$7.11 for 4 items at 7-11? Is this puzzle solvable?
AACMrMaze
A student asked this question:
A customer at a 7-11 store selected 4 items to buy,
and was told that the cost was $7.11. Hw was curious
that the cost was the same as the store name, so he
inquired as to how the figure was derived. The clerk
said that he had simply multiplied the prices for the
4 induvidual items.The customer protested that the
4 prices should have been ADDED, not MULTIPLIED.
The clerk said that that was okay with him, but the
results was still the same, exactly $7.11
I couldn't find an answer, and am guessing that the puzzle can't be
solved, but I'm willing to be proven wrong.
John
Volunteer Teacher
Academic Assistance Center
P.S. Teachers or others with tutoring/teaching experience who have been
on AOL at least six months can get free accounts. Investigate at keyword
"AAC", "AAC News and Information", "Be an Online Tutor".
Wolfram Kresse
Hi.
AACMrMaze wrote:
> A customer at a 7-11 store selected 4 items to buy,
> and was told that the cost was $7.11. Hw was curious
> that the cost was the same as the store name, so he
> inquired as to how the figure was derived. The clerk
> said that he had simply multiplied the prices for the
> 4 induvidual items.The customer protested that the
> 4 prices should have been ADDED, not MULTIPLIED.
> The clerk said that that was okay with him, but the
> results was still the same, exactly $7.11
>
A short program (three(!) nested loops to cover all possible cases) shows that there is no
solution.
After I finished that program I started thinking about the problem...:
The prime factors of 711 are: 3, 3 and 79. For a product of four numbers to become 711,
these factors have to be included - exactly *once* (the number '1' is also allowed). But
there is no way that a sum of the same 4 numbers can also result in 711.
The possible prime factor combinations (including '1') of 4 numbers that yield 711 are:
sum:
1 1 1 711 715
1 1 3 237 242
1 1 9 79 90
1 3 3 79 86
None of these sums equals 711.
Cheers,
Wolfram
PS: so the puzzle *is* solvable... there is just no solution ;-)
--
+-------+-----Wolfram Kresse---------------------------------------------+
| _ _ | wkr...@igd.fhg.de http://www.igd.fhg.de/~wkresse |
| +-------------------------+-----------------+--------------------+
| -O-O- |"Meeneemeeneemeenee" | CU l8r, LE g8r! |
| > |"Yes,that's right,Twiki."+-----------------+
| _____ +-----+-----+-------------+
| U | 8^) | :u) |
+-------+-----+-----+
Matthew Daly
Yes, but the idea is that the sum and product don't equal 711 but 7.11
-- the difference makes for a problem that DOES have a solution.
SPOILER
The individual items cost $1.20, $1.25, $1.50, and $3.16.
-Matthew
--
Matthew Daly I feel that if a person has problems communicating
mwd...@kodak.com the very least he can do is to shut up - Tom Lehrer
My opinions are not necessarily those of my employer, of course.
--- Support the anti-Spam amendment! Join at http://www.cauce.org ---
AACMrMaze
I originally wrote:
>>> A customer at a 7-11 store selected 4 items to buy,
>>> and was told that the cost was $7.11. Hw was curious
>>> that the cost was the same as the store name, so he
>>> inquired as to how the figure was derived. The clerk
>>> said that he had simply multiplied the prices for the
>>> 4 induvidual items.The customer protested that the
>>> 4 prices should have been ADDED, not MULTIPLIED.
>>> The clerk said that that was okay with him, but the
>>> results was still the same, exactly $7.11
Two people answered:
>> None of these sums equals 711.
>> This puzzle can be solved by saying it has no solution.
I, too, thought that there was no solution, and had already written back
to the student that I couldn't find the answer.
I am delighted to be proved wrong, and have already sent Matthew's answer
to the student.
Thank You, Matthew for finding an answer.
John <www.MAZES.com> Knoderer
AACMrMaze
Here is what another of our teachers found.
In a message dated 97-09-17 13:46:31 EDT, Tpmath2 writes:
<< A customer at a 7-11 store selected 4 items to buy, and was told that
the cost was $7.11. He was curious that the cost was the same as the store
name, so he inquired as to how the figure was derived. The clerk said that
he had simply multiplied the prices for the 4 induvidual items.The customer
protested that the 4 prices should have been ADDED, not MULTIPLIED. The
clerk said that that was okay with him, but the result was still the same,
exactly $7.11 What were the prices of the four items? >>
Converting the $7.11 to 711 cents, we are left with a massive
Diophantine problem of finding four factors of 711,000,000 that add up to
711. In order to produce the six zeros in the final product, the result of
any individual multiplication must end in zero. Since the other two figures
could not furnish more than 5 zeros, the first figure of such a product
then not being 2 or 5, and since the unit digits must add up to 1; the only
possibilities are 0, 0, 0, 1 or 0, 0, 5, 6. The factors of 711,000 are
(2)^6, (3)^2, (5)^6, and (79), and in order to fulfill the above conditions
79 must be multiplied by 4, since if multiplied by 5, the four factors
would have to add up to more than 711, because even if the other three
factors could be equal, which gives the smallest sum, they would add up to
more than 711 minus 395. This derives from the general relation that the
factors of a number add up to the least amount when they are equal. We
therefore know that one factor is 316 and the other three factors which end
in 5, 0, and 0 respectively add up to 395. Since the cube root of the
remaining product is 131, none of the three factors can depart much from
this average. This leaves only a few combinations to try before arriving at
the figures of 120, 125, and 150.
Therefore, the costs of the four items were $3.16, $1.50, $1.25, and
$1.20 which when added together total $7.11 and when multiplied together
total $7.11 also.
Confusing? Perhaps, and I have wrestled with the explanation many
times, and still feel that something is missing or not exactly clear.
I take no credit for the solution. It appears in the book "INGENIOUS
Mathematical Problems and Methods" by L.A. Graham, published by Dover
Publications, Inc., New York, 1959. The problem, and solution, was
submitted by Mr. David A. Grossman, New York, and first appeared in the
"Private Corner for Mathematicians" section of the then widely circulated
"Graham DIAL," which circulated to well over 25,000 engineers and
production executives over an 18 year period. The book is an excellent
source of some great problems and I can heartily recommend it to everyone.
I believe I have seen the problem in another of my references and am
trying to find it in the hope that it might offer another
explanation/solution. I would be happy to send it to anyone if they are
interested, if, and when, I do find it.
As a math problem buff, I have periodically attempted to derive
another, perhaps more easily understood, solution, but time, and ingenuity
to be honest, have been my enemies so far. But, there is always tommorrow.
Perhaps some of our more formally trained mathematicians would like to take
a crack at it. Good luck.
I hope you have found this of some interest.
AACTchWill
AAC Staff
Michael Reeve
On 18 Sep 1997 04:16:14 GMT, aacm...@aol.com (AACMrMaze) wrote:
>A student asked this question:
>
> A customer at a 7-11 store selected 4 items to buy,
> and was told that the cost was $7.11. Hw was curious
> that the cost was the same as the store name, so he
> inquired as to how the figure was derived. The clerk
> said that he had simply multiplied the prices for the
> 4 induvidual items.The customer protested that the
> 4 prices should have been ADDED, not MULTIPLIED.
> The clerk said that that was okay with him, but the
> results was still the same, exactly $7.11
>
>I couldn't find an answer, and am guessing that the puzzle can't be
>solved, but I'm willing to be proven wrong.
>
>John
>Volunteer Teacher
>Academic Assistance Center
I found the following information in a file that I've had knocking
about on my computer for a few years. I hope it helps.
Myk.
==> arithmetic/7-11.p <==
A customer at a 7-11 store selected four items to buy, and was told
that the cost was $7.11. He was curious that the cost was the same
as the store name, so he inquired as to how the figure was derived.
The clerk said that he had simply multiplied the prices of the four
individual items. The customer protested that the four prices
should have been ADDED, not MULTIPLIED. The clerk said that that
was OK with him, but, the result was still the same: exactly $7.11.
What were the prices of the four items?
==> arithmetic/7-11.s <==
The prices are: $1.20, $1.25, $1.50, and $3.16
$7.11 is not the only number which works. Here are the first 160 such
numbers, preceded by a count of distinct solutions for that price.
Note that $7.11 has a single, unique solution.
1 - $6.44 1 - $7.83 2 - $9.20 3 - $10.89
1 - $6.51 1 - $7.86 1 - $9.23 1 - $10.95
1 - $6.60 3 - $7.92 1 - $9.24 2 - $11.00
1 - $6.63 1 - $8.00 1 - $9.27 1 - $11.07
1 - $6.65 1 - $8.01 2 - $9.35 1 - $11.13
1 - $6.72 1 - $8.03 3 - $9.36 1 - $11.16
2 - $6.75 5 - $8.10 1 - $9.38 1 - $11.22
1 - $6.78 1 - $8.12 5 - $9.45 2 - $11.25
1 - $6.80 1 - $8.16 2 - $9.48 2 - $11.27
2 - $6.84 2 - $8.19 1 - $9.54 1 - $11.30
1 - $6.86 1 - $8.22 1 - $9.57 1 - $11.36
1 - $6.89 1 - $8.25 1 - $9.59 1 - $11.40
2 - $6.93 3 - $8.28 2 - $9.60 2 - $11.43
1 - $7.02 3 - $8.33 1 - $9.62 2 - $11.52
1 - $7.05 1 - $8.36 2 - $9.63 2 - $11.55
2 - $7.07 1 - $8.37 1 - $9.66 2 - $11.61
1 - $7.08 2 - $8.40 1 - $9.68 1 - $11.69
1 - $7.11 1 - $8.45 2 - $9.69 1 - $11.70
1 - $7.13 2 - $8.46 1 - $9.78 1 - $11.88
2 - $7.14 1 - $8.52 2 - $9.80 1 - $11.90
3 - $7.20 5 - $8.55 1 - $9.81 1 - $11.99
1 - $7.25 1 - $8.60 1 - $9.87 1 - $12.06
1 - $7.26 4 - $8.64 4 - $9.90 1 - $12.15
2 - $7.28 1 - $8.67 1 - $9.92 1 - $12.18
1 - $7.29 1 - $8.69 2 - $9.99 1 - $12.24
3 - $7.35 1 - $8.73 1 - $10.01 1 - $12.30
1 - $7.37 2 - $8.75 1 - $10.05 1 - $12.32
1 - $7.47 1 - $8.76 2 - $10.08 1 - $12.35
1 - $7.50 1 - $8.78 1 - $10.17 2 - $12.42
1 - $7.52 5 - $8.82 1 - $10.20 1 - $12.51
4 - $7.56 1 - $8.85 2 - $10.26 1 - $12.65
1 - $7.62 1 - $8.88 3 - $10.29 2 - $12.69
4 - $7.65 2 - $8.91 3 - $10.35 1 - $12.75
1 - $7.67 1 - $8.94 2 - $10.44 1 - $12.92
2 - $7.70 1 - $8.96 1 - $10.53 1 - $12.96
3 - $7.74 3 - $9.00 1 - $10.56 1 - $13.23
1 - $7.77 1 - $9.02 1 - $10.64 1 - $13.41
1 - $7.79 2 - $9.03 2 - $10.71 1 - $13.56
2 - $7.80 1 - $9.12 3 - $10.80 1 - $14.49
1 - $7.82 2 - $9.18 1 - $10.85 1 - $15.18
There are plenty of solutions for five summands. Here are a few:
$8.28 -- at least two solutions
$8.47 -- at least two solutions
$8.82 -- at least two solutions
--
Mark Johnson ma...@microunity.com (408) 734-8100
There may be many approximate solutions, for example: $1.01, $1.15,
$2.41,
and $2.54. These sum to $7.11 but the product is 7.1100061.
NickGard
aacm...@aol.com (AACMrMaze) wrote:
>A student asked this question:
>
> A customer at a 7-11 store selected 4 items to buy,
> and was told that the cost was $7.11. Hw was curious
> that the cost was the same as the store name, so he
> inquired as to how the figure was derived. The clerk
> said that he had simply multiplied the prices for the
> 4 induvidual items.The customer protested that the
> 4 prices should have been ADDED, not MULTIPLIED.
> The clerk said that that was okay with him, but the
> results was still the same, exactly $7.11
>
[SPOILER]
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f
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The items cost $1.20, $1.25, $1.50 and $3.16.
1.2 + 1.25 + 1.5 + 3.16 = 7.11
1.2 x 1.25 x 1.5 x 3.16 = 7.11
Nick.
!!emailid!!
In <19970918041...@ladder02.news.aol.com>, aacm...@aol.com (AACMrMaze) writes:
>A student asked this question:
>
> A customer at a 7-11 store selected 4 items to buy,
> and was told that the cost was $7.11. Hw was curious
> that the cost was the same as the store name, so he
> inquired as to how the figure was derived. The clerk
> said that he had simply multiplied the prices for the
> 4 induvidual items.The customer protested that the
> 4 prices should have been ADDED, not MULTIPLIED.
> The clerk said that that was okay with him, but the
> results was still the same, exactly $7.11
>
>solved, but I'm willing to be proven wrong.
>
>John
**SPOILER**
**SPOILER**
This problem has a solution: $1.20, $1.25, $1.50, and $3.16. Unless
you argue that the product has units $^4 and therefore cannot equal
$7.11.
--
Mark Thornquist
Mike Williams
In message <19970918041...@ladder02.news.aol.com> AACMrMaze wrote:
> A student asked this question:
>
> A customer at a 7-11 store selected 4 items to buy,
> and was told that the cost was $7.11. Hw was curious
> that the cost was the same as the store name, so he
> inquired as to how the figure was derived. The clerk
> said that he had simply multiplied the prices for the
> 4 induvidual items.The customer protested that the
> 4 prices should have been ADDED, not MULTIPLIED.
> The clerk said that that was okay with him, but the
> results was still the same, exactly $7.11
>
> I couldn't find an answer, and am guessing that the puzzle can't be
> solved, but I'm willing to be proven wrong.
Spoiler...
$1.20 $1.25 $1.50 $3.16
Abhay
Possible one solution
If we assume two items of $1 each, then the problem reduces
to
A*B = 7.11
A+B = 5.11
which should be solvable.
Abhay
--
****************************************************************
-- Please REMOVE "nospam" from the reply address --
-- Sorry for the inconvienence. --
****************************************************************
Matthew Daly
Abhay <ab...@accesscom.com.nospam>, if that is your REAL name, said:
>Possible one solution
>
>If we assume two items of $1 each, then the problem reduces
>to
> A*B = 7.11
> A+B = 5.11
>which should be solvable.
No doubt, but it needs to be solvable with A and B such that 100A and
100B are both natural numbers.
What the full puzzle requires is a solution to the Diophantine system
A+B+C+D = 711
ABCD = 711000000
and (according to the archive) the solution {A,B,C,D} =
{120,125,150,316} is unique.
Abhay
Possible one solution
If we assume two items of $1 each, then the problem reduces
to
A*B = 7.11
A+B = 5.11
which should be solvable.
Abhay
Thomas Kiss
AACMrMaze wrote:
>
> A student asked this question:
>
> A customer at a 7-11 store selected 4 items to buy,
> and was told that the cost was $7.11. Hw was curious
> that the cost was the same as the store name, so he
> inquired as to how the figure was derived. The clerk
> said that he had simply multiplied the prices for the
> 4 induvidual items.The customer protested that the
> 4 prices should have been ADDED, not MULTIPLIED.
> The clerk said that that was okay with him, but the
> results was still the same, exactly $7.11
>
> I couldn't find an answer, and am guessing that the puzzle can't be
> solved, but I'm willing to be proven wrong.
>
> Volunteer Teacher
> Academic Assistance Center
>
> on AOL at least six months can get free accounts. Investigate at keyword
> "AAC", "AAC News and Information", "Be an Online Tutor".
is a sampling. a,b,c,d represent items. I don't see why this is
a massive Diophantine problem.
a: 2.05 b: 0.84 c: 1.54 d: 2.68
a: 2.03 b: 0.84 c: 1.55 d: 2.69
a: 2.01 b: 0.84 c: 1.56 d: 2.70
a: 1.99 b: 0.84 c: 1.57 d: 2.71
a: 1.97 b: 0.84 c: 1.58 d: 2.72
a: 1.95 b: 0.84 c: 1.59 d: 2.73
a: 1.93 b: 0.84 c: 1.60 d: 2.74
a: 1.49 b: 0.84 c: 2.20 d: 2.58
a: 1.49 b: 0.84 c: 2.21 d: 2.57
a: 1.49 b: 0.84 c: 2.22 d: 2.56
a: 1.49 b: 0.84 c: 2.23 d: 2.55
a: 1.48 b: 0.84 c: 2.24 d: 2.55
a: 1.93 b: 0.84 c: 1.60 d: 2.74
a: 1.92 b: 0.84 c: 1.61 d: 2.74
a: 1.90 b: 0.84 c: 1.62 d: 2.75
a: 1.88 b: 0.84 c: 1.63 d: 2.76
a: 1.87 b: 0.84 c: 1.64 d: 2.76
a: 1.86 b: 0.84 c: 1.65 d: 2.76
a: 1.84 b: 0.84 c: 1.66 d: 2.77
a: 1.57 b: 0.86 c: 1.88 d: 2.80
a: 1.57 b: 0.86 c: 1.89 d: 2.79
a: 1.56 b: 0.86 c: 1.90 d: 2.79
a: 1.55 b: 0.86 c: 1.91 d: 2.79
a: 1.55 b: 0.86 c: 1.92 d: 2.78
a: 1.54 b: 0.86 c: 1.93 d: 2.78
a: 1.54 b: 0.86 c: 1.94 d: 2.77
a: 1.53 b: 0.86 c: 1.95 d: 2.77
a: 1.38 b: 0.88 c: 2.58 d: 2.27
a: 1.38 b: 0.88 c: 2.59 d: 2.26
a: 1.38 b: 0.88 c: 2.60 d: 2.25
a: 1.39 b: 0.88 c: 2.61 d: 2.23
a: 1.39 b: 0.88 c: 2.62 d: 2.22
a: 1.39 b: 0.88 c: 2.63 d: 2.21
a: 1.39 b: 0.88 c: 2.64 d: 2.20
a: 1.40 b: 0.88 c: 2.65 d: 2.18
a: 1.40 b: 0.88 c: 2.66 d: 2.17
a: 1.40 b: 0.88 c: 2.67 d: 2.16
a: 1.22 b: 0.96 c: 2.51 d: 2.42
a: 1.22 b: 0.96 c: 2.52 d: 2.41
a: 1.22 b: 0.96 c: 2.53 d: 2.40
a: 1.22 b: 0.96 c: 2.54 d: 2.39
a: 1.18 b: 1.00 c: 2.26 d: 2.67
a: 1.18 b: 1.00 c: 2.27 d: 2.66
a: 1.17 b: 1.00 c: 2.28 d: 2.66
a: 1.17 b: 1.00 c: 2.29 d: 2.65
and thousands more.
Mike Naylor
AACMrMaze wrote:
> A customer at a 7-11 store selected 4 items to buy,
> and was told that the cost was $7.11. Hw was curious
> that the cost was the same as the store name, so he
> inquired as to how the figure was derived. The clerk
> said that he had simply multiplied the prices for the
> 4 induvidual items.The customer protested that the
> 4 prices should have been ADDED, not MULTIPLIED.
> The clerk said that that was okay with him, but the
> results was still the same, exactly $7.11
Thomas Kiss <t...@abc.ca> wrote:
>There are thousands of solutions to this 7.11 problem. Here
>is a sampling. a,b,c,d represent items. I don't see why this is
>a massive Diophantine problem.
>a: 2.05 b: 0.84 c: 1.54 d: 2.68
(The product of these numbers is about 7.1070384. Many more examples with
products that round off to 7.11 snipped.)
I have a deal for you, Tom. If you'll explain to me what "Diophantine"
means, in return I'll explain to you what "exactly" means.
Mike Naylor - myfirstname...@mail.serve.com
--
Play Five by Five Poker at http://www.serve.com/games/
David Conrad
NickGard (nick...@aol.com) wrote:
> aacm...@aol.com (AACMrMaze) wrote:
>
> > [A customer at 7-11 buys four items such that the prices have $7.11
> > as both their product and sum.]
>
> [A solution, which was also offered by another poster (Matthew?).]
I couldn't solve it, so I wrote a program to search for possible
solutions. (I often wimp out in this manner.) Unfortunately, the
program used floating point, so it also spat out a lot of near
misses, such as numbers that have something like 7.110143281 as
their product.
Anyway, when I finally found the right solution, I tried to check it
using bc, but I feared that maybe it was rounding off the number and
I had only found another 'almost' solution.
I turned up the scale (increased the precision), but I still didn't
have a warm, fuzzy feeling. And then I realized how I could be sure
the answer wasn't rounded off. I multiplied the numbers together
as whole numbers of cents, and they came to 711000000.
And then it hit me how to solve it analytically. Just factor
7.11 x 10^8 (which is 2,2,2,2,2,2,3,3,5,5,5,5,5,5,79) and then
find ways to combine the factors into four groups such that
the products of the groups sum to 711.
I don't know if that will seem obvious to the rest of you, but
it was quite a revelation for me. A real "Eureka!" moment.
--
David R. Conrad, con...@detroit.freenet.org PGP key on home page
http://detroit.freenet.org/staff/conrad/ GDFN {Hard,Soft}ware Cmte
No, his mind is not for rent to any god or government.
Abigail
David Conrad (con...@grfn.org) wrote on 1482 September 1993 in
<URL: news:603n7i$j1p$1...@newsserv.grfn.org>:
++
++ I couldn't solve it, so I wrote a program to search for possible
++ solutions. (I often wimp out in this manner.) Unfortunately, the
++ program used floating point, so it also spat out a lot of near
++ misses, such as numbers that have something like 7.110143281 as
++ their product.
++
++ Anyway, when I finally found the right solution, I tried to check it
++ using bc, but I feared that maybe it was rounding off the number and
++ I had only found another 'almost' solution.
No, bc will be as precise as you want it to be. The standard
precision of 'bc -l' is 20 places behind the floating point;
precise enough for multiplying 4 numbers with 2 digits after
the point.
++ I turned up the scale (increased the precision), but I still didn't
++ have a warm, fuzzy feeling. And then I realized how I could be sure
++ the answer wasn't rounded off. I multiplied the numbers together
++ as whole numbers of cents, and they came to 711000000.
++
++ And then it hit me how to solve it analytically. Just factor
++ 7.11 x 10^8 (which is 2,2,2,2,2,2,3,3,5,5,5,5,5,5,79) and then
++ find ways to combine the factors into four groups such that
++ the products of the groups sum to 711.
That would miss any potential solution that has a item
costing 1 cent.
Abigail
Charles Bryant
In article <19970918041...@ladder02.news.aol.com>,
AACMrMaze <aacm...@aol.com> wrote:
> A customer at a 7-11 store selected 4 items to buy,
> and was told that the cost was $7.11. Hw was curious
> that the cost was the same as the store name, so he
> inquired as to how the figure was derived. The clerk
> said that he had simply multiplied the prices for the
> 4 induvidual items.The customer protested that the
> 4 prices should have been ADDED, not MULTIPLIED.
> The clerk said that that was okay with him, but the
> results was still the same, exactly $7.11
>
>solved, but I'm willing to be proven wrong.
There is a flaw in the puzzle. Any cash registers I have seen do not
use decimal points, but have a special "00" button for amounts which
are a whole number of dollars (or pounds). Consequently, it would be
impossible to implement the normal solution (which multiples amounts
as dollars with non-zero cents). Perhaps some readers have actually
worked in a 7-11 and can tell us if the registers used there have
decimal points and a multiply function.
Matthew Daly
Charles Bryant <n186912...@chch.demon.co.uk>, if that is your REAL
name, said:
>There is a flaw in the puzzle. Any cash registers I have seen do not
>use decimal points, but have a special "00" button for amounts which
>are a whole number of dollars (or pounds). Consequently, it would be
>impossible to implement the normal solution (which multiples amounts
>as dollars with non-zero cents). Perhaps some readers have actually
>worked in a 7-11 and can tell us if the registers used there have
>decimal points and a multiply function.
I don't presume that it's supposed to be a reality-based puzzle. It's
just set in a 7-11 to give a tie-in to the answer. I mean, the fact
that the clerk is telling the customer a cost based on the product of
the prices at all should be a clue that this wouldn't actually happen.
We don't harrass our census-takers in the US as badly as the puzzles
suggest either. :-)
That being said, virtually all electronic 10-key adding machines that
I've seen do support the mutiplication function. They also have decimal
points, although you can easily set the default to have 150 be
recognized as 1.50 to save a keypress. And those adding machines do
show up in stores when the cash register is broken, so other than the
quirky clerk, it's not completely unrealistic.
Mike Naylor
AACMrMaze wrote:
>A customer at a 7-11 store selected 4 items to buy,
>and was told that the cost was $7.11. Hw was curious
>that the cost was the same as the store name, so he
>inquired as to how the figure was derived. The clerk
>said that he had simply multiplied the prices for the
>4 induvidual items.The customer protested that the
>4 prices should have been ADDED, not MULTIPLIED.
>The clerk said that that was okay with him, but the
>results was still the same, exactly $7.11
Wolfram Kresse <wkr...@mindless.com> wrote:
>A short program (three(!) nested loops to cover all possible cases) shows
>that there is no solution.
>After I finished that program I started thinking about the problem...:
>
>The prime factors of 711 are: 3, 3 and 79. For a product of four numbers to become 711,
>these factors have to be included - exactly *once* (the number '1' is also allowed). But
>there is no way that a sum of the same 4 numbers can also result in 711.
>The possible prime factor combinations (including '1') of 4 numbers that yield 711 are:
>
> sum:
>1 1 1 711 715
>1 1 3 237 242
>1 1 9 79 90
>1 3 3 79 86
>
>None of these sums equals 711.
>
>Cheers,
>
>Wolfram
>
>PS: so the puzzle *is* solvable... there is just no solution ;-)
Wolfram, I just want to nit-pick a little here in the interest of absolute
accuracy. What I wanted to point out is that when you say that there is no
solution, some people might feel more comfortable if you amend that
slightly to say, "Aside from the numbers $1.20, $1.25, $1.50, and $3.16,
there is no solution." However, if one sets that trivial matter aside, you
are absolutely correct.
Joseph W. DeVincentis
In article <34215efb...@news.mcmail.com>,
Michael Reeve <myk....@mcmail.com> wrote:
<On 18 Sep 1997 04:16:14 GMT, aacm...@aol.com (AACMrMaze) wrote:
<> A customer at a 7-11 store selected 4 items to buy,
<> and was told that the cost was $7.11. Hw was curious
<> that the cost was the same as the store name, so he
<> inquired as to how the figure was derived. The clerk
<> said that he had simply multiplied the prices for the
<> 4 induvidual items.The customer protested that the
<> 4 prices should have been ADDED, not MULTIPLIED.
<> The clerk said that that was okay with him, but the
<> results was still the same, exactly $7.11
<==> arithmetic/7-11.p <==
<A customer at a 7-11 store selected four items to buy, and was told
<that the cost was $7.11. He was curious that the cost was the same
<as the store name, so he inquired as to how the figure was derived.
<The clerk said that he had simply multiplied the prices of the four
<individual items. The customer protested that the four prices
<should have been ADDED, not MULTIPLIED. The clerk said that that
<was OK with him, but, the result was still the same: exactly $7.11.
The funniest part of this whole problem is that the original problem
statement is word-for-word the same (with changes only in spacing, spelling,
and OK written as okay, etc.) as the problem statement from the archive.
/dev/joe
Mike Naylor
Charles Bryant <n186912...@chch.demon.co.uk> wrote:
>There is a flaw in the puzzle. Any cash registers I have seen do not
>use decimal points, but have a special "00" button for amounts which
>are a whole number of dollars (or pounds). Consequently, it would be
>impossible to implement the normal solution (which multiples amounts
>as dollars with non-zero cents). Perhaps some readers have actually
>worked in a 7-11 and can tell us if the registers used there have
>decimal points and a multiply function.
The puzzle doesn't mention a cash register or calculating machine of any
kind. It simply says that the clerk did the math and got the stated
results. For all we know, he worked it out with a pencil and paper or a
Chinese abacus, or he might even have done it in his head. The point is
the puzzle states that the results were exact and we are not entitled to
stipulate otherwise.