Easy math puzzles
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43 members have voted
October 21st, 2021 at 6:46:19 AM
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None.Quote: WizardAny objections to:
beers_i_owe_charlie+=1.0; ?
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October 21st, 2021 at 7:00:21 AM
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Quote: WizardQuote: ChesterDog
The sheep started grazing 4.5 days after the grass started growing.
Let t stand for the time starting at t = 0 when the sheep first start grazing in that field. And let L be the length of a uneaten grass blade. Then L = a + bt, where a is the length at t = 0, and b is the growth rate of the blade in inches per day.
Call the eating rate of the sheep, expressed as pounds per day, M. It is related to their eating rate expressed as acres per day by:
M = K( a + bt ) dA/dt, where K is some constant related to the number of blades per acre.
Rearranging the equation:
dA = M/[K( a + bt )] dt
Integrating the equation from 0 days to 3 days gives:
1 acre = M/(bK) ln [( a + 3b ) / a]
Integrating from 3 days to 8 days gives:
1 acre = M/(bK) ln [( a + 8b ) / ( a + 3b )]
Equating and dividing both sides by the constants gives:
ln [( a + 3b ) / a] = ln [( a + 8b ) / ( a + 3b )]
Exponentiating both sides to the power e:
( a + 3b ) / a = ( a + 8b ) / ( a + 3b)
Solving:
( a + 3b )2 = a2 + 8ab
a2 + 6ab + 9b2 = a2 + 8ab
9b2 = 2ab
9b = 2a
The length of a grass blade is L = a + bt = a + (2a/9) t.
To achieve the initial length a, the blade would have had to grow 9/2 = 4.5 days.
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I agree!
I think this one is beer worthy. Thus,
(int)(beers I owe Charlie)++;
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Quote: WizardAny objections to:
beers_i_owe_charlie+=1.0; ?
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I am not a lawyer, but as a sometimes Public Offender, I humbly object on behalf of ChesterDog that apart from all this discussion, you seem to be advancing the wrong variable???
beers_i_owe_ChesterDog+=1.0; ?
(NOTE: I don't know "C++" so maybe variable names only go by the first two letters & CHarlie and CHesterDog are the same.)
(NOTE: I am not implying any sock puppetry...too much of that lately)
October 21st, 2021 at 8:01:17 AM
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Channeling your inner Marty McFly, you travel one week back in time in an attempt to win the lottery. It’s worth $10 million, and each ticket costs a dollar. Note that if you win, your ticket purchase is not refunded. All of this sounds pretty great.
The problem is, you’re not alone. There are 10 other time travelers who also know the winning numbers. You know for a fact that each of them will buy exactly one lottery ticket. Now, according to the lottery’s rules, the prize is evenly split among all the winning tickets (i.e., not evenly among winning people).
How many tickets should you buy to maximize your profits?
Last edited by: Gialmere on Oct 21, 2021
Have you tried 22 tonight? I said 22.
October 21st, 2021 at 8:58:05 AM
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Let x be the number of tickets I buy.
Then my win is given by:
win(x) = x*(9999990 - x)/(10+x)
Taking the derivative and setting the numerator to 0 gives a quadratic with positive solution x = 9990 tickets.
Plugging this back in gives a maximized profit of $9,980,010.
My calculus is rusty, so surely I made an error somewhere. I am also not sure if the final prize pool is 9999990-x or 10M - x, based on the problem description.
Then my win is given by:
win(x) = x*(9999990 - x)/(10+x)
Taking the derivative and setting the numerator to 0 gives a quadratic with positive solution x = 9990 tickets.
Plugging this back in gives a maximized profit of $9,980,010.
My calculus is rusty, so surely I made an error somewhere. I am also not sure if the final prize pool is 9999990-x or 10M - x, based on the problem description.
Last edited by: teliot on Oct 21, 2021
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October 21st, 2021 at 9:47:02 AM
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Quote: DieterI don't think 6 decimal places of partial beers is what they usually mean by "microbrews".
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Okay, that was funny and nerdy. My kind of site.
Order from chaos
October 21st, 2021 at 10:21:45 AM
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I agree with Eliot's answer. Hopefully I can get some of the credit by showing a more complete solution.
Let n = tickets purchased.
Let f(n) = Net profit = 10,000,000*(n/(n+10)) - n
f'(n) = 10,000,000*((n+10)*1 - n*1)/(n+10)^2 - 1 = 0 (using the quotient rule)
A little cancelling and moving things around gives us:
100,000,000 = (n+10)^2
10,000 = n+10
n = 9,990
I'd like to request that Eliot and I split a beer, on G.
Last edited by: Wizard on Mar 21, 2022
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
October 21st, 2021 at 10:26:38 AM
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Quote: Wizard
I agree with Eliot's answer. Hopefully I can get some of the credit by showing a more complete solution.
Let n = tickets purchased.
Let f(n) = Net profit = 10,000,000*(n/(n+10)) - n
f'(n) = 10,000,000*((n+10)*1 - n*1)/(n+10)^2 - 1 = 0 (using the quotient rule)
A little cancelling and moving things around gives us:
10,000,000 = (n+10)^2
10,000 = n+10
n = 9,990
I'd like to request that Eliot and I split a beer, on G.
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Mike, I would request that my 1/2 beer be donated to The Animal Foundation -- https://animalfoundation.com/
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October 21st, 2021 at 1:58:33 PM
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This problem is inspired by the Squid Game, which I highly recommend to anybody not squeamish about blood. Mrs. Wizard couldn't take it and had to leave the room.
In one of the games there is a glass bridge. The bridge is divided into glass panels in an 18 by 2 grid. Let's explain it as 18 rows and two columns. In each row, one piece of glass is tempered, meaning a player could easily stand in it without the glass breaking. The other piece was ordinary glass that would break under the weight of a player.
The players must advance, one at a time, in a predesignated order. There are 16 total players. The rules are a little different on the show, but let's keep it simple here.
Assuming random guessing at each new row, what is the expected number of players to cross safely?
In one of the games there is a glass bridge. The bridge is divided into glass panels in an 18 by 2 grid. Let's explain it as 18 rows and two columns. In each row, one piece of glass is tempered, meaning a player could easily stand in it without the glass breaking. The other piece was ordinary glass that would break under the weight of a player.
The players must advance, one at a time, in a predesignated order. There are 16 total players. The rules are a little different on the show, but let's keep it simple here.
Assuming random guessing at each new row, what is the expected number of players to cross safely?
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
October 21st, 2021 at 2:02:01 PM
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How many players?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
October 21st, 2021 at 2:04:09 PM
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Quote: unJonHow many players?
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Sorry, I forgot that. I amended the question to say there are 16 players and 18 rows.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
