What did each item cost.?

Question

A guy walks into a store called 7/11. He selects 4 different items and takes them to the cashier to pay for them. The cashier checks them and asks for $7.11. The guy says thats unusual that the total is the same as the store name. He checks his reciept and discovers that the cashier has multiplied the items instead of adding them, he shows this to the cashier but the cashier says it does not make any difference as the total is still $7.11. What were the 4 prices?

A+B+C+D= $7.11
A*B*C*D= $7.11

Answer 1

Since none of the prices would (presumably) include fractions of a cent, what we're really looking for is a set of four integers with the following properties:

A/100 + B/100 + C/100 + D/100 = 7.11
(A/100)(B/100)(C/100)(D/100) = 7.11

These equations can be simplified:

A + B + C + D = 711
A * B * C * D = 711,000,000

Now, for the sum to work, all the numbers have to be < 709, and perhaps we can assume < 705, since all the items are different. What else can we determine?

711,000,000 factors as 2^6 * 3² * 5^6 * 79.

Each integer factor of 711,000,000 must be composed of those factors, which means there are a total of (6+1)(2+1)(6+1)(2) = 294 integer factors of 711,000,000, but the only ones < 705 are:

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 30, 32, 36, 40,
45, 48, 50, 60, 64, 72, 75, 79, 80, 90, 96, 100, 120, 125, 144,
150, 158, 160, 180, 192, 200, 225, 237, 240, 250, 288, 300,
316, 320, 360, 375, 395, 400, 450, 474, 480, 500, 576, 600,
625, and 632.

Whew.

We just have to find a set of four of those that adds to 711 while having 711,000,000 as a product. The good news is that one of them has to be a number with 79 as a factor, and the only ones that qualify are 79, 158, 237, 316, 395, 474, and 632.

Let's eliminate 632 = 2³ * 79 first, because the other numbers would have to be three factors of 1,125,000 (711,000,000 / 632) that add up to 79 (711 - 632). That's not gonna happen. We can eliminate 474 for the same reason (711,000,000 / 474 = 1,500,000, but 711 - 474 = 237), as well as 395 (711,000,000 / 395 = 1,800,000, but 711 - 395 = 316).

So now we start by trying Item #1 at 316. 711,000,000 / 316 = 2,250,000, and 711-316 = 395. We need three integer factors of 2,250,000 = 2^4 * 3^2 * 5^6 that add up to 395. Possible factors: 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 36, 45, 50, 60, 75, 90, 100, 125, 150, 180, 225, 250, 300, 375.

And... that's all I have time to give you. Perhaps there's a trick I'm not seeing, but if you can't find a working triplet for Item #1 at 316, then next you'd try Item #1 at 237 (so the other three would have to have 711,000,000/237 = 3,000,000 as a product and 711-237 = 474 as a sum), then 158, and lastly 79.

Hope that helped!

Answer 2

Well, assuming that there was no rounding when the prices were multiplied together, there is a very limited factorization.

Turning the prices into number of cents and multiplying would give 711000000 and this factors into
3^2 * 79 * 2^6 *5^6.

So, one of the items must be priced one of the following:
.79, 1.58, 2.37, 3.16, 3.95, 4.74, 5.53, or 6.32.

OK, I gave up on the analysis method and hit it with a non-linear integer optimization routine. It came up with:

.82, 2.15, 2.57, 1.57. Yes, these add to be 7.11, but they multiply to give 7.11352870 which rounds to 7.11 to the nearest cent.

Answer 3

I think that we need more information. If there are four variables to be solved for, we need at least four equations...unless there is something else in the problem that is missing.

Answer 4

there are many combinations that could work.

$1 * $ 1 * $1 * $7.11
or each item could be $1.63
or many more

Answer 5

It's sounds cheap, why do you care what each item cost?

Answer 6

1.2, 1.03, 1.99, 2.89


and it rounded to 7.11

Answer 7

$1.7775