = 1/3
3) a + c = 1/4
Subtracting 2) from 1) yields:
4) a - c = 1/6
Adding 3) and 4) yields:
2a = 5/12
a = 5/24 houses/day
1 / (5/24 houses/day) = 24/5 days/house = 4.8 days for Alice to paint a house, which is 4 days, 19 hours, and 12 minutes.
b = 1/2 - a = 1/2 - 5/24 = 7/24 houses/day
1 / (7/24 houses/day) = 24/7 days/house = 3 3/7 days for Bill to paint a house, which is about 3 days, 10 hours, and 17 minutes.
c = 1/4 - a = 1/4 - 5/24 = 1/24 houses/day
1 / (1/24 houses/day) = 24 days for Cindy to paint a house
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I agree!
Quote: ThatDonGuyYes, you did, sort of; you used 3, 4, 5 instead of 2, 3, 4
link to original post
It's such a good puzzle, it's worth repeating once in a while.
The expected value and standard deviation per trial is 3 and 1.5^5 respectively. For 600 trials it’s 1800 and 30. Using a z-table, the chance of finishing below (1844.5 - 1800) / 30 SDs above the mean is 0.93101Quote: AnotherBillQuote: gerback123Find an approximation for the probability that W is less than 1845.
link to original post
I figured out how to solve this problem, but doing the calculations by hand takes too long, so I wrote code in Python to do it.We can calculate occurrence of each possible sum for the die and the coin separately this way:Die:
(x¹ + x² + x³ + x⁴)⁶⁰⁰
Coin:
(x⁰ + x¹)⁶⁰⁰
The obtained powers and their coefficients give the sum and the occurrence respectively. For example, for 600 tosses of the coin we get:1·x⁶⁰⁰ + 600·x⁵⁹⁹ + 179700·x⁵⁹⁸ + 35820200·x⁵⁹⁷ ...
That means that there exist:
- 1 combination for the sum of 600,
- 600 combinations for the sum of 599,
- 179700 combinations for the sum of 598, and so on.
Then, we can just combine the above results to check their sums and calculate occurrences.
The final code:import fractions as frac
die = [1, 2, 3, 4]
coin = [0, 1]
def CombCounter(obj):
counter = {s: 1 for s in obj} # {sum: number of combinations}
for _ in range(599):
prev = counter
counter = {}
for val in obj:
for s, n in prev.items():
Sum = s + val
counter[Sum] = counter.get(Sum, 0) + n
# for k, v in counter.items(): print(k, v)
return counter
die_combs = CombCounter(die)
coin_combs = CombCounter(coin)
n_combs = 0
n_all_combs = 0
for die_sum, die_n_combs in die_combs.items():
for coin_sum, coin_n_combs in coin_combs.items():
n_all_combs += die_n_combs * coin_n_combs
if die_sum + coin_sum < 1845:
n_combs += die_n_combs * coin_n_combs
p_exact = frac.Fraction(n_combs, n_all_combs)
p_approx = float(p_exact)
print('Exact probability:', p_exact, sep='\n')
print()
print('Approximate probability:', p_approx, sep='\n')
The result is:Exact probability:
16629696692942443221311600805732571727882056503186445797071356619220143871973750380670420495722626540277141546853704453502562909306791685045031373467661131681245447148833330909962703142185937903212616044186346871542074272828072146756783959026590407852531117642742436997304638345486431845969028253743772232874927266087537986207158866084048753547876182678807964235979895578405803847502461039492476417081293332575410523848089271026552194744106232466060091481194107062426685104974467724051884406154898804316213028728752858074007202926320266624549/17862087144182552090100651130756164665976827862122256167423329247233898321830719666540870487544055009398369586776484355671625497973604947199022105534491506565630899537593179994034477830621111528653321226095994410794048389454474256755765855031213008497406701691127284482478696801343353279019313560663355819346556656789780042055905915034991279242393102960416490645747179216448162518823663813131314335790891510706123877326968206993303684221095374172364198108537519867527915452940344040267013916003084424614072759887985824821063642644988081209344
Approximate probability:
0.9310052380053759
link to original post
Quote: Ace2The expected value and standard deviation per trial is 3 and 1.5^5 respectively. For 600 trials it’s 1800 and 30. Using a z-table, the chance of finishing below (1844.5 - 1800) / 30 SDs above the mean is 0.93101
link to original post
Wow! Interesting solution, though, I don't quite understand probability math. Where does 1.5⁵ come from? A spreadsheet gives ≈1.2247 for a StD for a population (1, 2, 2, 3, 3, 4, 4, 5).
Quote: SkinnyTonyQuote: ThatDonGuyHere's one (well, two) from me:
1 + 2 + 3 = 6
1 x 2 x 3 = 6 as well.
1. Given two positive numbers a and b, it is always possible to find a positive number c such that a + b + c = abc, except for what pairs (a, b)?
2. Are there any sets of three integers a, b, c besides 1, 2, 3 where a <= b <= c and a + b + c = abc?
link to original post
The answer for 1 is...c exists for all pairs except when ab=1 (and never exists when ab=1)
Proof:
a+b+c = abc
a+b = (ab-1)c
c = (a+b)/(ab-1)
This exists unless ab = 1 (you can't divide by 0). So clearly you can find c when ab is not 1.
If ab = 1 then we get
abc = a+b+c
c = a+b+c
a+b = 0
which is a contradiction (since a and b must be positive). So c never exists when ab=1
link to original post
The solution to #1 is correct.
In all cases, (a, b) represent the numbers a and b, and n and k are positive integers
(1, 1) has no solutions as 2 + c = c
(1, 2): 1 + 2 + c = 2c
c = 3
(1, 2, 3) is a solution
(1, 3): 1 + 3 + c = 3c
c = 2, but b = 3 > c
(1, 3 + n): 1 + 3 + n + c = (3 + n) c
4 + n = (2 + n) c
c = (4 + n) / (2 + n) = 1 + 2 / (2 + n), which is not an integer when n is a positive integer
(2, 2): 2 + 2 + c = 4c
c = 4/3, which is not an integer
(2, 3): 2 + 3 + c = 6c
c = 1, but b = 3 > c
(2, 3 + n): 2 + 3 + n + c = (6 + 2n) c
5 + n = (5 + 2n) c
c = (5 + 2n) / (5 + n) = 1 + n / (5 + n), which is not an integer when n is a positive integer
(3, 3): 6 + c = 9c
c = 3/4, which is not an integer
(3, 3 + n): 6 + n + c = (9 + 3n) c
6 + n = (8 + 3n) c
c = (8 + 3n) / (6 + n) = 1 + (2n + 2) / (6 + n) < 1 + (2n + 12) / (6 + n), so c < a
(3 + k, 3 + k + n): 6 + 2k + n + c = (9 + k^2 + 6k + 3n + nk) c
6 + 2k + n = (8 + k^2 + 6k + 3n + nk) c
c = (8 + k^2 + 6k + 3n + nk) / (6 + 2k + n)
b = 3 + k + n = (6 + 2k + n)(3 + k + n) / (3 + k + n)
= (18 + 8k + 2k^2 + 6n + 2kn) / (3 + k + n)
= c + (10 + k^2 + 2k + 3n + nk) / (3 + k + n) > c, so c < b
Therefore, (1, 2, 3) is the only solution in positive integers
Quote: AnotherBillQuote: Ace2The expected value and standard deviation per trial is 3 and 1.5^5 respectively. For 600 trials it’s 1800 and 30. Using a z-table, the chance of finishing below (1844.5 - 1800) / 30 SDs above the mean is 0.93101
link to original post
Wow! Interesting solution, though, I don't quite understand probability math. Where does 1.5⁵ come from? A spreadsheet gives ≈1.2247 for a StD for a population (1, 2, 2, 3, 3, 4, 4, 5).
link to original post
1.5^.5 =1.2247
Quote: Ace2Quote: AnotherBillQuote: Ace2The expected value and standard deviation per trial is 3 and 1.5^5 respectively. For 600 trials it’s 1800 and 30. Using a z-table, the chance of finishing below (1844.5 - 1800) / 30 SDs above the mean is 0.93101
link to original post
Wow! Interesting solution, though, I don't quite understand probability math. Where does 1.5⁵ come from? A spreadsheet gives ≈1.2247 for a StD for a population (1, 2, 2, 3, 3, 4, 4, 5).
link to original post
1.5^.5 =1.2247
link to original post
Ah! Thank you for the explanation! You just missed the decimal point in you original post.
Quote: AnotherBillQuote: Ace2Quote: AnotherBillQuote: Ace2The expected value and standard deviation per trial is 3 and 1.5^5 respectively. For 600 trials it’s 1800 and 30. Using a z-table, the chance of finishing below (1844.5 - 1800) / 30 SDs above the mean is 0.93101
link to original post
Wow! Interesting solution, though, I don't quite understand probability math. Where does 1.5⁵ come from? A spreadsheet gives ≈1.2247 for a StD for a population (1, 2, 2, 3, 3, 4, 4, 5).
link to original post
1.5^.5 =1.2247
link to original post
Ah! Thank you for the explanation! You just missed the decimal point in you original post.
link to original post
FWIW you don't need a spreadsheet, or even a calculator, to figure this out.
To find variance, you look at each number, and take the distance from the average. Then you square those distances, and take the average of what you have left.
So in this case we start with 1,2,2,3,3,4,4,5
The average is 3, so we want to take the distance between each number and 3. We get:
2,1,1,0,0,1,1,2
Squaring each number we get:
4,1,1,0,0,1,1,4
And the average of those is 1.5 (there are 8 numbers and they add to 12). So the variance is 1.5
And then standard deviation is just the square root of variance, hence, sqrt(1.5). Taking that square root is the only step we really need a calculator for.
note I don't know the answer to the first one (second part) and found the solution to the second online..
(i) You have a stick which is 100cm long and randomly cut it into two pieces. What is the average length of the shorter piece.
The answer to this is relatively obvious; but I had actually misread the question and assumed you made two cuts. At this stage I have no idea what the answer to this is (except I might run a simulation sometime).
(ii) Which is bigger 2100! or 2100! ?
(iii) Nice property of 1/998001.
Quote: charliepatrickSometimes I see puzzles on facebook; some can be interesting and others just stupid. So here's a couple -
note I don't know the answer to the first one (second part) and found the solution to the second online..
(i) You have a stick which is 100cm long and randomly cut it into two pieces. What is the average length of the shorter piece.
The answer to this is relatively obvious; but I had actually misread the question and assumed you made two cuts. At this stage I have no idea what the answer to this is (except I might run a simulation sometime).
(ii) Which is bigger 2100! or 2100! ?
(iii) Nice property of 1/998001.
link to original post
(ii) 2100! = ( ... (((21)2)3)4 ... )100
(2100)! = 1·2·3·4· ... ·2100
Since, 2n > n, 2100! > (2100)!
(iii) What is "nice property"?
There are 100 people and 100 closed doors. Both the doors and people are numbered 1 to 100. When a person stops at a door, they will reverse it (in other words if it's open they close it and if it's closed they open it). Each person will stop at every door that is evenly divisible by his own number. For example, person 5 will stop at doors 5, 10, 15, ... 95, 100. After everyone has had their turn, which doors will be open?
Quote: charliepatrick(i) You have a stick which is 100cm long and randomly cut it into two pieces. What is the average length of the shorter piece.
The answer to this is relatively obvious; but I had actually misread the question and assumed you made two cuts. At this stage I have no idea what the answer to this is (except I might run a simulation sometime).
link to original post
I get 100/8 = 12.5 cm.
Solution available upon request. It's a clever mix of geometry and calculus (assuming I'm even right).
Note: updated 9:38 AM PST
Parms: sh:100000000 Time:19:30:6:626
Overall Result: T: 1111178163.0952969 Avg: 11.111781630952969
Parms: sh:100000000 Time:19:30:33:225
Quote: WizardThis one is an old classic that has probably been asked before.
There are 100 people and 100 closed doors. Both the doors and people are numbered 1 to 100. When a person stops at a door, they will reverse it (in other words if it's open they close it and if it's closed they open it). Each person will stop at every door that is evenly divisible by his own number. For example, person 5 will stop at doors 5, 10, 15, ... 95, 100. After everyone has had their turn, which doors will be open?
link to original post
Is it the Sieve of Eratosthenes?
Quote: AnotherBillQuote: WizardThis one is an old classic that has probably been asked before.
There are 100 people and 100 closed doors. Both the doors and people are numbered 1 to 100. When a person stops at a door, they will reverse it (in other words if it's open they close it and if it's closed they open it). Each person will stop at every door that is evenly divisible by his own number. For example, person 5 will stop at doors 5, 10, 15, ... 95, 100. After everyone has had their turn, which doors will be open?
link to original post
Is it the Sieve of Eratosthenes?
link to original post
No I think it is
Quote: unJonNo I think it is
perfect squares, which are the only numbers that have an odd number of factors.
link to original post
Agree.
Quote: WizardThis one is an old classic that has probably been asked before.
There are 100 people and 100 closed doors. Both the doors and people are numbered 1 to 100. When a person stops at a door, they will reverse it (in other words if it's open they close it and if it's closed they open it). Each person will stop at every door that is evenly divisible by his own number. For example, person 5 will stop at doors 5, 10, 15, ... 95, 100. After everyone has had their turn, which doors will be open?
link to original post
Associated problem: what if there are 100 doors but only 50 people, whose numbers are 1, 3, 5, ..., 99?
I now get 100/9.
I previously made an error when the first cut was within the middle third of the stick.
I plan to write up a solution.
Quote: charliepatrick^ I've finally got round to running a simulation and it gets a slightly different answer
So I'm guessing it's 11 1/9! I wonder whether it's 100/N2Parms: sh:100000000 Time:19:30:6:626
Overall Result: T: 1111178163.0952969 Avg: 11.111781630952969
Parms: sh:100000000 Time:19:30:33:225
link to original post
For something more challenging, prove the formula the expected length of the shortest piece for arbitrary N. After this, find a formula for the expected length of the k-th shortest piece or the k-th longest piece for arbitrary N and prove it. Based on my simulations, I have a conjecture for the expected length of the 2nd shortest piece, but anything beyond that is a mystery.
