Author |
Topic: Arithmetic puzzle (Read 1540 times) |
|
Gerd
Newbie
Gender: 
Posts: 26
|
 |
Arithmetic puzzle
« on: Mar 7th, 2003, 1:29am » |
 Quote  Modify
|
Hello there! 
I got a arithmetic puzzle (including the solution). My problem is, how can I solve such puzzles.
I would be very lucky, if someone could help me with my problem!
Here's the puzzle:
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?
Wish you a nice day
|
|
 IP Logged |
|
|
|
cho
Guest
|
In the CS forum there is a thread called 7-11 Riddle that shows you how to solve it with a computer program. If you want to solve it without a computer, it will take a little work.
First, as the other thread points out, it's easier to work with whole numbers, so you'll multiply each of your 4 items by 100, and you'll multiply 7.11 *100*100*100*100.
711=3*3*79
The remaining 3 100's add 6 2's and 6 5's as additional factors.
Your 4 items will have to be some combination of those 15 factors. That's a lot of possible combinations.
You'd best start by apportioning them as evenly as possible:
2*79=158
2*3*5*5=150
2*3*5*5=150
2*2*2*5*5=200
But that adds up to 658.
Now you just have to start shifting the numbers around until you find the right combination. I don't know of any way to lead you more directly to the right answer than that.
|
|
IP Logged |
|
|
|
cho
Guest
|
I would add one other hint, that any time you move farther away from that even distribution, the total gets higher. So if your total is too low, move factors towards the extremes (for example, swap a 5 in one of the lower numbers with a 3 in one of the higher number to make the low lower and the high higher). If your total is too high, you have to get a couple numbers back closer to the average.
|
|
IP Logged |
|
|
|
cho
Guest
|
One other thought: since 711 is odd, either one item or 3 items must have an odd price, so either one item is a multiple of 64, or you know exactly one item does not have 2 as a factor.
|
|
IP Logged |
|
|
|
Gerd
Newbie
Gender:
Posts: 26
|
|
:)Re: Arithmetic puzzle
« Reply #4 on: Mar 7th, 2003, 10:52am » |
Quote Modify
|
Thanks a lot, cho!
I think that I'm fine with your guidance.
I got a couple of these puzzles (always only 1 solution). Now I can do it.
I wish you a nice day
Gerd
P.S. Sorry for my bad English. 
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print
Remove