SOLUTION: Caroline bought some bones for her 7 dogs. Had she owned 8 dogs, she could have given each the same number of bones. As it was, she needed two more bones to give each dog the same

Algebra ->  Expressions-with-variables -> SOLUTION: Caroline bought some bones for her 7 dogs. Had she owned 8 dogs, she could have given each the same number of bones. As it was, she needed two more bones to give each dog the same       Log On


   



Question 1177011: Caroline bought some bones for her 7 dogs. Had she owned 8 dogs, she could have given each the same number of bones. As it was, she needed two more bones to give each dog the same number of bones. The number of bones she could have bought was ___.
Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i think 40.
if she had 8 dogs she could have given each dog 5 bones because 8 * 5 = 40.
she only has 7 dogs.
6 bones each would require 42.
since she only has 40, she would need 2 more to give each of her 7 dogs the same number of bones.

to be enough to give each of 8 dogs the same number of bones, she would need to buy bones in multiples of 8.

if she buys 8 bones, she can feed 7 dogs 1 bone each and have 1 left over.
she will be 6 short from feeding 7 dogs 2 bones each.

if she buys 16 bones, she can feed 7 dogs 2 bones each and have 2 left over.
she will be 5 short from feeding 7 dogs 3 bones each.

if she buys 24 bones, she can feed 7 dogs 3 bones each and have 3 left over.
she will be 4 short from feeding 7 dogs 4 bones each.

if she buys 32 bones, she can feed 7 dogs 4 bones each and have 4 left over.
she will be 3 short from feeding 7 dogs 5 bones each.

if she buys 40 bones, she can feed 7 dogs 5 bones each and have 5 left over.
she will be 2 short from feeding 7 dogs 6 bones each.

she more then likely bought 40 bones.
8 dogs would have had 5 bones each.
7 dogs would have 5 bones each with 5 bones left over.
with another 2 bones, she could feed each dog 6 bones each.

i believe this is the answer you are looking for, if i understand the problem correctly.





Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The number of bones she has is a multiple of 8, because she could have divided the bones equally among her dogs if she had had 8 dogs instead of 7:
N+=+8a

She would have needed 2 more bones to divide them equally among her 7 dogs, so the number of bones she has is 2 less than a multiple of 7:
N+=+7b-2

So we have

8a+=+7b-2

This is a Diophantine equation: two variables but only one equation and the variables must have whole number values. Here is a standard method for solving linear Diophantine equations.

(1) Solve for the variable with the smaller coefficient:

8a+=+7b-2
7b-2+=+8a
7b+=+8a%2B2

Perform the division as a quotient and a remainder:

7b+=+7a%2B%28a%2B2%29
b+=+a+%2B+%28a%2B2%29%2F7

a and b are integers, so %28a%2B2%29%2F7 has to be an integer. So a can be any integer that is 2 less than a multiple of 7.

POSSIBLE ANSWERS:
(1)               a = 5
    b = a + (a+2)/7 = 6
             N = 8a = 40
           N = 7b-2 = 40

(2)               a = 12
    b = a + (a+2)/7 = 14
             N = 8a = 96
           N = 7b-2 = 96

(3)               a = 19
    b = a + (a+2)/7 = 22
             N = 8a = 152
           N = 7b-2 = 152

And of course there is no end to the list of possible answers. However, only the smallest few are reasonable answers in terms of number of bones and number of dogs.