Seemingly Impossible Math Problem

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lisaz0224

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Okay, people. I've tried this problem over and over again, but I can't seem to find the answer. It seems impossible to me.

A guy walks into a 7-11 store and selects four items to buy. The clerk at the counter informs the gentle man that the total cost of the four items is $7.11. He was completely surprised that the cost was the same name as the store. The clerk informed the man that he simply multiplied the cost of each item and arrived at the total. The customer calmly informed the clerk that the items should be added, not multiplied. The clerk then added the items together and informed the customer that the total was exactly $7.11. What was the cost of each item??

Please explain steps. Thank you!
 
It is only for the frustrated that it seems impossible. Sadly, as you have posted in pre-algebra, I'm not quite sure how you are expected to proceed, except by LOTS of guessing. Let's have a little algebra lesson.

We have four prices, I'll call them a, b, c, and d, we get:

It should be obvious that 0 < a < 7.11 and 0 < b < 7.11 and 0 < c < 7.11 and 0 < d < 7.11

This is what we know.

Multiplied: a*b*c*d = 7.11
Added: a+b+c+d = 7.11

The problem is that there are only two relationship but an unreasonble four values to locate. All is not lost. Actually, this situation gives us MORE solutions. Let's just pick one or two.

I choose to start with a = 2

This gives the two equations this appearance

2*b*c*d = 7.11
2+b+c+d = 7.11

or (simplifying)

b*c*d = 3.56
b+c+d = 5.11

We're stiill stuck!

I choose to add with b = 2 (Again, no real reason. It just seemed like a nice value.)

This gives the two equations this appearance:

2*c*d = 3.56
2+c+d = 5.11

or (simplifying)

c*d = 1.78
c+d = 3.11

Now we're down to two equations and we have no more freedom to pick values out of the air.

Can you solve the last two equations for c & d?

I get c = 0.76 and d = 2.35 (Actually, it could have gone the other way, d = 0.76 and c = 2.35)

Let's check: 2*2*0.76*2.35 = 7.14 and 2+2+0.76+2.35 = 7.11 -- Okay, not quite, but with a few more decimal places, we're there. Rounding to two places is a sloppy business, but when delaing with cash amounts, that is an important limitation. Indeed, using just one more decimal place gets us rather close.
 
Hey TK....we put out an APB on you :shock:

For fun, did it this way:

u=711 : v=711000000 : a =< b =< c =< d = 4 items

a + b + c + d = u [1]

abcd = v : d = v / (abc)

[1] a + b + c + v / (abc) = u
a^2 bc + a(b^2 c + b c^2 - bcu) + v = 0
Let k = b^2 c + b c^2 - bcu ; then:
a^2 bc + ak + v = 0

a = [-k +- SQRT(k^2 - 4bcv)] / (2bc)

Only one integer solution: a = 120, b = 125, c = 150, d = 316
 
Woah... I get the last explanation, I think. Kind of weird that this type of problem appeared on my homework, when I'm in the sixth grade... Oh well, thanks!
 
Much more fun at a 8-10 store; 5 possibilities:
.50 2.40 2.50 2.70
.60 1.50 3.00 3.00
.75 1.20 2.40 3.75
.75 1.35 2.00 4.00
.80 1.25 2.00 4.05

All cheap Xmas presents...the kind I get :shock:
 
lisaz0224 said:
Woah... I get the last explanation, I think. Kind of weird that this type of problem appeared on my homework, when I'm in the sixth grade... Oh well, thanks!
Click to expand...

That is right. I must question the appropriateness of the question at this level, unless some rreally big hints were given.
 
Sorry...
my math teacher gave us this problem for homework and I was like "What...?"
She said it was hard, though...
 
This is what is known as a Diophantine equation.

Here is a nice discussion on this problem:

http://jwilson.coe.uga.edu/EMAT6680Fa05 ... 1/711.html


Here's one along the same lines:

"A man walks into a bank to withdraw X dollars and Y cents from his account. Instead, the teller mistakenly gives him Y dollars and X cents.

After spending $10, he notices he has twice what he intended to withdraw. How much did he intend to withdraw and how much was he given?".

amount intended: \(\displaystyle \underbrace{100X}_{\text{dollars converted to cents}}+Y \;\ cents\)

amount received: \(\displaystyle 100Y+ X \;\ cents\)

After spending $10 (1000 cents) he has: \(\displaystyle 100Y+X-1000\) left.

So, \(\displaystyle \overbrace{100Y+x-1000}^{\text{amount given minus 10 dollars}}=\underbrace{2(100X+Y)}_{\text{amount intended}}\)

\(\displaystyle \Rightarrow 98Y-199X=1000\)

\(\displaystyle 98Y-(2(98)+3)X=1000\)

\(\displaystyle 98(Y-2X)-3X=1000\)

Here is where the Euclidean algorithms come into play.

After all the computations, which I am too lazy to post at the moment(I will finish later if you would like to see it), we find he asked for 26.63 and was given 63.26.

Perhaps you can give it a go if you have the wherewithal to do so.
 
The original problem though - as posted - does not have a solution(Incorrect as noted below - I was trying to factor 711 as opposed to 7.11). 711 has following factors - 1, 3, 9, 79, 237 and nothing adds upto 711.
 
The original problem though - as posted - does not have a solution. 711 has following factors - 1, 3, 9, 79, 237 and nothing adds up to 711.
Click to expand...

Are you sure, SK?.

(3.16)(1.25)(1.2)(1.5)=7.11

3.16+1.25+1.2+1.5=7.11

These appear to be solutions. Am I misunderstanding you?.

Factoring 7.11, as opposed to 711, gives us \(\displaystyle \frac{3^{2}\cdot 79}{2^{2}\cdot 5^{2}}=\frac{711}{100}\)

Factors of \(\displaystyle 3.16=\frac{79}{5^{2}}, \;\ 1.25=\frac{5}{2^{2}}, \;\ 1.2=\frac{2\cdot 3}{5}, \;\ 1.5=\frac{3}{2}\)

\(\displaystyle \left(\frac{79}{5^{2}}\right)\left(\frac{5}{2^{2}}\right)\left(\frac{2\cdot 3}{5}\right)\left(\frac{3}{2}\right)=\frac{711}{100}=7.11\)

\(\displaystyle \left(\frac{79}{5^{2}}\right)+\left(\frac{5}{2^{2}}\right)+\left(\frac{2\cdot 3}{5}\right)+\left(\frac{3}{2}\right)=\frac{711}{100}=7.11\)
 
Cody, you are absolutely correct - was trying to factor 711 as opposed to 7.11------
 
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