• The TMF is sponsored by Clips4sale - By supporting them, you're supporting us.
  • >>> If you cannot get into your account email me at [email protected] <<<
    Don't forget to include your username

The TMF is sponsored by:

Clips4Sale Banner

math puzzle

milagros317

Verified
Joined
Jan 12, 2002
Messages
575,155
Points
63
The following math puzzle seems to have too little information to be solved, but that is not the case. There is a unique minimal solution to all the questions at the end.

Susan is a child who hates pennies. She hates pennies so much that she will neither receive any in change nor possess any to give to any cashier. Instead she refuses to buy any collection of items that would involve giving or receiving pennies. For example, if the total charge is $1.28, she will refuse to make the purchase because she would have to receive two pennies in change or hand over three pennies.

Susan loves candy of all kinds. She has some US coins in her pocket. Of course, she has no pennies among them. [For those not familiar with US coins, they are in denominations of 1, 5, 10, 25, 50, and 100 cents. They are called, respectively, a penny (or a cent), a nickel, a dime, a quarter, a half dollar, and a dollar.]

Susan goes into a candy store. There are several jars of candy on the counter. All of the candies in any one jar are identical and have the same price.

Each jar contains a different type of candy at a different price from the other jars, and all the prices are a positive whole number of cents.

Susan asks the price of two candies from each of the jars. In every case, the total cost of two pieces would require her to hand over a penny besides using some of the coins in her pocket, so she refuses, even though she has more than enough money to pay.

Susan asks the price of three candies form each of the jars. In every case, she has coins that would pay for the three candies but she would receive one penny in change, so she refuses.

Susan then asks the price of one candy from each of the jars. To her delight, the total cost of one candy from each jar is an amount she can pay exactly with some of the coins in her pocket.

You now have enough information to answer all of the following questions:

1) What is the fewest number of jars of candy that there could be on the counter?

2) What are the lowest prices possible for the candies in each jar, listed lowest to highest?

3) What is the least total of amount of money that Susan could have in her pocket?

4) Presume that Susan has as few coins as possible to make all of the above true. Exactly what coins does Susan have in her pocket?
 
To solve this problem requires no higher mathematics; indeed, it can be solved without even using high school algebra. All that is required is clear thinking and the ability to do arithmetic (with or without a calculator) with whole numbers under 100. Lots of clear thinking.
 
I'm taking a stab at it.... it may stab me. Dang, you shoulda posted this back in January, when we don't have to worry about doing yard work and such.... my mind's gettin old, haven't used it for anything mathematical for a long time (I have a BS in geophysics, so I know some of the higher math....this lower stuff tho...LOL). Don't be too quick with giving the answer.... maybe a year would work. Just joshin'. Probably a second grade problem anyways.....
 
I'm taking a stab at it.... it may stab me. Dang, you shoulda posted this back in January, when we don't have to worry about doing yard work and such.... my mind's gettin old, haven't used it for anything mathematical for a long time (I have a BS in geophysics, so I know some of the higher math....this lower stuff tho...LOL). Don't be too quick with giving the answer.... maybe a year would work. Just joshin'. Probably a second grade problem anyways.....
I will post a solution in a few weeks, in June.
 
1) What is the fewest number of jars of candy that there could be on the counter?

2) What are the lowest prices possible for the candies in each jar, listed lowest to highest?

3) What is the least total of amount of money that Susan could have in her pocket?

4) Presume that Susan has as few coins as possible to make all of the above true. Exactly what coins does Susan have in her pocket?

1) 837
2) $253.00, , $45.67i, $9>2,$467.00, $596.31, $145,352,533,323,535.00, 1.
3) -54,543,021.00
4) An Israeli Shekel, a Yemeni Real, a German Mark, a French Fry.
5) The doctor was his mother!
 
1) 837
2) $253.00, , $45.67i, $9>2,$467.00, $596.31, $145,352,533,323,535.00, 1.
3) -54,543,021.00
4) An Israeli Shekel, a Yemeni Real, a German Mark, a French Fry.
5) The doctor was his mother!

Darn it SharonP, I wanted to try and solve it, but you already posted the answer :laughhard:
 
1) 837
2) $253.00, , $45.67i, $9>2,$467.00, $596.31, $145,352,533,323,535.00, 1.
3) -54,543,021.00
4) An Israeli Shekel, a Yemeni Real, a German Mark, a French Fry.
5) The doctor was his mother!
:rowfull:
Yet a serious solution exists. :D
 
Losing some of my hinges as well, so I'll try just because it's fun:

1) What is the fewest number of jars of candy that there could be on the counter? one (1)

2) What are the lowest prices possible for the candies in each jar, listed lowest to highest? 3 cents, 5 cents, 6 cents

3) What is the least total of amount of money that Susan could have in her pocket? 2 nickels, or 1 dime ~ (10 cents)

4) Presume that Susan has as few coins as possible to make all of the above true. Exactly what coins does Susan have in her pocket? 2 nickels, or 1 dime

Ugh. :eek:
 
2) What are the lowest prices possible for the candies in each jar, listed lowest to highest? 3 cents, 5 cents, 6 cents
One condition was: "In every case, the total cost of two pieces would require her to hand over a penny besides using some of the coins in her pocket, so she refuses, even though she has more than enough money to pay."

Since she has no pennies in her pocket, the any coins from her pocket will be some multiple of five cents. The cost of two pieces must be one cent more than a multiple of five. 2x3=6, Yes. But 2x5=10, No, and 2x6=12, No.

Thank you for being the only person so far to make an actual effort to solve it. :D
 
One condition was: "In every case, the total cost of two pieces would require her to hand over a penny besides using some of the coins in her pocket, so she refuses, even though she has more than enough money to pay."

Since she has no pennies in her pocket, the any coins from her pocket will be some multiple of five cents. The cost of two pieces must be one cent more than a multiple of five. 2x3=6, Yes. But 2x5=10, No, and 2x6=12, No.

Thank you for being the only person so far to make an actual effort to solve it. :D

You're welcome but Ohhh NOOO... I am only answering for the possible price of the candy, not necessarily whether she has a penny or not, that makes it: 2 candies cost 3 cents each; 3 candies cost 3 cents each... etc. Anyway, I'll re-anlayze and will see. Hehehe. :D
 
So to meet the first condition, only prices ending in 3 or 8 would work. (3*2=6, 8*2=16) both require handing over a penny. Both also meet the second condition. (3*3=9, 8*3=24) both require getting a penny back. In order for one candy from each jar, the jars will need to be a multiple of 3 or 8 times 5 or 10.
So.
1) 5 jars
2) 3,8,13,18,23
3) 65 cents
4) 2 quarters, a dime, and a nickel

*drops mike*
 
Maestro, you did it. My solution was too long and included 10 jars with prices ending in 3 or 8. Good job.
 
So to meet the first condition, only prices ending in 3 or 8 would work. (3*2=6, 8*2=16) both require handing over a penny. Both also meet the second condition. (3*3=9, 8*3=24) both require getting a penny back. In order for one candy from each jar, the jars will need to be a multiple of 3 or 8 times 5 or 10.
So.
1) 5 jars
2) 3,8,13,18,23
3) 65 cents
4) 2 quarters, a dime, and a nickel

*drops mike*
Almost totally correct! Congratulations! :cheer:
She does need to have 70 cents to be able to buy three of the 23-cent candies and get a penny change, so make that two nickels.
 
Details are of the devil

If you don't mind answering properly, I would like to know why you said "Details are of the devil" in a mathematical thread, created under the General Discussion Forum...
 
So to meet the first condition, only prices ending in 3 or 8 would work. (3*2=6, 8*2=16) both require handing over a penny. Both also meet the second condition. (3*3=9, 8*3=24) both require getting a penny back. In order for one candy from each jar, the jars will need to be a multiple of 3 or 8 times 5 or 10.
So.
1) 5 jars
2) 3,8,13,18,23
3) 65 cents
4) 2 quarters, a dime, and a nickel

*drops mike*

You did that while holding somebody in mid-air?! :shock:

lol
 
Guess I didn't understand the conditions. So, if she selected 2 or 3 pieces of the candy, she would have been choosing from a single jar. If she chose one piece, then she have had to take one from each of the jars. Had I understood that, I might have done it. Ah, well, live and learn. Congrats to Maestro anywho.
 
What's New

3/28/2024
Stop by the TMF Welcome Forum and take a second to say hello!
Tickle Experiment
Door 44
NEST 2024
Register here
The world's largest online clip store
Live Camgirls!
Live Camgirls
Streaming Videos
Pic of the Week
Pic of the Week
Congratulations to
*** brad1701 ***
The winner of our weekly Trivia, held every Sunday night at 11PM EST in our Chat Room
Back
Top