How Many Animals Does Bert Have?

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Discussion Overview

The discussion revolves around a mathematical problem involving Bert's animals, specifically the number of cows, horses, and dogs he has, each represented by different prime numbers. The problem includes a specific equation relating these quantities.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the equation c*(c+h) = 120 + d can be used to find the values of c, h, and d, where c, h, and d are different prime numbers.
  • One participant suggests using a "guess and check" method, starting with c=11 and h=3, leading to a potential solution of c=11, h=3, and d=23.
  • Another participant notes that if both h and c are even, it is impossible to satisfy the conditions of the problem.
  • There is a discussion about the implications of setting h=2 and c as an odd prime, leading to different values for d.
  • One participant corrects a previous claim regarding the primality of d, stating that d can be expressed as (c-10)(c+12) and is prime when c=11.
  • Another participant questions the validity of the solution involving 2 cows, 59 horses, and 2 dogs, pointing out that the problem specifies that each prime must be different.
  • One participant notes the absence of additional solutions for primes less than 100,000.

Areas of Agreement / Disagreement

Participants express differing views on potential solutions and the implications of the conditions set by the problem. There is no consensus on a definitive solution, and some participants acknowledge missed conditions in their reasoning.

Contextual Notes

Participants highlight the importance of the condition that each of the three numbers must be a different prime number, which complicates the search for solutions.

K Sengupta
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Bert has some cows, horses and dogs, a different prime number of each.

If the number of cows (c) is multiplied by the total of cows and horses (c+h), the product is 120 more than the number of dogs (d), that is:
c*(c+h) = 120 + d.

How many cows, horses and dogs does Bert have?
 
Last edited:
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K Sengupta said:
Bert has some cows, horses and dogs, a different prime number of each.

If the number of cows (c) is multiplied by the total of cows and horses (c+h), the product is 120 more than the number of dogs (d), that is:
c*(c+h) = 120 + d.

How many cows, horses and dogs does Bert have?
I can't think of any "formula" for solving something like this but "guess and check" is an old, respected method for solving problems. 120 is about 11 squared so start by trying numbers around that. If c= 11, h= 3 c+h= 13, c(c+h)= 11(13)= 143= 120+ 23 and 23 is prime! c= 11, h= 3, d= 23 works.
 
Of course, h=even and c=even is impossible.

If h=2 , c=odd
c**2 + 2c = 120 + d
(c+1)**2 = 11**2 + d , so d is not prime.

If h=odd , c=odd, then d=2 , and c*(c+h) = 2*61 , which is impossible.

If c=2 and h=odd, then d=2. Then, c=2 and h=59.

:smile:

Edited:

Of course, I should have noted "c" and "d" cannot be 2 ( Remember: "a different prime number").
And then, I should have seen d=(c+12)(c-10) is prime if c=11, which implies d=23, and h=2.
 
Last edited:
Rogerio said:
(c+1)**2 = 11**2 + d , so d is not prime.
This is incorrect. It yields,
d = (c-10)(c+12)
which is prime if c=11. So h=2, c=11, d =23 is a solution (I assume this is the one HallsofIvy meant).
 
I have edited my post so I can pretend I didn't make that mistake!
 
What about 2 cows, 59 horses and 2 dogs?
 
Interestingly, no more solutions for primes < 100000.
 
Borek said:
What about 2 cows, 59 horses and 2 dogs?

Borek, the question says that each of the three primes is different.

Bert has some cows, horses and dogs, a different prime number of each.
 
Ah OK, somehow missed this condition.
 

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