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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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#1

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

02/12/2007 1:02 AM

The value of the unspent coin is 13 cents.

Associate

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#2

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

02/12/2007 2:22 AM

The unused coin is 10 cents. The value of the articles are 17 cents ( 14 + 3) and the other is 34 ( 13+12+ 9)

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#3
In reply to #2

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

02/12/2007 2:34 PM

I concur with the solution from NatNat. Althought the first solution may also be possible given a single purchase, NatNat's has the advantage of permitting two separate purchases presumably from separate vendors, using exact change. I go with the more elegant solution.

Anonymous Poster
#4

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

02/13/2007 12:26 AM

You are left with 10 cent coin. The first item was 17 (3+14) and the second 34=17x2 was paid with 9+12+13=34

Anonymous Poster
#5

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

02/13/2007 12:31 AM

Your are left with the 10 cent coin. Paid the first item 17 with the 14 and 3 coins. Paid the second 34 with the 9+12+13 coins.

Anonymous Poster
#6

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

02/13/2007 1:39 AM

The value of the coin left is 10 cents

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#7

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

02/13/2007 1:57 AM

The value of the coin kept is 10 cents

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#8

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

02/13/2007 3:50 AM

There are two solutions:

If you purchases are worth x and 2x, then your total expenditure is 3x, which gives us the constraint that your total outlay must be a multiple of 3.

Your total worth is 61c - which is (10x3)+1. Therefore, the coin remaining in your pocket or purse must have a value of (3y+1), and your outlay must therefore be 3(10-y)=x.

Two coins satisfy the 3y+1 criterion: 10c and 13c

Holding on to the 10c coin means you spend 51c, on items worth 17c & 34c, these can be paid for with 14c + 3c and 9c + 12c + 13c coins repsectively.

Holding on to the 13c coin means you spend 48c, on items worth 16c and 32c; you have to a) pay for these in one transaction, or b) handle other demoninations in change between to the two transactions (the question does not preclude this!).

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#13
In reply to #8

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

02/13/2007 10:02 AM

All answers are correct if you allow partial cents. Therefore, I assumed only whole cents are allowed. Given that, I came up with the exact same solution as English Rose (i.e. keeping the 10c piece allows one item at 17c and one at 34 and keeping the 13c piece allows for one item at 16c and one 32c). Again, given the assumption that change is allowed or both purchases being placed at the same place and time.

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#42
In reply to #8

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

02/14/2007 6:20 AM

I think this is a simple question with simple catch in it. There mathematically 2 answeres 10c and 13c. But physically 13c is not possible. So the answer is 10c only

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#44
In reply to #42

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

02/14/2007 7:17 AM

The point is to spot the catch and identify its implications. It is not explicitly stated in the question whether the items are purchased separately or together, so both solutions are valid and physically possible, with the caveats as previously stated. Your limiting the solution to only the 10c coin is imposing a condition that is not set out in the question.

Usually these ambiguities prompt a lot of discussion and lead to distinct families of solutions - this one has got three families:

a) separate purchases (10c left)

b) combined purchases (10C or 13C left)

c) fractional coin values allowed (all coins are possible)

It's all part of the fun and point of these questions

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Anonymous Poster
#95
In reply to #8

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

02/16/2007 7:51 AM

These 10 cent or 13 cent solutions also depends on the country's method of selling items. Do they have individual vendors (like apple carts) or do they have supermarkets?

If one shop or street vendor sells the 17 cent item and another vendor sells the 34 cent item then you can only satisfy the 10 cent in your pocket option. (since you need the 13 cents to make a 16 cent sale and the 32 cent sale is not possible with out the 13 cents as well and having spent the 13 cents twice you don't have it in your pocket at the end.

However if the country has a market place where you can purchase multiple items in one transaction, then either the 10 cent (3+14=17 and 9+12+13=34) or the 13 cent (3+9+10+12+14=48 where one item is 16 cents and the second is 32 cents) options will work. Because they are purchased together and exact amounts for either is not required.

In either case (multiple vendors or supermaket) the 10 cent remaining works. The 13 cent remaining works only in supermarket areas therefore because this is a remote country and probably does not have supermarkets, the 10 cent in your pocket is most likely.

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#9

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

02/13/2007 3:59 AM

More rules are required.

Both 10 and 13 yield a answer of rounded cents (17-34 and 16 -32)

the other coins can also give a correct answer!

3 cents coin == prices of 19 1/3 and 38 2/3

9 cent coin == prices of 17 1/3 and 34 2/3

12 cent coin == prices of 16 1/3 and 32 2/3

etc.

take your pick!

I think the coinage system is to blame for the state of the economy of that country.

The next question should be to name the country.

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#10
In reply to #9

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

02/13/2007 4:12 AM

Nice one! I can remember 1/2p here in the UK and before my time there was the 1/4d (farthing) and there were one-third, one-half and one-quarter farthings at one time...

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Anonymous Poster
#11

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

02/13/2007 8:31 AM

The coin unspent is 10 cents. One item was purchased for 17 cents (14+3). The other item was purchased at 34 cents (9+12+13)

Anonymous Poster
#12

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

02/13/2007 8:48 AM

9 cents

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#14

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

02/13/2007 10:07 AM

You will keep the 10 cents coin - all other coins can be devided into 3.

Oded

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#15
In reply to #14

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

02/13/2007 11:02 AM

Just because it is snowing out side and I am bored, I have to point out that 13 and 14 are NOT divisible by 3, the only saving grace is that 27 is divisible.

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Anonymous Poster
#16

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

02/13/2007 12:14 PM

I came up with

Tc = Total of coins 61 cents

Lc = Value of coin left

X = value of item one

Y = value of item two

Y = (Tc-Lc)/3

X = ((Tc-Lc)/3)2

Through process of elimination answer can be either 10 or 13.

61 minus 13 equals 48. 3+9+10+12+14=48

61 minus 10 equals 51. 3+9+12+13+14=51

Anonymous Poster
#17

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

02/13/2007 12:36 PM

3 + 9 + 12 = 24

10 + 14 = 24

13 cents is left.

Anonymous Poster
#19
In reply to #17

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

02/13/2007 12:59 PM

one is to cost twice as many cents as the other item

Anonymous Poster
#18

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

02/13/2007 12:52 PM

you kept the 10 cent coin.

you bought two object of worth

17cents(14+3) and the other of

34cents(12+9+13)

the other option of 16 and 32 not working..............

Anonymous Poster
#20

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

02/13/2007 1:15 PM

x is the value of the cheaper item. 2x is the value of the other item.

Total value of all coins is 61 cents. Value of spent coins, is 61 minus the value of the retained coin, and is also 3x (x+2x=3x). Assuming items are priced only in whole cents, and that we don't allow "manager's discretion" concerning fractional cents; the total value of the coins spent must be evenly divisible by three. I.e., spending any combination of coins, where the total value of the spent coins is evenly divisible by three, will yield a correct answer.

(61-3)/3, (61-9)/3, (61-12)/3, and (61-14)/3 all give fractional cents for the value of x. (61-10)/3 and (61-13)/3 both give whole numbers for the value of x. Therefore, you can keep either the 10 cent coin or the 13 cent coin.

Anonymous Poster
#21

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

02/13/2007 1:30 PM

13 cents left over
3, 9, 10 , 12 and 14 equal 48- the items cost 16 and 32

Anonymous Poster
#22

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

02/13/2007 2:10 PM

The correct answer is 13 cents.

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#23

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

02/13/2007 2:27 PM

you kept the dime-

i sure hope that is correct as i must say-though this is actually my first time responding- that the questions posed are a far cry from those with which i was first attracted to this feature. am i mistaken in this? this seems to be a high school logic question....

Anonymous Poster
#24
In reply to #23

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

02/13/2007 2:59 PM

You're right. The questions certainly have gotten simpler lately - More Jr. High math, less Engineering creativity. On the bright side though, I'm now able to solve a few of them.

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#59
In reply to #23

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

02/14/2007 9:41 AM

Prove it!

Trial and error is not a brain teaser. If instead of having six coins and two purchases you have 600 coins and 200 purchases, you will not be able to solve the problem observation only.

The idea of math challenges is to solve problems by giving a definite, unquestionable, general answer. The only way to do this is to use your math and logic skills.

If I ask you what is the color of the sky (when it is not snowing), you may answer that it is blue. This is a boring answer. If besides telling me that the sky is blue you prove it, then you will see that even if the question asked is simple, the answer gave your brain a little exercise and you will fill happy.

Anonymous Poster
#94
In reply to #59

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

02/15/2007 11:17 PM

anonymous...........and if I fill happy, how will I feel?

Anonymous Poster
#70
In reply to #23

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

02/14/2007 11:20 AM

And, just like a high school logic question, it is the debate provoked which is of interest!

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#77
In reply to #70

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

02/14/2007 4:17 PM

Absolutely well said!!!

Anonymous Poster
#25

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

02/13/2007 3:32 PM

10 cents coin

Anonymous Poster
#26

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

02/13/2007 3:52 PM

So, you have 61 cents in your pocket. The articles you are buying cost x and 2x. Thus, you must spend an amount that is divisible by three. Once 61 is not divisible by three, you can't keep the coins of 3, 9 and 12 cents. Subtracting the other three coins from 61 you will find that the only possible result that is divisible by three are 61 - 10 = 51 an 61 - 13 = 48. In the second case one paper would cost 16 cents, and there's no possible way to combine your coins to that price. Thus, you keep your 10 cents coin, pay 17 cents for one article (3 + 14) and 34 cents for the other one (9 + 12 + 13)

Anonymous Poster
#27

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

02/13/2007 4:53 PM

For the your question below:

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?

This is my answer:

Since one article costs twice as many as the other, the total cost for the 5 coins should be multiple of 3. As the sum of the six coins is 61, taking out 10 or 13 will make the remaining to be multiple of 3. Thus, the value of the coin kept can be either 10 or 13. Cheers.

Anonymous Poster
#35
In reply to #27

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

02/13/2007 10:19 PM

Sorry, you can't keep the 13.

Item 1 cost = x. Item 2 cost = 2x. Total cost = 3x.

Kept coin = y.

Total of all coins = 61=3x+y.

Therefore, (61-y)/3=x

Assuming only whole numbers of cents only y=10 or y=13 are acceptable. But if y=13, then x=16. Only combination of coins to buy item 1 is 13+3=16. So you use the 13 cent coin to keep the 13 cent coin in your pocket. Not possible.

Choose y=10, so that x = 17. Then item 1 cost 17=14+3 while item 2 cost 34=9+12+13.

Anonymous Poster
#28

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

02/13/2007 5:10 PM

10 cents Left

Item 1 34 cents 12+13+9

Item 2 17 cents 14+3

Anonymous Poster
#29

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

02/13/2007 6:28 PM

kept the coin valued at 10 cents

Anonymous Poster
#30

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

02/13/2007 6:41 PM

you have 13 cents

Associate

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#31

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

02/13/2007 8:21 PM

it is the 10c coin you kept. the only totals of 5 of the 6 coins that are divisible by 3 are 48 and 51. The total must be divisible by 3 because no matter how much you spend on one item, call it x, it has to add with the other item, which costs twice as much, that is, 2x, which means you spend 3x. 48 cannot be it because its 1/3 divisor is 16 and there is no way to get 16 and 32 from the combination that makes 48, which is 3, 9, 10, 12, and 14. i'm sure there is a more elegant way of putting or proving this. james h waters phd

Anonymous Poster
#32

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

02/13/2007 8:53 PM

12+13+9=34

3+14=17

I'm left with a thin dime..

Mike

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#33

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

02/13/2007 9:17 PM

the value of the remaining coin is 10 cents. (3 & 14 = 17). (9 & 12 & 13 = 34). figured it by hit and miss. still working on the equation.

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#34

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

02/13/2007 9:57 PM

10 cents

Simple question simple answer. Don't make it out to be more than it is.

Anonymous Poster
#36

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

02/14/2007 12:37 AM

13 cent willremain with me.

as sum of all coins exept 13 cent coin is 48 . the cost of 1st article is 16 cent and other of 32 cent. and 32 is twice of 16.

Anonymous Poster
#37

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

02/14/2007 2:34 AM

I kept the coin with a value of 13 cents.

3+9+10+12+14 cents = 48 cents

-> the other item costs 16 cents and the other 32 cents.

Anonymous Poster
#38

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

02/14/2007 3:33 AM

Owenmeany you are right.. this 'Challenge' Question is nothing short of crap! A primary school maths quiz question, not even an exercise in lateral thinking which would at least provide some exercise for the brain cells!

So what if we can actually solve the puzzles now, I like to be enlightened when I receive the answer to something I don't know, not bored by the original question!

Come on CR4! Stop this dumbing down of what used to be an interesting blog!

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#39
In reply to #38

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

02/14/2007 4:08 AM

To Owenmeany & Guest #38 - quit whinging and submit some of your own.

As for the quality of the question, whilst these are not up to the usual standard, at least they're new...I was getting fed up of the recycled ones (yes guys, some of us noticed that questions from 2 - 3 years ago were making a second appearance!).

Jorrie was submitting questions for a while - is he ok and still around CR4?

So there we go - there's the new challenge - write your own questions in the style you'd like to see them and give the hardworking CR4 staff a break.

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#69
In reply to #39

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

02/14/2007 11:11 AM

did not intend to whine. Yet i did notice that you concur....

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#71
In reply to #69

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

02/14/2007 11:33 AM

No arguing with facts but like beauty they're somtimes skin-deep...a number of other posts here point out why the question isn't as trivial as may be first thought, even if the maths behind it is straight-forward. Maybe what we're all saying is that we like the concept style questions (your plane crashed/you go to the cabin...) better than the apparantly rational and logical/mathematical style questions.

And I apologise, you didn't whinge (memo to self - double check ), so Guest #38 takes the full rap!

And I'm not complaining about the CQs coz I don't get off my a\$\$ and write any; I'm grateful to those who do take the time!

Please keep contributing!

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Anonymous Poster
#68
In reply to #38

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

02/14/2007 10:50 AM

Such limited thinking!! Bravo, English Rose on your excellent response!

Anonymous Poster
#40

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

02/14/2007 4:13 AM

I sure am glad I have all you who understand how to do this... all I have to do is look at your test paper for the solutions. He he ... I became a musician instead because of questions like these and never regretted that choice :-) Maybe someday I will say Aha!!!

Anonymous Poster
#41

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

02/14/2007 4:20 AM

That's cool! I keep 10 cents and buy articles of 17 cents (14+3) and 34 cents (13+12+9).

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#43

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

02/14/2007 7:05 AM

It is only when you have to pay for each item individually with a whole number of coins that there is one answer. Unless in this weird country you can cut up coins with tin-snips to make change.

If you can pay for both items together in a single purchase then any coin can remain.

ie., I kept the 3 cent coin. One item was 19.333 cents and the other was 38.666 a total of 58 cents.

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Anonymous Poster
#57
In reply to #43

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

02/14/2007 9:23 AM

Sorry, I get a total of 57.999... not correct

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#60
In reply to #57

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

02/14/2007 9:46 AM

We were not asked how much the items were. So my answer is still correct because I still only have a 3 cent coin left over.

It looks as though I have overpaid.

But in terms of engineering correctness how do we denote/enter 'recurring' decimals in CR4.

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#96
In reply to #57

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

02/16/2007 9:14 AM

Use fractions, not decimals!

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Anonymous Poster
#45

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

02/14/2007 7:18 AM

Well, it appears that there are 2 answers. the total value of the coinage is 61 cents. So if you held back 13 cent coin you would have 48 cents thus one article would cost 16 cents and the other 32.

However, if you held back a dime, you would have bought a 17 cent article and a 34 cent article.

Anonymous Poster
#46

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

02/14/2007 7:22 AM

If you bought the articles at different stores you could have used the (14 + 3) coins and at another the (13 + 12 + 9).

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#47

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

02/14/2007 7:42 AM

Just leave the damned 10 cents as a tip. You ain't gonna retire on the dime anyway ! :)

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Anonymous Poster
#48

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

02/14/2007 8:02 AM

Its a no brainer the answer is 13 cents.

Anonymous Poster
#49

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

02/14/2007 8:35 AM

10 cents

spent coin values:

3 + 14 = 17

9 + 12 + 13 = 34

Anonymous Poster
#50

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

02/14/2007 8:35 AM

Six coins

The six coins add to 61 cents. Trere are two possible answers:

You buy 17 + 34 = 51, keeping 10 cents, or

you buy 16 + 32 = 48, keeping 13 cents.

L. Maia (maia@inf.ufsc.br)

Anonymous Poster
#51

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

02/14/2007 8:37 AM

3+9+10+12+13+14=61

16+32+13=61 ...answer=13

or

17+34+10=61...answer=10

Anonymous Poster
#52

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

02/14/2007 8:45 AM

Two possible answers 10 or 13 cent coins left

George

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#53

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

02/14/2007 8:48 AM

3+9+10+12+13+14=61

16+32=48+13=61....answer=13

or

17+34=51+10=61....answer=10

Anonymous Poster
#54

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

02/14/2007 8:50 AM

This is grade school math, right up my alley; (No strange formulas required, just the process of elimination...)

3+14=17

9+12+13=34

The remaining coin is the 10 cent coin.

Rob W

Anonymous Poster
#55

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

02/14/2007 9:02 AM

if you kept coin with the value of 13 cents
you have left
3+9+10+12+14 cents = 48 cents and claim that
one item costs 16 and the 32, good so far... but
which coins do you use to get 16 and 32?
It seems to me that the only answer is 10 when
you use 3+14=17 and 9+12+13=34 -> 34=2*17
cheers

Anonymous Poster
#56

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

02/14/2007 9:20 AM

So just how many people will answer this 4th grade math challenge question? Geez people.

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#66
In reply to #56

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

02/14/2007 10:20 AM

Why are you coming back to this thread without registering and/or logging on? Geez!

Seriously, not everyone will access this through the website. A lot of the guests simply send their answer via the link in Specs & Techs newsletter, so they haven't read your pearls of wisdom.

And the number who do give an interesting insight into the flexibility of engineers' thinking...the "too trivial" brigade to which you belong; the unique solution lot, the slightly flexible lot and the very outside the box few. That in itself is worth something, as it the mild debate that's going on as to whether both the 10c and 13c solutions are allowed, etc.

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Anonymous Poster
#67
In reply to #56

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

02/14/2007 10:43 AM

Silly! It's not a math question, but a study of how the human mind works :-)!

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#76
In reply to #56

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

02/14/2007 3:40 PM

The solution of the problem is easy to find: just spend some minutes looking a two subsets wth the following conditions: (1) the sum of cents in one set will be twice as big the sum of the cents in the other set, and (2) the number of elements in both sets has to be five. It is not difficult to do this by trial and error, but the elegance of the solution (put your brain in high gear) is to establish a general solution using the most beautiful intellectual tools ever invented: mathematics (including logic and set theory)!

As I said, with only five coins it is easy to solve the problem by trial and error (I agree with you that a fourth grader may be able to do it). How about if we have the same problem with 600 coins and 200 purchases? Could you think this also is a trivial problem. Use your brain and use it often!

Anonymous Poster
#58

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

02/14/2007 9:36 AM

The fact that one item is twice as expensive as the other means nothing. You've spent a total of five coins thats it, no total, so I would keep the \$0.14 coin cause I am cheap and want to spend as little as possible.

Anonymous Poster
#61

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

02/14/2007 9:49 AM

I get two possible answers:

One item costs twice as much as the other, so the items cost n cents and 2n cents. In total they cost 3n cents. So the total of the combination of coins used must be divisible by 3.

There is one coin left and the total of all coins, 61 cents, is not divisible by 3. Since the 3, 9, and 12 are alone divisible by 3, we have proven that those coins are used (since they cannot be left behind). That leaves the 10, 13 and 14 cent values. Both 10 + 14 and 13 + 14 are divisible by three. In the first case, the total purchase is 48 cents, leaving a 13 cent coin and so the two items were 16 and 32 cents. In the second possible case we have the 10 cent coin in our pocket and the two items were 17 and 34 cents.

Anonymous Poster
#62

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

02/14/2007 9:50 AM

13 cents, article 1 cost 16 cents, article 2 cost 32 cents.

Anonymous Poster
#63

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

02/14/2007 10:05 AM

Kept coin 10 cents. the total cents spent needs to be divisable by 3.

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#64

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

02/14/2007 10:06 AM

The six coins add up to 61. The two articles can be expressed as x + 2x = 61 - y where y is the value of the coin you kept. So .... 3x = 61 - y. For (61-y) to be divisible by 3, y must be odd. This leaves 3, 9 or 13. 13 is the only coin when subtracted from 61 giving an evenly divisible quantity, i.e., 48. The two articles cost 16 and 32 and the coin that is left is 13.

Anonymous Poster
#65

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

02/14/2007 10:15 AM

I was able to make both of my purchases at one merchant. I paided \$0.16 for a length of white ribbon and \$0.32 for a red gift bow. I kept the rare 1907 13 cent coin, placed it in a box, adorned with the ribbon and bow and gave it to my niece for Valentines Day.

Truly, Seth MacMurray

Anonymous Poster
#72

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

02/14/2007 12:32 PM

You have 13 cents left. The items cost 16 cents and 32 cents.

Anonymous Poster
#73

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

02/14/2007 12:39 PM

You would be left with the 10 cents. It is not possible to come up with a solution for the 13 cents because the coins that you have do not add up to 16 or 32 cents and for 1 item to be double the cost of the other you would have to get change from one of the items you buy.

Anonymous Poster
#74

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

02/14/2007 1:48 PM

The consensus that there are two possible answers has been well documented, however, no one has addressed how to actually pay for the purchase that allows you to keep the 13 cent coin with the coin denomination available:

Purchase #1 = 16 cents - pay with the 14+12, and get a 10 cent coin as change.

Purchase #2 = 2 x 16 cents = 32 cents - 2 x 10, 9 and 3 cent coins = 32 cents for the second purchase

You are left with the 13 cent coin.

RJ

Anonymous Poster
#75

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

02/14/2007 3:08 PM

Using the 12cent 13 cent and 9 cent coins you can buy the more expensve 34 cent item.

Then with 2 of your remaining coins you can buy the 1/2 priced 17cent item with the 14 and 3 cent coins.

That leaves you with the lowly 10 cent coin.

Altenatively If you buy both items at the same time, and have slightly lower standards you can buy 48 cents worth of items. a 16 cent item and a 32 cent item, leaving you with 13 cents in your pocket. Which, if you like money in your pocket more could be the desireable result. However without a unique change system these items cannot be bought independently

Anonymous Poster
#78

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

02/14/2007 6:40 PM

You keep the 10c coin:

Item 1 = 16c

Item 2 = 32c

Subtotal = 48c

Tax @ 6% = 2.88c - but rounded to 3c (the gov't keeps the round up, taking more than the stated rate!)

Total = 51c

leaving you the 10c coin...

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#83
In reply to #78

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

02/15/2007 12:33 AM

slick, tax!

Anonymous Poster
#79

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

02/14/2007 6:40 PM

if you add up 5 coins and they come to an even number and divide by three and the answer is an even number then one item is twice as much as the other .the remaining no is 13 .this only works out if you assume to use any coins other than the 13 to pay for the 2 items and recieve change on one to pay for the other eg 10 cents

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#80

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

02/14/2007 9:38 PM

The answer is 10. 17 (3+14) and 34 (9+ 12+13). We're not talking multiple purchases or cutting coins in half. What are you going to do with half a coin anyway?

Anonymous Poster
#81

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

02/15/2007 12:01 AM

the unspent coin is 10cents

since the remaining coins left are 3, 9, 12, 13 and 14.

and you can buy two articles with following denominations:

article1 :- 14+3=17

article2 :- 9+12+13=34

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#82

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

02/15/2007 12:20 AM

.61 total

2x+x=?

take a coin away from .61 until you get a number divisible by 3,

.61-.13=48

(2x+x=48)

answer: .13

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#84

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

02/15/2007 12:49 AM

I have to agree with removing denomination amounts from 61 until a number divisible by 3 is found, but then the numbers must be checked by the denominations of the coins available for the task. The coin left is the 10 cent coin.

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#86
In reply to #84

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

02/15/2007 5:47 AM

Ah, maybe the key is whether both were bought at the same time or during two separate transactions. But, since .10 seems to cover more possibilities, including the calculation of a 6% sales tax as was interestingly pointed out, I agree it seems to be the best answer.

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#87
In reply to #86

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

02/15/2007 7:42 AM

The tax point (and it's neat!) only works if the country adds it's sales tax at the point of purchase. In Europe, all retail prices are shown inclusive of sales tax (known as value added tax). I was quite confused in Canada when my bill came to 6% more that the ticket price - and the shop assistant thought I was mad when I told her the UK includes in in the ticket price!

So this falls into the caveat section.

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Anonymous Poster
#85

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

02/15/2007 3:43 AM

There two answers for this. One set is 3,9,10,12and 14. and the other is 3,9,12,13 and 14.

Since the cost of two articles is to be divided by three ( since one costs double the other).

Anonymous Poster
#88

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

02/15/2007 10:09 AM

It appears you will have either 10 or 13 cents left over as either of these values fits the parameters. BCC

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#89

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

02/15/2007 1:00 PM

10 or 13 cents.

Someone was wondering what country this was... It HAS to be the US because 61 cents is all I have left after the government took the rest of my \$100,000 salary to help build a multi-billion dollar seaport in North Dakota

Bill

For those non- US people, North Dakota is about as far from the ocean as one can get in the US. But this is typical of the government.

Anonymous Poster
#90

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

02/15/2007 3:16 PM

Keep te 13 cents coin.

x + 2x = 3x = 3+9+12+10+24=48

x=16

2x=32

Anonymous Poster
#91

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

02/15/2007 4:45 PM

10 or 13

Anonymous Poster
#92

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

02/15/2007 5:31 PM

I understand RJ's explanation (entry #74), which allows for you to keep the 13-cent coin. However, as explained you need to get a 10-cent coin back as change so that two 10-cent coins plus 9cents plus 3cents can be used to complete the second purchase. Now you've used a total of six coins in the two transactions.

I assumed the problem was to use the coins we were given, and no change. That's why I could not understand how anyone would continue to think the 13-cent coin could be correct. Thanks to the lively discussions and RJ's response (and the problem not stating specifically to use only your ORIGINAL coins), I agree there are two answers. Answers requiring fractions of a coin are a little too far outside the box for me to accept, though.

Anonymous Poster
#93

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

02/15/2007 6:29 PM

10

17 cent + 34 cent = 51 with 10 cent left over

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#97

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

02/16/2007 2:00 PM

I agree with the methodology that several people have used to come to the conclusion that there are two answers where the 10 or 13 cent coin is left but everybody has forgotten about the tax on each item. If we factor in all the taxes and say that they come to 33 ⅓% then the total cost of the value must be divisible by 3 times 133⅓% or 4 which means the only possible answer becomes a total expenditure of 52 cents with the 9 cent coin remaining and the items costing 13 and 26 cents before a tax of 33 ⅓%

Seriously though judging by the challenge questions of late there is always a snippet of information missing or red herring thrown in that confuses the issue. Since the answer is usually meant to be unique and there is most likely a little fact like the two transactions are separate. If that is the case then the answer would be the 10 cent coin remains with one item cost 17 cents paid with the 14 and 3 cent coins and the other is 34 cents paid for with the 9, 12 and 13 cent coins.

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#101
In reply to #97

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

02/19/2007 4:40 AM

Masu - tax has already been covered: see post 86 & 87.

It all depends on whether tax is included in the ticket price or added to the ticket price at the "till".

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#103
In reply to #101

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

02/19/2007 10:37 AM

Hi English Rose,

I missed that post some how but in any case the 6% doesn't give an exact answer where the 33 ⅓% dose so I like my tax solution better. Anyway I only put the tax thing in as a joke, I don't really think it's part of the answer.

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#104
In reply to #103

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

02/19/2007 11:41 AM

ROFL...I was looking at it as trying to work out what would be on the missing bits of the jigsaw, or what colour table the jigsaw is being assembled on! Not serious, but part of the fun!

<tickle>

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