Comments on: Brain Game: Quick Change Artist

By: jeff

jeff — Thu, 26 Mar 2009 21:14:00 +0000

when .50 cent was a teenager… .20 cent baby!! just had to throw that in there

By: Ryan

Ryan — Thu, 26 Mar 2009 20:46:00 +0000

No, because the total number of coins is unknown, and grammatically speaking it is correct to use the plural there.

Plus, “all but two is a quarter” sounds bad and kind of ruins the puzzle. Maybe if this were Sesame Street and not Mental Floss… :)

By: WillieO

WillieO — Thu, 26 Mar 2009 19:21:00 +0000

I got the answer correct but take exception with the plurality of the words “quarters, nickles and pennies”. Doesn’t this imply that there was more than one quarter, one nickel and one penny?

By: Will

Will — Thu, 26 Mar 2009 18:28:00 +0000

Going to have to agree with Bill and Matt on this one; that was my thinking as well.

By: john

john — Thu, 26 Mar 2009 17:36:00 +0000

There’s a riddle like this that goes: if you have two coins and they total 30 cents, what are they if one of them is not a quarter. The answer is a nickel and a quarter. The nickel is the coin that is “not a a quarter.”

That said, I don’t think you can say that you have two dimes in your pocket and say that every coin in your pocket is a nickel except two of them (the two dimes). Saying that you have a group of something except part of it implies that you have something from the first group to begin with. Otherwise, there is no “exception.” The two dimes make up the entire contents of your pocket.

By: Matt

Matt — Thu, 26 Mar 2009 16:50:00 +0000

$0.20 is also a correct answer. If he has two dimes, it is perfectly correct to say that every coin in his pocket is a nickel except for two of them. Misleading, duh, but 100% true.

By: Sandy Wood

Sandy Wood — Thu, 26 Mar 2009 15:28:00 +0000

True… but saying “all but two of them were nickels” implies that there were nickels in his pocket, and so on with quarters and pennies.

I don’t know the actual theory behind it, but it seems like you couldn’t assume that other variables were involved if they weren’t mentioned in the problem. I mean, for that matter, he could have had foreign coins… but that’s just an assumption.

By: Eli

Eli — Thu, 26 Mar 2009 15:09:00 +0000

Tom you’re right with the equation, but you forgot some variables. The equation should be P + N + D + Q + H + $ = X

We know how many pennies nickels and quarters he has but we know nothing about dimes half-dollars or dollar coins, so we also have valid answers where P,N,Q = 0 and D+H+$=2

So the correct answers would be .31 (penny, nickel and quarter)
.20 (2 dimes)
.60 (dime and half-dollar)
1.00 (2 half-dollars)
1.10 (dime and dollar)
1.50 (half-dollar and dollar)
2.00 (2 dollars)

By: Tom in GA

Tom in GA — Thu, 26 Mar 2009 14:25:00 +0000

No, 31 cents is the only acceptable answer.
Here’s the “mathy” way of doing this.

Let x be the total number of coins he has. So N+Q+P=x.
Now since all but 2 of the coins are nickles, we can say that N=x-2. Similarly Q=x-2 and P=x-2.

Putting N, Q, and P into the first equation gives you (x-2)+(x-2)+(x-2)=x which solves for x=3. Thus there must be one of each coin.

Apropos reCaptcha: and publish

By: bill

bill — Thu, 26 Mar 2009 14:12:00 +0000

couldn’t he have also had 20 cents?