How Many Animals Does Bert Have?

[K Sengupta]

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?

 

[HallsofIvy]

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.

 

[Rogerio]

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.



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.

 

[gunch]

(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).

 

[HallsofIvy]

I have edited my post so I can pretend I didn't make that mistake!

 

[Borek]

What about 2 cows, 59 horses and 2 dogs?

 

[Borek]

Interestingly, no more solutions for primes < 100000.

 

[uart]

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.

 

[Borek]

Ah OK, somehow missed this condition.