If x,y,z,w represent the amounts of the four items in cents then we
What can we say so far?
The factors : 711 = (3)(3)(79)
If there is to be an exact solution then the product xyzw must be divisible by 3, 9, 79.
The exact sum of the 4 items is 7.11 so the cost of each item is less than $7.11.
After exploring these two equations on scratch paper I came up with two scenerios.
Since the clerk did not know that total meant to add, he must not have known how to round to the nearest penny (or even that you should round). Suppose the customer selected four items which cost $.79, $2.00, $1.76, and $2.56. The total cost of these 4 items is exactly $7.11. However, the product of the 4 items is 7.118848. Thus the clerk rounds incorrectly or doesn't round at all and quotes the amount $7.11.
The clerk does know how to round to the nearest cent. Suppose the customer selected four items which cost $.79, $1.75, $2.00, $2.57. The total cost is $7.11. However, the product is 7.10605.
The clerk rounds correctly and quotes the amount $7.11.
Are there other amounts that would fit into one of the two scenerios mentioned?
Are there other scenerios?
What about Dr. Wilson's comment?
READ the problem statement again.
... the total cost of the four items is $7.11.
... simply multiplied the cost of each item and arrived at the total.
Neither of these phrases say anything about "exactly" or "exact". Thus the clerk could have rounded or truncated the product to arrive at his quote.
... then added the items together and informed the customer that the total was still exactly $7.11.
When he added the 4 amounts he did get an exact total. Thus the customer was told he "still owed exactly $7.11".
... exact costs of each item?
In both scenerios the cost of each item was "exact".
Is this response to Dr. Wilson's comment accurate?
Now let's seriously consider the question "is there an 'exact' solution"?
The answer: yes.
If the customer purchased four items worth $3.16, $1.50, $1.20, and $1.25, the the total cost and the product would be exactly $7.11 .
How did this answer come about? It began by finding the first two "estimate" solutions. I then used a spreadsheet to make the variations of the amounts and the calculations easier. I let x = the amount divisible by 79. So x = 79, 158, 316. All of the calculations with x = 79 or 158 were estimates (did I miss any "exact" ones?). Finally, with x = 316 I was able to arrive at the solution.
Alas there is an "exact" solution. BUT ...... Are there anymore???