Solution for #17

Let a be the number of red squeegies you buy. Let b be the number of yellow squeegies you buy. Let c be the number of blue squeegies you buy. We know two equations:

             a + b + c = 100
        6a + 3b + 0.1c = 100 

By multiplying the first equation by 6 and then 3, then subtracting these two equations from the second, we can come up with two more equations:

        6a + 6b + 6c = 600
        3a + 3b + 3c = 300
                  3b = 500 - 5.9c
                  3a = 2.9c - 200 

Normally, two equations isn't enough to solve for three variables. But we know that a and b are nonnegative integers. So, if 3b ≥ 0, then 500 - 5.9c ≥ 0. This means c ≤ 84.75. Also, if 3a ≥ 0, then 2.9c - 200 > 0. This means c ≥ 68.97.

However, since buying blue squeegies is the only way to spend a fraction of a dollar, the number of blue squeegies we buy must cost an even dollar amount. The only two numbers of blue squeegies we can buy between 68.97 and 84.75 that satisfy this condition are 70 and 80.

If we substitute 80 for c in the last two equations listed above, we can solve for a, which equals 10.67, and b, which equals 9.33. But the values of a and b must be integers, so we know this is not the solution.

If we substitute 70 for c in the last two equations listed above, we can solve for a, which equals 1, and b, which equals 29.

So one red squeegie must be bought for $6.00, 29 yellow squeegies must be bought for $87.00, and 70 blue squeegies must be bought for $7.00.