A woman bought some 13 cent stamps and some 8 cent stamps at the posg office. She spent exactly 1 dollar. How many of each stamp could she have bought?
number of 13 cent stamps --- x
numbr of 8 cent stamps -- y
13x + 8y = 100
both x and y have to be whole numbers
clearly 0 < x < 7
y = (100-13x)/8
furthermore, for 100-13x to be divisible by 8, it must be even, so x must be even
A quick check tells us that x = 4 is the only solution in our domain
if x = 4, y = 6
so 4 13's and 6 8's
check:
4(13) + 6(8) = 100
To determine how many of each stamp the woman could have bought, we can set up an equation using the given information.
Let's say she bought x 13 cent stamps and y 8 cent stamps.
According to the problem, she spent exactly 1 dollar, which is equal to 100 cents. Therefore, we can form the equation:
13x + 8y = 100
Now, we need to find the values of x and y that satisfy this equation. To evaluate the possible combinations, we can use a method called trial and error.
Let's start with a reasonable assumption that she bought only 13 cent stamps, which means y = 0. Substituting this into the equation, we get:
13x + 8(0) = 100
13x = 100
x ≈ 7.69
Since x represents the number of stamps, it should be a whole number. So, we cannot have a fraction of a stamp. Therefore, this combination is not valid.
Next, let's try the assumption that she bought only 8 cent stamps, which means x = 0. Substituting this into the equation, we get:
13(0) + 8y = 100
8y = 100
y = 12.5
Again, we have a fraction of stamps, which is not possible. Hence, this combination is also invalid.
Now, we must consider combinations that involve both 13 cent stamps and 8 cent stamps. Rather than trying every possible combination, let's use a systematic approach:
If we assume x = 1, we can substitute it into the equation:
13(1) + 8y = 100
13 + 8y = 100
8y = 100 - 13
8y = 87
y ≈ 10.88
Again, we have a fraction, which is invalid. So, this combination is not suitable.
Let's continue this process and try with x = 2:
13(2) + 8y = 100
26 + 8y = 100
8y = 100 - 26
8y = 74
y ≈ 9.25
Once again, we have a non-whole number, rendering this combination invalid.
We can continue this process by incrementing the value of x and determining suitable values of y, or we can try a different strategy. In this case, since both x and y should be whole numbers, we can use a method called "integer programming" or "brute force" to check every possible combination to find the solution.
Using this method, we can determine that the woman could have bought 6 13 cent stamps and 4 8 cent stamps:
13(6) + 8(4) = 78 + 32 = 100
Thus, the woman could have bought 6 13 cent stamps and 4 8 cent stamps while spending exactly 1 dollar.