Logic Puzzle Forums can anyone explain the logic please?
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#1
12-16-2013, 02:09 PM
 stanstar Senior Member Join Date: May 2012 Location: Leith, Edinburgh, Scotland at present UK Posts: 127
can anyone explain the logic please?

I was stuck and asked for a hint and it read: CLUE #8: If Avery does not equal 129, and Art and Pre-Calculus can only be paired with either Avery or 114, then Art cannot be equal to 129. Mark the highlighted cell as FALSE.

To my way of thinking that is not logical. Why is Art not equal to 129?

And now the next hint;
CLUE #8: If Avery does not equal 129, and Art and Pre-Calculus can only be paired with either Avery or 114, then Pre-Calculus cannot be equal to 129. Mark the highlighted cell as FALSE

Similarly why is pre Calculus not equal to 129

Stan

Last edited by stanstar; 12-16-2013 at 02:15 PM.
#2
12-16-2013, 02:18 PM

Hi Stan -

Remember that every item in every category must be used once and only once for each of the people/places/things in the puzzle. Therefore Art must have a room #, and Pre-Calculus must have a room #. Clue #8 says that one of those two (either Art or Pre-Calc) is in room #114, and the other is Avery (who you already know isn't room #129). So any way you slice it, Art is either in room #114 or in Avery's room (which definitely isn't 129). The same also goes to Pre-Calc. It is impossible for either Art or Pre-Calc to be in room #129.

When you see a clue like that - i.e. Of A or B, one is C and the other is D - always be on the lookout for data already on the grid that tells you what C can or cannot be in relation to D's category, and vice versa. You can often swap this data into the original clue to make additional TRUE/FALSE conclusions.
#3
12-16-2013, 02:37 PM
 stanstar Senior Member Join Date: May 2012 Location: Leith, Edinburgh, Scotland at present UK Posts: 127

Thank you Sir, the way you describe it is much easier to understand. I get the drift now.

Stan
#4
12-16-2013, 04:41 PM
 stanstar Senior Member Join Date: May 2012 Location: Leith, Edinburgh, Scotland at present UK Posts: 127

can the administrator that helped me with the one above help me again please. This hint has me stumped too.

CLUE #3: If Greg does not equal rare book, and Oakdale and \$7.75 can only be paired with either Tea Set or Greg, then Oakdale cannot be equal to rare book. Mark the highlighted cell as FALSE.

And next hint: CLUE #3: If \$7.75 does not equal Cherryfield, and tea set and Greg can only be paired with either Oakdale or \$7.75, then tea set cannot be equal to Cherryfield. Mark the highlighted cell as FALSE. (+120
I cannot understand this logic.

Stan

PS When you explained it, it was easy to follow.

Last edited by stanstar; 12-16-2013 at 04:45 PM.
#5
12-16-2013, 04:56 PM

Hi Stan -

This follows the very same logic from your first post at the top of the thread.

"Rare book" and "Tea Set" are both in the same category (not sure what that category is, as I'm going from memory, but let's say they are "Gifts" for the sake of this explanation).

Clue #3 says that Oakdale is either Greg's gift or the tea set, and \$7.75 is either Greg's gift or the teaset. If you know that Greg's gift isn't the rare book, then you know Oakdale isn't the rare book, and \$7.75 isn't the rare book.

Here's another way to look at it - the two logic statements we're using are (color coded by category):

1. Of Oakdale and \$7.75, one is the tea set and the other is Greg.

2. Greg does not equal the rare book.

Take #2 and use it to replace "Greg" in #1, like so:

-- Of Oakdale and \$7.75, one is the tea set and the other does not equal the rare book.

We inherently know, because tea set and rare book are different items in the same category, that:

3. Tea set is not equal to the rare book.

... so now let's replace "tea set" in our modified Clue #3:

-- Of Oakdale and \$7.75, one does not equal the rare book (originally: Tea Set) and the other does not equal the rare book (originally: Greg).

Now the logic should be clear - none of the two options given for Oakdale and \$7.75 (i.e. Tea Set and Greg) can be equal to rare book. Therefore, since Oakdale must be either "tea set" or "Greg", and \$7.75 must be either "Tea Set or "Greg", Oakdale does not equal rare book, and \$7.75 does not equal rare book.
#6
12-16-2013, 05:02 PM
 stanstar Senior Member Join Date: May 2012 Location: Leith, Edinburgh, Scotland at present UK Posts: 127

Thank you Sir again! I will read through your explanation and try to understand. You are so helpful and thank you.

Stan
#7
12-16-2013, 08:34 PM
 BrightSky9 Junior Member Join Date: Apr 2011 Posts: 4
Does this call for an EASY MODERATE?

I don't know if Stanstar agrees, but I need an EASY MODERATE. I'm bored in EASY and flummoxed in MODERATE by the same circumstances and hints as Stan.
#8
12-19-2013, 04:51 PM
 stanstar Senior Member Join Date: May 2012 Location: Leith, Edinburgh, Scotland at present UK Posts: 127

Yes I agree. I find the challenging ones a wee bit too difficult. An in between one after medium and before challenging would certainly be better for me. Even the hints on challenging challenge me somewhat.

Stan
#9
12-20-2013, 04:50 AM
 Leftylucy1115 Junior Member Join Date: Dec 2010 Posts: 5
Another Strategy

So, once i have read through the clues a couple of times and have a lot of squares filled in, I look at the puzzle.

Example:

Let's say I have filled in a good amount of squares to the point that I am stuck. I will look for one attribute that has two empty squares- call it A. Then look across the board (from left to right in the top part of the grid only) to find another attribute (call it B) that has those same two spaces marked as false. Which logically means A cannot be B.

I'll explain...

Into the puzzle at this point, I am confident and can say it is correct. So, if Paula is either \$1750 or \$1780, I will use her as Attribute A. Scroll across the board looking for any attribute that is not \$1750 or \$1780, and calling it B.

Since I know A is one of two things (and nothing else). Any attribute that is not true for both of the possibilities, will never be true for A.

To check yourself, imagine Paula being \$1750. If she was, would she be equal to B? Imagine her as \$1780, would she be equal to B? No! Because B will never be \$1750 or \$1780 no matter what.

It works as long as the puzzle is correct up to this point!

This helps narrow down choices left. It can also add a" false" and leave only one empty square in a row or column (Yay!). And it is often a deduction that is not written in the clues.

?? Make sense?

Hope this helps!

P.S. The tricky part is when Attribute A has empty squares that are not right on top of each other........

Last edited by Leftylucy1115; 12-20-2013 at 05:09 AM.
#10
12-20-2013, 09:10 AM
 Snoopy Junior Member Join Date: Dec 2013 Posts: 2

"Clue #8 says that one of those two (either Art or Pre-Calc) is in room #114, and the other is Avery (who you already know isn't room #129). So any way you slice it, Art is either in room #114 or in Avery's room (which definitely isn't 129)."

I finally got it when I started thinking of it as : Avery is not #129, AND #114 is not #129... (so BOTH art AND pre-calc can't be #129...

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