"Let x = number of cows farmer A had. Then farmer B would have had 30-x cows.
Let y = farmer A’s selling price for each cow and let p= farmer B’s selling price for each cow.
Then, xy =p(30-x)…….they get the same money for their cows…equation 1
Also, px = 320 which implies that p= 320/x……eqn 2
And y(30-x) = 245 which implies that y= 245/(30-x)…eqn 3
Substituting values of p and y from equations 2 and 3 into equation 1 and simplifying, we get
x^2 -256x +3840 = 0 which gives x = 240 or 16. Since x is less than 30, x = 16. Farmer A owned 16 cows.
"
Kevin, Australia
Monday, September 1, 2025
"Farmer A owned 16 cows and sold the cows for 17.50 pounds each. Farmer B owned 14 cows and sold the cows for 20 pounds each.
🐮 🐮 🐮"
Chris, Scotland
Monday, September 1, 2025
"Pleased that the 1885 puzzle was within my grasp!
Farmer A has a cows, selling each for £x.
Farmer B then has (30-a) cows, selling each for £y.
Three facts:
Equal income: ax = (30-a)y
Switch price and Farmer A earns 320: ay = 320
Switch price and Farmer B earns 245: x(30-a) = 245
If we take these last two equations and rearrange then we get x and y in terms of a: y = 320/a and x = 245/(30-a)
Substituting these into the equal income equation we end up with 245a/(30-a) = 320(30-a)/a
This can be simplified to get a quadratic 75a^2 - 19200a + 288000 = 0
Dividing through by 75 we get a^2 - 256a + 3840 = 0
Factorise (a - 240)(a - 16) = 0
This gives two solutions: a = 240 or a = 16
But there were only 30 cows in the first place so Farmer A must've had 16 cows (selling each at £17.50) and Farmer B must've had 14 cows selling at £20.
Thanks. for the amoooosing puzzle.
"
Rick, United States
Tuesday, September 2, 2025
"Here is my answer to the September 2025 puzzle. It appears to be four equations with four unknowns.
Let A = number of cows owned by farmer A.
Let B = number of cows owned by farmer B.
Let X = price of cows owned by farmer A.
Let Y = price of cows owned by farmer B.
[1] A+B=30
[2] AX=BY
[3] AY=320
[4] BX=245
From equation [3]
[5] Y=320/A
And from equation [4]
[6] X=245/B
Substituting the values for X and Y into equation [2]
[7] 245A/B=320B/A
From equation [1]
[8] B=30-A which can be incorporated into equation [7].
[9] 245A/(30-A)=320(30-!)/A)
Since I am lazy, I let CoPilot solve this and received two answers to the quadratic equation (16 and 240) but since 240 is not in the range of 1 to 30, the answer is 16 cows owned by A.
"
Leonard, United States
Tuesday, September 2, 2025
"Here’s my take on September’s Puzzle of the Month.
Focusing in on integer solutions, I considered the money received after swapping prices. Considering the prime factorizations of 320 (2,2,2,2,2,2,5) and 245 (5,7,7) respectively, I created the following table in MSExcel.
Unfortunately, none of the rows were solutions for both criteria. I took this to mean that the prices might not be integers (certainly the number of cows were). The easiest “path forward” seemed to be to add a factor of 2 to the 245 amount. The new table looked like this:
Further investigation of the (16,14) and (20,10) combinations showed the Farmer A having 16 cows was indeed a solution. Here is my final analysis:
Mala, New Zealand
Monday, September 1, 2025
"Let x = number of cows farmer A had. Then farmer B would have had 30-x cows.
Let y = farmer A’s selling price for each cow and let p= farmer B’s selling price for each cow.
Then, xy =p(30-x)…….they get the same money for their cows…equation 1
Also, px = 320 which implies that p= 320/x……eqn 2
And y(30-x) = 245 which implies that y= 245/(30-x)…eqn 3
Substituting values of p and y from equations 2 and 3 into equation 1 and simplifying, we get
x^2 -256x +3840 = 0 which gives x = 240 or 16. Since x is less than 30, x = 16. Farmer A owned 16 cows. "
Kevin, Australia
Monday, September 1, 2025
"Farmer A owned 16 cows and sold the cows for 17.50 pounds each.
Farmer B owned 14 cows and sold the cows for 20 pounds each. 🐮 🐮 🐮 "
Chris, Scotland
Monday, September 1, 2025
"Pleased that the 1885 puzzle was within my grasp!
Farmer A has a cows, selling each for £x.
Farmer B then has (30-a) cows, selling each for £y.
Three facts:
Equal income: ax = (30-a)y
Switch price and Farmer A earns 320: ay = 320
Switch price and Farmer B earns 245: x(30-a) = 245
If we take these last two equations and rearrange then we get x and y in terms of a: y = 320/a and x = 245/(30-a)
Substituting these into the equal income equation we end up with 245a/(30-a) = 320(30-a)/a
This can be simplified to get a quadratic 75a^2 - 19200a + 288000 = 0
Dividing through by 75 we get a^2 - 256a + 3840 = 0
Factorise (a - 240)(a - 16) = 0
This gives two solutions: a = 240 or a = 16
But there were only 30 cows in the first place so Farmer A must've had 16 cows (selling each at £17.50) and Farmer B must've had 14 cows selling at £20.
Thanks. for the amoooosing puzzle. "
Rick, United States
Tuesday, September 2, 2025
"Here is my answer to the September 2025 puzzle. It appears to be four equations with four unknowns.
Let A = number of cows owned by farmer A.
Let B = number of cows owned by farmer B.
Let X = price of cows owned by farmer A.
Let Y = price of cows owned by farmer B.
[1] A+B=30
[2] AX=BY
[3] AY=320
[4] BX=245
From equation [3]
[5] Y=320/A
And from equation [4]
[6] X=245/B
Substituting the values for X and Y into equation [2]
[7] 245A/B=320B/A
From equation [1]
[8] B=30-A which can be incorporated into equation [7].
[9] 245A/(30-A)=320(30-!)/A)
Since I am lazy, I let CoPilot solve this and received two answers to the quadratic equation (16 and 240) but since 240 is not in the range of 1 to 30, the answer is 16 cows owned by A. "
Leonard, United States
Tuesday, September 2, 2025
"Here’s my take on September’s Puzzle of the Month.
"
Focusing in on integer solutions, I considered the money received after swapping prices. Considering the prime factorizations of 320 (2,2,2,2,2,2,5) and 245 (5,7,7) respectively, I created the following table in MSExcel.
Unfortunately, none of the rows were solutions for both criteria. I took this to mean that the prices might not be integers (certainly the number of cows were). The easiest “path forward” seemed to be to add a factor of 2 to the 245 amount. The new table looked like this:
Further investigation of the (16,14) and (20,10) combinations showed the Farmer A having 16 cows was indeed a solution. Here is my final analysis: