Academic challenge for advanced-users

[BrooksTech]

Heres an academic challenge. Unfortunately there's no prize other than Kudos and receiving credit for answering the challenge correctly:

1. CREATE A SPREADSHEET THAT WILL ANSWER THE FOLLOWING:

Youre trying to buy an item at a Vending Machine that costs 95 cents. The machine requires EXACT CHANGE ONLY. You do NOT have exact change and the change you have adds up to $1.15 (too much.)

Q. WHAT ARE THE NUMBER AND DENOMINATIONS OF COINS THAT YOU HAVE THAT RENDER $1.15 BUT YOU WILL NOT BE ABLE TO GET 95-CENTS FROM?<?xml:namespace prefix = o ns = "urn:schemas-microsoft-comfficeffice" /><o></o>

(This challenge was originally introduced by CarTalk on 12/24) Good Luck!

 

[xenou]

From the car talk site:

RAY: Last week, we had just finished the show, and Tommy and I were leaving the building. We were walking by the official NPR-sanctioned vending machines, with granola bars and the like. I had a hankering for a candy bar. I reached into my pocket, and, luckily, I had a dollar bill.

As I approached the machine with my crisp dollar bill, I notice the thing says, "Exact Change Only." The candy bar I wanted was 95 cents.

I turn to Tommy, and I say, "Can you break a buck?"

Tommy reaches into his pocket, and says, "I know I have six coins. The six coins add up to $1.15. But I can't make change for a dollar."

I say, "Huh? Can you make change for half a dollar?"

Tommy says, "No."

I then ask, "Can you make change for a quarter?"

Tommy says, "No, I can't do that either."

"How about a dime?"

"No."

"A nickel?"

"No."

So, I say, "Can you just buy me the candy bar?" To which Tommy replies, "No, I can't do that, either."

The question is, what were the six coins in Tommy's pocket?

 

[BrooksTech]

Please read above-correction, may be trick question

Sorry again. Ive put in some time on this and due to the corrected facts that youre not able to make basic change as stated, Im starting to think this is a trick-question relayed to me.

Tune into CarTalk on NPR or SiriusXM in your area on Saturday 12/31 to discover if this is a trick-question:

http://www.cartalk.com/content/no-change-ray-0?question

My intentions were good here, but Im not convinced this is spreadsheet solveable.

 

[Chasden]

Re: CORRECTION AND LINK Re: Academic challenge for advanced-users

If I understand this correctly - if you can't make change for $1.00, $.50, OR $.25 cents, that means you can't have any quarters, half-dollars, or dollar coins. Is that right?

You can have quarters, but the total number of quarters and nickels has to be odd to avoid making change for $1.00 and $.50. Specifically, it has to be odd and less than 2 (so 1, basically).

To avoid making change for $.25, the total number of nickels has to be even (meaning you have to have an odd number of quarters).

To avoid making change for a dime, you must have fewer than 2 nickels. Since it must be even, you must have 0 nickels.

To make $1.15 with only 1 quarter, and no nickels, you'd need a Quarter, a Half Dollar, and 4 Dimes.

Unless i'm not thinking this through properly...this actually sounds like a neat thing to do on a spreadsheet!

 

[BenMiller]

Re: CORRECTION AND LINK Re: Academic challenge for advanced-users

You can have quarters, but the total number of quarters and nickels has to be odd to avoid making change for $1.00 and $.50.

To avoid making change for $.25, the total number of nickels has to be even (meaning you have to have an odd number of quarters).

Unless i'm not thinking this through properly...this actually sounds like a neat thing to do on a spreadsheet!

Odd number of quarters isn't enough; you can have a maximum of one; otherwise you have change for $.50 ... I used the solver add-in and I got this: One quarter, one half dollar, and four dimes. Is that ok? Or can we not use half-dollars?

 

[Chasden]

Re: CORRECTION AND LINK Re: Academic challenge for advanced-users

Odd number of quarters isn't enough; you can have a maximum of one; otherwise you have change for $.50 ... I used the solver add-in and I got this: One quarter, one half dollar, and four dimes. Is that ok? Or can we not use half-dollars?

I edited as you were replying!

 

[BenMiller]

Re: CORRECTION AND LINK Re: Academic challenge for advanced-users

To avoid making change for a dime, you must have fewer than 2 nickels. Since it must be even, you must have 0 nickels.

Why wouldn't you be able to have, say one nickel, one dime, and one quarter? Why are you saying that you can't have any nickels?

 

[BrooksTech]

Ben got it, but SPREADSHEET CHALLENGE STILL ON

I think Ben got it. (Yes, as far as I know any coins, even Susan-B's are fair game.)

Of course my point was not simply who could get the answer, but who could produce an automated spreadsheet that would produce this correct answer.

For those of you wanting a CarTalk T-Shirt or Credit at their Shameless-Commerce-Division, feel free to register your answer at their website, and your name may be drawn from a pool of correct answers.

I wish to emphasize however that the challenge for producing a spreadsheet on this is still ON. Thanks again.

 

[Ron Coderre]

Re: CORRECTION AND LINK Re: Academic challenge for advanced-users

1 half-dollar + 1 quarter + 4 dimes = 1.15
BUT...you can use 1 half-dollar + 1 quarter + 2 dimes to get 0.95

I suspect that no combination of current coins satisfies both criteria.
If 3-cent coins (last minted in 1889 per wikipedia) are allowed then
5 3-cent coins + 1 silver-dollar = 1.15
and no combination equals 0.95

 

[Chasden]

I'm trying to find a way to show "all solutions" assuming theoretical coins of any value are allowed (For example, coins of value 0.04, 0.09, 0.12, 0.17, 0.24, 0.49 would be another solution).

I couldn't figure it out using the Solver Add-in and a list of all values from 0.01 to 1.14.

Is there a way to do this?
Set up a 109C6 combination in Excel (using all "coin" values from 0.01 to 1.09), and get all combinations that have a sum of 1.15, then, using this solution set, find all sums of 6C5, 6C4, 6C3, 6C2, and 6C1 that are not equal to 0.05, 0.10, 0.25, 0.50, and 1.00.

I chose 1.09 because you must have 6 coins, so 1.11 and above are impossible as a single coin. 1.10 is impossible because then you'd need 5 pennies, which would be change for a nickel.

Of course, 109C6 with repetition allowed yields (114*113*112*111*110*109)/(6*5*4*3*2*1) = 2,666,926,108 possibilities! Couldn't list them even if we wanted to!

 

[Chasden]

Re: CORRECTION AND LINK Re: Academic challenge for advanced-users

Why wouldn't you be able to have, say one nickel, one dime, and one quarter? Why are you saying that you can't have any nickels?

I assumed that (quarters + nickels) had to be odd, neglecting that 1 quarter and 1 nickel would avoid making change for $0.50