Easy math puzzles

Quote: Ace2

Agree. 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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Quote: Wizard

I 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?
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Quote: ChesterDog

questions-and-answers/math/34502-easy-math-puzzles/104/#post962824]link to original post

Quote: Wizard

Quote: ChesterDog

questions-and-answers/math/34502-easy-math-puzzles/104/#post962824]link to original post

Show Spoiler

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.

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Quote: Wizard

I may have asked this one before

Quote: ThatDonGuy

Here'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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Quote: gerback123

Find an approximation for the probability that W is less than 1845.
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Die:
(x¹ + x² + x³ + x⁴)⁶⁰⁰

Coin:
(x⁰ + x¹)⁶⁰⁰

1·x⁶⁰⁰ + 600·x⁵⁹⁹ + 179700·x⁵⁹⁸ + 35820200·x⁵⁹⁷ ...

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')

Exact probability:
16629696692942443221311600805732571727882056503186445797071356619220143871973750380670420495722626540277141546853704453502562909306791685045031373467661131681245447148833330909962703142185937903212616044186346871542074272828072146756783959026590407852531117642742436997304638345486431845969028253743772232874927266087537986207158866084048753547876182678807964235979895578405803847502461039492476417081293332575410523848089271026552194744106232466060091481194107062426685104974467724051884406154898804316213028728752858074007202926320266624549/17862087144182552090100651130756164665976827862122256167423329247233898321830719666540870487544055009398369586776484355671625497973604947199022105534491506565630899537593179994034477830621111528653321226095994410794048389454474256755765855031213008497406701691127284482478696801343353279019313560663355819346556656789780042055905915034991279242393102960416490645747179216448162518823663813131314335790891510706123877326968206993303684221095374172364198108537519867527915452940344040267013916003084424614072759887985824821063642644988081209344

Approximate probability:
0.9310052380053759

Quote: ChesterDog

Show Spoiler

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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Quote: ThatDonGuy

Yes, you did, sort of; you used 3, 4, 5 instead of 2, 3, 4
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Quote: AnotherBill

Quote: gerback123

Find 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.

Show Spoiler

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

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Quote: Ace2

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 0.93101
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Quote: SkinnyTony

Quote: ThatDonGuy

Here'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...

Show Spoiler

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

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Quote: AnotherBill

Quote: Ace2

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 0.93101
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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).
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Quote: Ace2

Quote: AnotherBill

Quote: Ace2

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 0.93101
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).
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1.5^.5 =1.2247
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Quote: AnotherBill

Quote: Ace2

Quote: AnotherBill

Quote: Ace2

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 0.93101
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).
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1.5^.5 =1.2247
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Ah! Thank you for the explanation! You just missed the decimal point in you original post.
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