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Six Coins: Newsletter Challenge (02/13/07)

Posted February 11, 2007 5:01 PM
Pathfinder Tags: challenge questions
User-tagged by 1 user

The question as it appears in the 02/13 edition of Specs & Techs from GlobalSpec:

You have six coins with the following values (you live in a remote country of course!): 3 cents, 9 cents, 10 cents, 12 cents, 13 cents, and 14 cents. You use five of them to buy two articles. One costs twice as many cents as the other. What is the value of the coin you kept?

(Update 9:15 Am EST 02/20/07) And the Answer is....

If a, b, c, d, and e are the five coins used to buy the articles, then, their sum can be expressed as

a + b + c + d + e = 2x + x = 3x

where x is the cost of the less expensive article. From the above equation we see that the sum of these five coins is a multiple of 3.

Now, the sum of all the coins is: 3 + 9 + 10 + 12 + 13 + 14 = 61 which is 20 * 3 + 1. In other words, 61 is 20 + 1 (mod 3).

Therefore, the coin left in your hands has to have a value of 1(mod 3).

The only two coins in the group that give a value of 1(mod 3) are 10 and 13.

Let's look at the two cases:

(a) Assume that the coin you kept is 13. Then the sum of the other five must be 3x, where x is less expensive product bought. This, however is not possible because 3 + 9 + 10 + 12 + 14 = 48 = 16 * 3, but no combination of coins in this group will give 16.

(b) Now assume the coin left is 10. Then 3 + 9 + 12 + 13 + 14 = 51 = 17 * 3. The combination of coins to buy the products will be: 14 + 3 = 17 = x and 13 + 12 + 9 = 34 = 2x.

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Participant

Join Date: Feb 2007
Location: Raleigh, NC,USA
Posts: 3
#98

Re: Six Coins: Newsletter Challenge (02/13/07)

02/17/2007 3:40 PM

Five of the six coins must sum to an amount which has 3 as a factor since one of the items has twice the cost of the other. The 6 coins total 61 cents. The unused coin may not be 3, 9, or 12 cents since these would leave a sum not divisible by 3, just as 61 is not. The 14 cent coin would leave 47 cents, also not divisible by 3. The 13 cent coin would leave 48 cents one third of which is 16 cents for the value of the cheaper item, but without the 13 cent coin there is no way to pay for it (I assume that each item was paid for in a separate transaction; otherwise there are two solutions to the puzzle).

The solution, therefore, is: one item cost 14+3=17 cents, the other cost 13+12+9 = 34 cents = 2x17, and I (or you) keep the 10 cent coin. QED

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ajk
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Power-User

Join Date: Jan 2006
Posts: 377
Good Answers: 2
#99

Re: Six Coins: Newsletter Challenge (02/13/07)

02/17/2007 5:32 PM

The answer could be the 13 cent coin because of the discount given for purchasing both items at the same time.

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Anonymous Poster
#100

Re: Six Coins: Newsletter Challenge (02/13/07)

02/18/2007 10:36 AM

I kept with 10 cents coin.

if was 13, I can keep wich two coins 3 + 10 or only one, 13 cents

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Guru

Join Date: Jan 2007
Location: Edinburgh, Bonnie Scotland
Posts: 1337
Good Answers: 23
#102

Re: Six Coins: Newsletter Challenge (02/13/07)

02/19/2007 4:46 AM

Interesting - everyone assumes that this place uses base 10 - they may use base 18 for all we know.

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Anonymous Poster
#105

Re: Six Coins: Newsletter Challenge (02/13/07)

02/20/2007 2:07 PM

The question does not say that the two articles are bought separately, suggesting that the two solutions are both possible.

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Power-User

Join Date: Jan 2007
Location: Timmins Ontario Canada
Posts: 128
Good Answers: 1
#106

Re: Six Coins: Newsletter Challenge (02/13/07)

02/20/2007 2:21 PM

That answer seems a little complicated. The 6 possible coin combinations of 5 coins have sums of: 47, 48, 49, 51, 52, and 58. Considering one purchase is twice the other (x+2x=3x), only sums divisable by 3 can be possible combinations. That leaves only the combinations minus either the 10cent, or 13cent coin. Of these two combinations, it is only possible to arrange the remaining coins into 17 and 34 cent values - not the 16 and 32 cent required by removing the 13cent coin. Therefore the 10cent coin is the one left in your pocket.

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Anonymous Poster
#107

Re: Six Coins: Newsletter Challenge (02/13/07)

02/20/2007 2:48 PM

While I understand the way you have expressed the equation, I arrived at the answer by just adding all coin values together and finding a number when doubled gave me a remainder that fit the odd coin. Does this make sense? I came up with ten as the coin left.

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Active Contributor

Join Date: Feb 2007
Posts: 11
#108

Re: Six Coins: Newsletter Challenge (02/13/07)

02/20/2007 3:31 PM

With the before mentioned tax issue, if said country uses coins in these weird denominations, then they probably don't have the economic impact and stable government to impose a tax. Logically speaking, the 10c coin remains in your pocket as no matter the situation, it will always remain a possibility. Putting any sort of constraint, just allows more possibilities.

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Anonymous Poster
#109

Re: Six Coins: Newsletter Challenge (02/13/07)

02/20/2007 11:22 PM

Sir

It is just 10 + 2 standard question. Please do ask enineering level question to challenge Engineers.

MOHD. SAEED KHAN

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Anonymous Poster
#110

Re: Six Coins: Newsletter Challenge (02/13/07)

02/21/2007 10:30 AM

Nowhere in the problem does it say that the items were both an exact (or exact sum of) coin value. The justification above regarding the the 13 cent coin is not relevant to the problem and asumes information not given. Therefore the coin left over could be either the 10 cent coin or the 13 cent coin.

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Anonymous Poster
#111

Re: Six Coins: Newsletter Challenge (02/13/07)

02/21/2007 11:18 AM

Both the 13 cent coin and the 10 cent coin would be correct answers for the problem as stated. The problem only stated that you bought 2 articles, it did not specify 2 seperate transactions. Also even if you were to assume that it were two transactions it did not specify that exact change was required for both transactions. You could use change from the first transaction to make up for the second transaction

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Member

Join Date: Feb 2007
Posts: 8
#112

Re: Six Coins: Newsletter Challenge (02/13/07)

02/21/2007 1:56 PM

I answered too late without reading your answer - a similar but little different approach;

61-y=3x

y=unused coin

61-y has to be evenly divided by 3, so it has to be 51 (3 x 17)

thus 10 is the unused coin

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Associate

Join Date: Feb 2007
Location: Brampton Canada
Posts: 50
#113

Re: Six Coins: Newsletter Challenge (02/13/07)

02/27/2007 7:52 PM

10 cents. As (3+14=17) and (9+12+13=34) double of 17.

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Participant

Join Date: Feb 2007
Posts: 2
#114

Re: Six Coins: Newsletter Challenge (02/13/07)

02/28/2007 8:51 AM

The sum has to be an even number because of the value of the articles. There are six coins which have 3 even and 3 odd values. There can only be an even number of "odd numbered coins" used, so we use 10,12,14 for starters giving us 36.

We must use two and only two of the odd ones. 3,9,13. We must use either 3+9,3+13 or 9+13. This gives us solutions of 36+12=48, 36+16=52, 36+22=58. The only one to give a whole number solution to x+2x is 48. The coins used are 3,9,10,12,14, leaving 13 in your pocket

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Associate

Join Date: Feb 2007
Location: Brampton Canada
Posts: 50
#115

Re: Six Coins: Newsletter Challenge (02/13/07)

03/03/2007 1:27 PM

Now find is the final answer, 10 cents

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Anonymous Poster
#116

Re: Six Coins: Newsletter Challenge (02/13/07)

08/19/2009 5:42 AM

tom

what are accountants for anyway?

ask it to your wife. then you will have the right answer. with no further questions to argue with.

they rules and they enslave.

them we govern.

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