# Math - Linear algebra

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Cats cost \$1, dogs cost \$15, and mice cost only \$0.25. I want a herd of 100 animals, and I have \$100, so being a cat lover, it's tempting to get only cats and mice. But I also want a dog, and my cat(s) want mice to "play" with, so the numbers of dogs, cats and mice are each positive. And no cash left over. What's the solution?

• Math - Linear algebra -

C = cats
D = dogs
M = mice

1(C) + 15D + .25M = 100 or
4C + 60D + M = 400 (1)

Also, C + D + M = 100 (2)

Subtracting (1) from (2) yields 3C + 59D = 300

Dividing through by the lowest coefficient yields
C + 19D + 2D/3 = 100

Let 2D/3 = an integer "k" making D = 3k/2

Substituting, 3C + 88.5k = 300 yielding C = 100 - 29.5k

"k" must be even and can only be 2 making C = 100 - 59 = 41 from which D = 3 and m = 56.

41 + 45 + 14 = 100 animals
41(1) + 45(15) + 56/4 = \$100

• Math - Linear algebra -

Another way:

Will use the same notations, C,D and M for cats, dogs and mice, and with the same constraints where C,D and M must be integers.

From the cost,
C+15D+M/4=100....(1)
from the number of animals,
C+D+M = 100......(2)

(1)-(2)
14D=(3/4)M, or
56D=3M
Since 56 and 3 are coprime, we can only have 3 dogs, 56 mice, or any multiple thereof.
However, 6 dogs and 112 mice would exceed the budget, so we go with
3 dogs, 56 mice, and are left with 41 cats.
Check:
3+56+41=100
3*15+56/4+41=100
So the solution is correct.

• Math - Linear algebra -

Thank you, that was the answer I got too, but in a slightly different method, but I want to learn for how you figure this out, with this problem. I feel like you just you the same method but I am having a little problem with this one:

"Suppose you want to make \$5 using exactly 100 common US coins.Easy, you say: just use nickels (5 cents each). But I say, no - we have NO nickels, only pennies (1 cent), dimes (10 cents) and quarters (25 cents) - is it possible to make \$5 now with exactly 100 of these coins? Why or why not?"

• Math - Linear algebra -

These problems belong to a class called diophantine equations. If these are what you are doing in class, they will be solved in a different way.

For linear algebra, I will solve it similar to the previous problem, as follows.

Let integers
p=number of pennies
d=number of dimes
q=number of quarters
then again,
p+10d+25q=500...(1) (number of cents)
p+d+q=100 ....(2) (number of coins)
Subtract (2) from (1),
9d+24q=400 or
3d+8q = 400/3 ...(3)
It is evident that (3) has no solution in integers because we cannot have the sum of two integers to be a mixed fraction.

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