@@icodestuff6241 here on the Philippines we have 1 peso coin,but we also have centavos.So in a sense we can just give 67 centavos for a total of 2/3 of a coin🙂
@@icodestuff6241 Could work with the pre-decimal pound system. If a coin is 1 shilling, and since there are 12 pence in a shilling, 20/3 coins would be 6 shillings, 8 pence.
@@PuzzleAdda , "I" could be Teresa's daughter or any man that had a daughter with Teresa's daughter (could be son-in-law, could be random one night stand, could even be some incestuous weirdness like Teresa's husband, father, son, etc.). Not a great riddle given that there is no definitive answer. And, yes, I am intentionally ruining your riddle because you are randomly advertising your own video on another channel's video.
A basket contains 5 apples. Do you know how to divide them among 5 kids so that each one has an apple and one apple stays in the basket? Answer - ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-5qWsYMYAeRs.html
@@bificommander I suspect the original problem was both simpler and more complicated, in as much as there probably were coins worth 1/4 and 1/6 of a given amount, so both farmers made "whole coin" amounts even after the swap. For example, pre-decimalisation British currency (£1 = 20 s = 240 d)
Eggzactly. Farmer A's eggs cost 1/4 shilling or 3d. Farmer B's would have been 1/6 shilling or 2d. 20/3 would have been written as 6'/8 (6 shilling, 8 pence) back in the day. For farmer A to have sold all his eggs at a higher price, he was either set up at the front of the market with Farmer B further back, or it was a busy day and many wanted eggs, and therefore both sold out.
@@Tiqerboy I doubt it would have been shillings and pence with Euler being Swiss (and living in Russia and Prussia), but non-decimal currency in his time was the norm.
That awkward moment you accidentally make yourself do an extra 15 lines of working out because you make an extra variable you need to solve out instead of just X and (100-x)...
I think it's cleaner as ax = by. You then also have ay = 15 and bx = (20/3). Then you do ay/bx = 45/20 = 9/4, and multiply the left hand side by 1 in the form of ax/by, giving you (a^2)/(b^2) = 9/4, therefore a/b = 3/2. Therefore 3x = 2y = 2(100 - x), which re-arranges to 5x = 200 and so x = 40, y = 60.
Nice problem and thanks for sharing! The algebra is a little bit easier if we work with these variables: x = number of eggs sold by 1st farmer y = number of eggs sold by 2nd farmer Condition: x+y=100 n/x = price of eggs of 1st farmer (instead of a) n/y = price of eggs of 2nd farmer (instead of b) So when they sell the eggs, both earn n coins. The other two conditions of the problem are y·n/x=15 and x·n/y=20/3. Isolating the quantity y/x of these two equations we get 15/n=3n/20 ⇒ n²=100 ⇒ n=10, so x/y=2/3. Hence, if x and y are in proportion 2:3 and add up tp 100, they must be x=40 and y=60.
U just did what he did in the video, but instead of a and b u used other notation. But the way he did it in the video is much better way, bcz it is easly understandable what he did and why.
Well didn't need the quadratic formula but my way ended up being a quadratic anyway. From ax=by, I formed two equations by multiplying both sides by x and y respectively: ax^2=bxy and ayx=by^2. Subbing in values for bx and ay gives ax^2=20y/3 and 15x=by^2. Rearranging gives ax=20y/3x and by=15x/y. Now you can equate the equations since ax=by: 20y/3x=15x/y. This will give you a quadratic where 20y^2=45x^2. Solving for x/y ratio gives x/y=2/3, which means x=40 and y=60.
@@shehannanayakkara4162 yeah, i did it pretty similarly, except i was dense enough to apply a binomial with (3x^2-2y^2)(3x^2+2y^2)=0. though technically x/y=-2/3 would satisfy that equation as well. giving one farmer 300 eggs and the other one -200.
A basket contains 5 apples. Do you know how to divide them among 5 kids so that each one has an apple and one apple stays in the basket? Answer - ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-5qWsYMYAeRs.html
Was quite easy, but I solved it in a different way; A = eggs of first farmer B = eggs of second farmer C = price of first farmer D = price of second farmer First line tells us that AC = BD Second line tells us BC = 15 Third line tells us AD = 20/3 BC * AD = (15 * 20/3), or 100. As they got the same amount of coin, 100 is the square of how much they got (BC * AD is the same as AC * BD) So they got 10 each. If BC is 15, it means the second farmer has 50% more eggs. From this point it's easy to figure it out, 60 and 40. (x + 1.5x = 100, so 2.5x = 100, and x = 40).
Another puzzle that is easy but fun. It is the kind of puzzle where you apply simple knowledge every step of the way, to gradually unveil the solution, and that much is really fun, even if not very challenging.
It’s not a riddle, it’s a word problem that gives 4 equations and 4 unknowns. There’s no trickery involved in a system of equations (unless you’re really into matrices)
Sounded like a joke too, when you realise the second farmers anger at making less money and inflection of voice while he thought about how many coins he had made....
Great, when you brought in the money/coins, you didn't actually really use the analogy best, there's no such thing as 20/3 coins. You've essentially just got a maths problem here and dressed it up as egg salesmen.
I solved it a bit differently (after some useless tangents), and unnecessarily used four variables instead of three. I named the egg quantities x and y, and the prices a and b, giving me x+y=100, ax=by, ay=15, and bx=20/3. I then multiplied ay and bx, giving me 100. We can rearrange that to ax*by=100. Since ax=by, they also equal the square root of 100, ie 10. Since the first farmer would have gotten 15 coins instead of 10 had he sold the other farmer's eggs, we get that y=15x/10=3x/2. Then we just solve x+3x/2=100, which gives us x=40. Since x+y=100, y=60.
What was Euler (supposedly) trying to show with this riddle? This seems like less of a riddle from a master mathematician and more a problem assigned in the third week of algebra class.
The ratio of eggs sold is the reciprocal of the ratio of the egg price, which is necessary for the total sales to be equal. Switching the eggs around, this bias is applied twice instead of cancelled, so the ratio of the new sales is the square of the egg ratio, inversed, since the one with the lower sales total is the one with the lower price and more eggs initially, and vice versa. Then it's just splitting up 100 eggs into that ratio.
This problem just requires algebra, no crazy ideas. There are four unknown variables: number of eggs from farmer 1, number of eggs from farmer 2, price of each egg from farmer 1, and price of each egg from farmer 2. This means you need (not by coincidence) four equations which relate these variables. Find the system of equations and solve it! My work is below. Of course, the quadratic polynomial in Presh’s video is a completed square from the beginning, so that solution will appear more elegant than the one I show below. x + y = 100 ax = by bx = 15 ay = 20/3 So... y = 100 - x x = 100 - y a(100 - y) = b(100 - x) 100a - 20/3 = 100b - 15 Also... x = 15/b y = 20/(3a) 15/b + 20/(3a) = 100 3a + 4b/3 = 20ab Now... 100b = 100a + 25/3 b = a + 1/12 3a + 4(a + 1/12)/3 = 20a(a + 1/12) 3a + 4a/3 + 1/9 = 20a^2 + 5a/3 0 = 20a^2 - 8a/3 - 1/9 0 = 180a^2 - 24a - 1 a = -1/30 or a = 1/6 Assume positive solution (it is the price of a single egg from one of the two merchants) Then... y = 20/(3/6) = 40 x = 60
The square root of the ratio of price when they exchanged is equivalent to the ratio of eggs. Then set that ratio equivalent to 100. The ratio of price was 9:4, take square root which equaled 3/2. So the ratio of eggs was 1:1.5 which gave you 40 and 60.
if the eggs are in ratio x:y then their selling price has to be in ratio y:x since initially equal gain. then selling other guys eggs at your price would give us y^2=15a and x^2 = 20a/3 therefore (x/y)^2 = 4/9 or x:y =2:3 hence eggs are 40,60
Another way of solving it (more or less the same, but with two interesting points): 1) We can deduce each farmer's gain by multipliing a, b, x and (100-x) since a(100-x)=15 and bx=20/3, we know that abx(100-x)=15*20/3 = 100 since ax = b(100-x), we know that abx(100-x) = ax*b(100-x) = (ax)^2 = (b(100-x))^2 = (each farmer's gain)^2 and we can deduce that each farmer got sqrt(100) = 10 coins 2) we can deduce the ratio of the two prices since the second farmer got 10 coins, but would have gotten 15 coins with the price of the first farmer, we know that the price of the second farmer (b) is 2/3 of the price of the first farmer (a). Hence b/a = 2/3 (we can do the same deduction using the equation with 20/3) since ax = b(100-x), we know that x = (100-x)*b/a = (100-x)*2/3. Hence 3x = 2(100-x) and x = 40
It makes it easier just noticing that the geometric mean of 20/3 and 15 is 10. It means that the number of coins they each got was 10. Then knowing that, the ratio of 15 to 10 means the price ratios are 3 to 2.
I think the point of the puzzle is to think a little. We are given ax=by so that x/y=b/a. We are also told ay=15 and bx=20/3. Thus (bx)/(ay)=x^2/y^2=(20/3)(1/15)=4/9 from which x/y=2/3 or x=(2/3)y. Then (2/3)y+y=100 yields y=60, x=40
I didn't use the proportions (ratio) short cut. I did the algebra, got a quadratic, (without simplifying, it should be 25x^2 + 4000x - 200000 = 0), completed the square, and arrived at x = -80 +/- 120. Since you cannot have negative eggs, x = -80 + 120 = 40.
When I heard the farmers say "15 coins" and "20/3 coins", I was confused and thought you were referring to pennies, dimes, quarters, etc. (possibilities to be determined). The problem would be clearer if they had said dollars, euros, pounds, or some other monetary units instead of coins.
For those who need a simple breakdown: Let number of eggs of 1st farmer be A costing x cents each. Let number of eggs of 2nd farmer be B costing y cents each. A + B = 100 ---- (1) Ax = By => (x/y) = (B/A) ---------- (2) Bx = 15 ---------- (3) Ay = 20/3 ------- (4) Take equation (3) divided by (4): (Bx/Ay) = 15 / (20/3) => (B/A) (x/y) = 9/4 ----- (5) Substitute (2) into (5): (B/A)(B/A) = 9/4 => B/A = 3/2 (negative value rejected as A & B are positive integers) => B = 3A/2 ----- (6) Substitute (6) into (1): A + (3/2)A = 100 => A = 40 => B = 60
Continuing from that solution: The first farmer claimed that if he sell his eggs (60 of them) for the second one's price, he'll get 15 coins. So the second farmer's price is 1/4 coin per egg (or a coin per 4 eggs). The second farmer claimed that if he sell his eggs (40 of them) for the first one's price, he'll get 20/3 coins. So the first farmer's price is 1/6 coin per egg (or a coin per 6 eggs). The first farmer sold his 60 eggs for a coin per 6 eggs. The first farmer sold his 40 eggs for a coin per 4 eggs. Each earned 10 coins.
The wording made me think there was more complexity than you intended. “15 coins” makes it sound like there’s a mix of denominations-quarters, dimes, nickels, etc-and that we need to figure out the mix.
I have a great problem, based on a paper coffee cup: the frustrum of a cone, of height H and base radius of r and top radius of R. The volume is (r^2+R^2+rR)H pi/3 = V the problem is how to calculate new h so to get V/2 (how far down to drink to drink just half).
x = number of eggs sold by the first farmer at price p y = number of eggs sold by the second farmer at the price q if x> y, being x * p = y * q by hypothesis, it will be p
At 1:36, instead, assume that the 2nd farmer has y eggs; then x + y = 100 (1) ax = by (2) ay = 15 (3) bx = 20/3 (4) It is crucial to see: "(3)*(4)" & (2) --> aybx = (ax)-squared = 100 --> ax = by = 10. (5) Now, the rest is easy. "(3)/(5)" --> y/x = 3/2 (6) (1) and (6) --> x = 40 eggs, y = 60 eggs. Finally, from (5) we get: a = 1/4 coin, b = 1/6 coin. (Each farmer received 10 coins.)
60 and 40. The puzzle becames simple by the following: Assume first farmer sold X eggs for M coins each and second farmer sold Y eggs for N coins each. Then X + Y = 100 (1) X*M = Y*N (2) X*N = 15 (3) Y*M = 20/3 (4) (2) might be written as X/Y * M/N = 1 (5) Dividiing (3) by (4) we get X/Y * N/M = 45/20 (6) Finally, multiplying (5) and (6) N and M disappear so we get X^2/Y^2 = 45/20 so 2*X = 3*Y (7), X+ Y = 100 and 2X =3Y hence X = 60 and Y = 40
I used a needlessly complicated method, and got the right answer but had the amounts reversed. Went over my math, made sure I got everything right...then I realized I misrepresented the problem in the first two equations. I accidentally set Farmer 1's eggs at Farmer 2's price equal to 15, and Farmer 2's eggs at Farmer 1's price equal to 20/3.
@@bzboii It is four variables and equations actually. The fourth one is x + y = 100 which is already solved for y to begin with and substituted in to the other equations. 😊
I wasted the pages 😂😂, and thought this problem require number theory, and some use of diophantine equations, but I again read the question , and my reaction was😶
Tbh I was surprised this euler problem was just a simult eq question... And for the curious, first farmer sold each egg at 1/4 coins, second farmer sold each at 1/6 coins (a=1/4, b=1/6, x=40)
What if the farmers are allowed to have negative eggs? In that case, I got that the first farmer sold -200 eggs and the second sold 300 eggs. But in this case the second farmer was giving the customers 1/30 of a coin with each egg.
I solve by creating equations. We call the number of eggs farmer 1 and 2 sold is x,y (eggs) We also call the price of each eggs farmer 1 and 2 sold is z,t (coin/eggs) We have : x + y = 100 xz = yt yz = 15 xt = 20/3 This could now be easily solved.
The other possibility is the first farmer had 300 eggs and he paid someone 10 coins to dispose of them; while farmer two had -200 eggs and paid someone ten coins to satisfy his debt.
Can you please do a video explaining The Verhoeff algorithm. I'm completely lost. I would appreciate the help. Plus, it is interesting. Thank you - F. Roach
I actually had a different solution. I set it up with a, b, x, and y. ax = by We can multiply by ab since both are positive. ax^2y = bxy^2 Substituting the values for ay = 15 & bx = 20/3 15x^2 = (20y^2)/3 9x^2 = 4y^2 Since x & y are positive we don’t have to worry about square roots. 3x = 2y x = 2y/3 Substitute back into x + y = 100 and y = 60 x = 40 a = .25 b = 1/6 It’s probably convoluted, but I liked the symmetry of multplying by xy and getting squares of each.
i did it a little differently; I used another variable for the number of farmer b's eggs. I then had 4 equations and 4 unknowns, and I substituted until I was home. along the way, I had 45x^2 = 20y^2, i took the square roots and reduced them, was left with sqrt 5 on both sides, divided that out, and got 2x = 3y. Made a final substitution of x = 100 - y, and that was it. Lots of fun.
The second farmers voice is the icing on the cake here 😅 " I would have GotTen....." sounds like he was telling the first farmer to be happy with what he did get...
One sold 40 eggs at 0.25 apiece, while the other sold 60 eggs at 0.16666667 apiece. Answer 40 eggs and 60 eggs I approach it differently: let y= your egg price, let m=my egg price, let n=numbers of your eggs sold, let 100-n= numbers of my eggs sold. I got this equation 15m^2= 20/3y^2 y^2/m^2= 20/15 y/m= 1.5 (this means 1 price was one and one-half greater than the other price). y=1.5 x 20/3 or y=10/n substitute into 100-n=ny/m to get 100-n=30n/20 or 2000-20n=30n or 2000=50n n = 40 and n-100 = 60 (Answer) PS 0.25 is one and one-half greater than 0.16666667 (or 1/6). 0.25 cents x 40 eggs = 10cents , and 0.16666667 x 60 eggs = 10 cents, so they make the same amount of money (10 cents) when they together sold 100 eggs. But if each was selling at the other seller's price one would earn 15 cents (0.25 x 60 eggs) and the other 20/3 cents (0.16666667 x 40 eggs) instead of each making 10 cents each.
Arranging the algebra differently, you get a polynomial for which Wolfram alpha says that the 2nd farmer has either 40 or -200 eggs is also a solution, meaning the first farmer would have 300 eggs
I first only read the right parameters, and came up with an equation that I simply could not solve. I then watched the solution and saw the answer started with "since the total is 100 eggs.." I made a giant facepalm and entered the rest. Was quite simple after that.
So both farmers got 10 coins each. Also one of the farmers sold eggs only in quantities divisible by 3, otherwise he would not have been able to get exactly 10 coins.
I tried it, before making a reasonable guess of 60 eggs for the first farmer and 40 eggs for the second farmer. I worked out the fractions of a coin they would receive per egg if this were the case, and farmer A's price would be 1/6 coins per egg, while farmer B's price would be 1/4 coins per egg. If they each had the different farmer's price, the prices do align, so my answer is that farmer A had 60 eggs, while farmer B had 40 eggs. My answer is wrong because I assumed that A would be selling his own eggs at B's price, not B's eggs at his price. My mistake.
the ship is twice older now than the captain was when the ship was as old as the captain now. Together they are 140 years old. how old is not the captain and the ship?
(1) A and B are positive whole numbers (2) A+B = 100 (3) A*Ap = B*Bp (4) Ap and Bp are not equal to each other. (5) B*Ap = 15 (6) A*Bp = 20/3 what is A and B? first we rearrange (5) and (6) so we get: (7) Ap = 15/B (8) Bp = 20/(3A) then in (3) we substitute using (7) and (8) (15A)/B = (20B)/(3A) multiply both sides by 3AB 45*A^2 = 20*B^2 divide both sides by 20 2.25*A^2 = B^2 take the square root of both sides sqrt(2.25*A^2) = sqrt(B^2) apply sqrt(X*Y)=sqrt(X)*sqrt(Y) to the left side sqrt(2.25)*sqrt(A^2) = sqrt(B^2) resolve the square roots, keeping in mind (1) to resolve the sign. (9) 1.5*A=B (2) with (9) gives 2.5*A = 100 (10) A=40 (2) with (10) gives 40+B=100 (11) B=60 (10) and (11) gives us our answer: A=40 B=60
You are given $100 to start a farm. In order for you to be successful you have to spend all your money and buy at least one of each of the following animals at the given prices. Cows $10 Sheep $3 Chickens $.50 How many of each do you buy to spend exactly $100?
That was somewhat easy... I just took the geometric mean of the two numbers they would've gotten if they traded prices (basically the actual money they each got lies between the two numbers). This is sqrt (15 x 20/3 = 100) which is 10. So they had actually got 10 coins each. Farmer B would've made 50% more profit (15 up from 10) if he sold his eggs at farmer A's price meaning that Farmer A's price is 50% higher than his. That also means that since they both made 10 coins. farmer B sold 50% more eggs than farmer A; in short the total is 100 eggs so farmer A sold 40 eggs for 1/4 coins per egg and farmer B sold 60 for 1/6 coins per egg (1:1.5 ratio in opposite ways). Cross checking our answer, farmer A would've gotten 60 x 1/4 = 15 coins and farmer B would've gotten 40 x 1/6 = 20/3 coins. Which is exactly what is mentioned in the problem. No need for complicated algebra here, folks!
At first I was thinking 20/3 coins doesn't make sense, but thinking of pounds, shilling and pence, 20/3 shillings does make sense. That would be 6 shilling 8 pennies (or pence). There are 12 pennies to a shilling. And it's reasonable to think a farmer back then would have gotten about that amount of money by selling his eggs at market back in the day. And given how old the problem was, that probably was the intent. Even though Euler lived in Germany (or was it Switzerland?), I have to think he was familiar with the British monetary system based on 12.
I got Farmer A is 60 eggs and Farmer B is 40 eggs. ... edit: I misread what the farmers said. I thought it was, "If I sold MY eggs at YOUR price..." Like, why would you even try to sell another person's eggs? I get changing the price, but not what you have.
I solved it way easier than Presh; so easy I didn't even write anything down to get the answer. Let the ratio of the number of eggs brought by the first farmer to the number brought by the second be a. Then, as both actual sales brought the same amount of money, the ratio of the prices is 1/a. The ratio of the prices in the hypothetical situations is therefore a². That hypothetical ratio is 15/(20/3) = 45/20 = 9/4, so the original ratio is the square root, 3/2. The first farmer brought 3/5 of the eggs = 60, the second 2/5 of the eggs = 40.
Man these first eggs are egg-spensive! He is egg-sploiting his customers... And the second farmer is just asking for chicken feed. If you were wondering what the prices are (spoiler!): The first farmer sells 4 eggs for 1 coins and the second farmer sells 6 eggs for 1 coin.
I solved it very differently. x = number of eggs farmer 1 y = number of eggs farmer 2 m = farmer 1's price n = farmer 2's price x + y = 100 xm = yn xn = 20/3 ym = 15 Solving the last two equations for m and n and substituting them back into equation 2 yields: 15x^2 = (20/3)y^2 12x^2 - (20/3)y^2 = 0 Multiply by the and factor out the five: 45x^2 - 20y^2 =0 9x^2 - 4y^2 = 0 Factor: (3x-2y)(3x+2y) Yields two possible simultaneous sets of equations: [1 1][x] = [100] [3 2][y] = [0] or [1 1][x] = [100] [3 -2][y] = [0] Solving for both yields two possible solutions: (-200, 300) or (40, 60) Since we can assume that both farmers brought a positive number of eggs, the first farmer must have brought 40 eggs and the second farmer must have brought 60 eggs.
xa=(100-x)b xa=100b-xb x(a+b)=100b x=100b/(a+b) xb=20/3 x=20/3b (100-x)a=15 100-x=15/a x=100-15/a 20/3b=100-15/a b=15-9/4a x=100(15-9/4a)/(15-9/4a+a) idk, i just didn’t want to take a piece of paper so i wrote it here down, don’t know what i was getting to 😂
