Tricky Algebra Question at 7-11?

Question

I went to the 7-11 store, I bought four items. The cashier multiplied my prices together and told me the total was $7.11. I told her she messed up and that she should add the prices together. Strangely, she totaled the sum and also got $7.11!!!! How much was each of the four items? PLEASE SHOW SOLUTION

Answer 1

Sorry, my last solution was nonsense because
I fell into a trap! Here's a better one:
As Northstar pointed out, if we turn the problem
into pennies, we have to solve the system.
xyzw = 711000000 = 79*3²*2^6* 5^6. (*)
x + y + z + w = 711,
where x, y, z and w are positive integers.
From the factorisation of (*), we may assume
that w is divisible by 79. So w = 79*c, where
c = 1,2,3,4,5,6 or 8, since if c>= 9, x+y+z+w >= 711,
which is impossible.
I will give the detailed solution for c = 1 ,
summarise the results for c = 2, 3, 5, 6 and 8
and then give the solution for c = 4, which is
the only solvable case.
I). c = 1.
In this case the system reduces to solving
xyz =9000000
x+y+z = 632.
Since 625 is the highest power of 5 less than
632, it is the highest possible power of 5
dividing x,y or z.
But if 625 divides z, say, we
get xy = 14400, x+y = 7,
which is clearly impossible.
So the highest possible power of 5 dividing x, y or z is 3
and precisely two of x,y,z, say y and z must be divisible by
5³ = 125 . This follows from 5^6 | xyz and the fact that not
all of x,y and z are divisible by 5.
So let y = 125a, z = 125b and we get
xab = 576 = 3² * 8².
x + 125a +125 b = 632.
This gives x = 7(mod 125) and x | 576.
But the only positive integers less than 632
satisfying x = 7(mod 125) are 7,132, 257, 382
and 507 and none of these divide 576, so this
case is impossible.

II). c = 2.
Here we have to solve
xyz = 4500000 = 2^5* 3² * 5^6.
x + y + z = 553.
By the same sort of reasoning as in case I, y and
z are both divisible by 125
and we have to solve
xab = 288 = 2^5* 3²,
x + 125a + 125b = 553.
So x = 53(mod 125)
and x = 53 or 178. Since neither of these
divides 288, this case is impossible.
By exactly the same method we may eliminate
the cases c = 3, 5, 6 and 8.

III). The case c = 4.
Now we have to solve
xyz = 2250000 = 3² * 2^4 * 5^6,
x + y + z = 395.
Here it is not hard to see that we must have
5 || x, 5² || y and 5³ || z.
Setting x = 5d, y = 25a, z = 125b, gives
dab = 144
5d + 25a + 125b = 395.
or
dab = 144,
d + 5a + 25b = 79.
Thus d | 144 and d = 4(mod 5).
So the only possibilities for d are
d = 4, 9 and 24.
If d = 4, we get ab = 36, a + 5b = = 15. That
gives 5 | a, which is impossible.
If d = 9, we get ab = 16, a + 5b = 14, which can
easily be proved impossible.
Finally, if d = 24 we get ab = 6, a + 5b = 11,
which yields a = 6, b = 1 and
we have our only solution
x = 120, y = 150, z = 125, w = 316.

Answer 2

We have four items with prices x, y, z, and w.

x + y + z + w = 7.11
xyzw = 7.11

To solve this let's turn dollars into cents. Then convert back.

x + y + z + w = 100*7.11 = 711
xyzw = (100^4)*7.11 = 711,000,000

Factor 711,000,000.

711,000,000 = (2^6)*(3²)*(5^6)*(79)

The four numbers must use up all the factors of 711,000,000.

2³ * 3 * 5 = 120
5³ = 125
2 * 3 * 5² = 150
2² * 79 = 316

If you total up the factors you will see they use up all the factors for 711,000,000.

Now convert back to dollars.

1.20
1.25
1.50
3.16

They both add and multiply to exactly $7.11.

Answer 3

Get the receipt. All the items aren't the Sam route u moron.

Answer 4

There is no answer that fits your criteria and also uses real money.

Because 711 factors out to 3*3*79, you would have to have something along the lines of 1 * 3 * 3 * .79= 7.11, or some other combination of those numbers, like, 1 * 1 * .9 * 7.90.

But, you also know that adding 4 odd numbers will always give you an even number. 7.11 isn't even. So, there is no way to do it. Did you misstate the problem?

Answer 5

regrettably, i don't have time to respond to all of them yet i will clarify the thanks to do it because inquiring for solutions under no circumstances fairly helps study. On a graph the horizontal line is the X axis and the vertical line is the Y axis. So the numbers in parentheses are the criteria. the first # is how a concepts it is going on the X axis. a good # is going correct and a unfavorable # is going correct. The second # is how severe it is going on the Y axis. Positve # is up and unfavorable is going down. understanding this you sould positioned both #'s at the same time to get a diploma. they furnish you with 2 factors so that you want to be sure how a lot larger up and more advantageous aside they are. it really is upward thrust/run. So the area up or down over the area left or correct. it is a fragment and if it would properly be simplified then you really could accomplish that. that's all i can assist so i'm hoping it helps. If no longer sorry, yet i attempted. i'm no longer a instructor, only a sensible tenth grade student. =)

Answer 6

You CAN solve it algebraically, but I don't think there is a solution.

If a, b, c and d are the four prices of the items you bought in cents then
a*b*c*d = 711c; and
a+b+c+d = 711c.

(a,b,c,d) must be factors of the number 711 for them to multiply up to 711.

The factors of 711 are:
1,3,3,79 (1*3*3*79 = 711)

So (a,b,c,d) can have the following values (all possible combinations of positive integers that multiply to 711):
(1,1,1,711)
(1,1,3,237)
(1,1,9,79)
(1,3,3,79)

The sums of these, respectively, are given by
714, 242, 90 and 86

None of these is 711 (the required sum) so I don't think there is a solution, unless you can pay part of a cent for something or the shop pays you for some stuff.

Answer 7

This is a well-known brain teaser - the answer is: $1.20, $1.25, $1.50, $3.16 - you need to use a computer program though or waste a lot of time trying combos manually

Answer 8

We need a little bit more info, unless you want us to sit here and waste our time by going thru all the different scenarios... which, we could, but we have better things to do...

Answer 9

well 1st u gotta break it down. but this would take a while to do though. however, why do u care so much?

Answer 10

$1.78