Puzzle Wizards? - DPChallenge Forums

Olympus OM-D E-M1 Mark III

03/30/2007 08:44:33 AM · #26

I kinda doubt it :)

I did try the 737 and 730 combinations, but I wasn't able to find anything at those spots. I think it'll be more straight-forward, without the ambiguity of that solution.

Originally posted by Lowcivicman99:

I would say that it might be something like this;

1561616616546516506894651651/98156098198198/198164.5465434631891/6816873
743541819864/735198468/465765468498/1984/86435/465/484/85/465/4654818916
8168435753418168498465489741685165410897489/746510894894/896413213132498
/6410321984956/4986489/56419+84198498//419865419841986510/98641981984535
/7654984108/08765423/46574654/1

Does that make any sense???

HAHAHA Just kidding, I just really needed to do that, I'm frustrated this morning because my kids are driving me insane.

colorcarnival

Olympus 45mm f/1.8 M.Zuiko Digital

03/30/2007 08:46:03 AM · #27

Just to get my brain wrapped around this, what did you do to solve the line of Text that was before this grid? Is the text related to these three digits?

dacrazyrn

Nikon D300

Nikon AF-S DX NIKKOR 18-200mm f/3.5-5.6G ED VR II

03/30/2007 08:56:30 AM · #28

Is it a UTM coordinate? Easting and Northing?
A certain point Kilometers N and E from a certain point?

03/30/2007 09:01:40 AM · #29

No, it'll be in Lat/Lon hdddÂ° mm.mmm' format.

I have the longitude, and this much of the latitude, but I still need 40 34.xxx

Originally posted by dacrazyrn:

Is it a UTM coordinate? Easting and Northing?
A certain point Kilometers N and E from a certain point?

03/30/2007 09:01:47 AM · #30

I hope this right

3 5 1 5 3 / 1 4 1 5 / 5 5 2 4

4 1 5 1 3 / 2 2 5 1 / 5 1 4 3

I'm not sure that I'm approaching this right but here goes

If this is a mathematical puzzle then it seems very vague - almost too vague to solve since there are a million mathematical opeartions that you could apply, each giving equally plausible results and no good reason to choose any one above another

I am struck by the fact that the first part of the grid has 2x5 digits while the rest have 2x4 - this seems to make a mkaths code unlikely

But if it is a straight forward letter code, given that 5 with 1 below appears twice in the first block then that block must represent seven, and 5 with one below is the letter e

this gives

? e ? e ? / ? ? ? e / ? e ? ?

this makes seven/five/zero a strong candidate and the presence of 1 over 5 in the first and second blocks in the correct position for a v seems to confirm this - so 750

let us know what you find !

Message edited by author 2007-03-30 09:04:40.

03/30/2007 09:03:14 AM · #31

Originally posted by colorcarnival:

Just to get my brain wrapped around this, what did you do to solve the line of Text that was before this grid? Is the text related to these three digits?

Each of the other parts were different types of puzzles. We had to figure out a puzzle for each of the other parts of it. That's why I didn't initially even mention the rest of it... all we're focusing on is this part of it.

03/30/2007 09:06:30 AM · #32

Originally posted by thelobster:

I hope this right

Honestly, that still sounds a little too "out-of-the-box" to me. I dunno...

The rest of the puzzles were such that once you recognized what they were, you pretty much KNEW they were right. We had solved everything else, and I went out and found one of the hidden caches (with yet more clues in it) yesterday. But our first sets of numbers were right on the money, and we were pretty much 100% confident in that before I went out traipsing through the woods :)

Message edited by author 2007-03-30 09:08:30.

03/30/2007 09:06:36 AM · #33

just bumping because our posts overlapped and you may not notice

03/30/2007 09:09:18 AM · #34

Originally posted by thelobster:

just bumping because our posts overlapped and you may not notice

(Note my post before yours)

03/30/2007 09:19:13 AM · #35

I think I can justify it - but with one small guess.

Write out the alphabet in order. Becsause this is based on 5s and there are 26 letters you have to leave one out. I left out K but it doesn't really matter provided the letter you leave out is between J and M - maybe the compiler ran I and J together.

Then write numbers above the letters. Start by putting 5 1s above the first 5 - then 5 2's

The 2s begin at F, threes at L, 4s at Q (having left out K or J or L or M), 5s at V

This gives you the numbers from the bottom row of your grid

Then number them 1, 2, 3, 4 ,5 , repeating that for each block of 5

E comes out as 5 over 1 - the fifth one of the 1s

Z comes out as 5 over 5 - the fifth and last of the 5's etc -

The other letters all work too

1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5 5 5 5

a b c d e f g h i j l m n o p q r s t u v w x y z

(When I wrote this they all came out under each other but the justification of the text has spoiled the arrangement)

Message edited by author 2007-03-30 22:26:13.

Olympus OM-D E-M1 Mark III

03/30/2007 09:35:29 AM · #36

I just realised how long winded that was. It's a grid. A 5x 5 grid with a - z in it and numbersed 1 to 5 on the horizontal and vertical.

I'll put hashes in to try and keep the spacing

####1#2#3#4#5
##1#a#b#c#d#e
##2#f#g#h#i#j
##3#l#m#n#o#p
##4#q#r#s#t#u
##5#v#w#x#y#z

Thje code is now just the co-ordinates of the letters in the grid - unless I've miscalculated badly all the letters work and I'd say this has to be right (probably).

Message edited by author 2007-03-30 10:00:24.

purpleflutterby13

Canon EOS-5D Mark III

Canon EF 50mm f/1.4 USM

03/30/2007 09:51:17 AM · #37

This isn't going to be right, because it's too obscure, but...

Assuming that the numbers in the grid are actually written in base 5 rather than base 10, you start with the numbers:

31513 / 1415 / 5524
41513 / 2251 / 5143

And when you convert them to base 10, you end up with the numbers:

2553 / 235 / 764
2758 / 330 / 673

You subtract the top from the bottom, and end up with:
205+95-91 = 209.
Which would be my best guess.

The problem with a mathematical solution is that there's thousands of them you can come up with. (why not multiply, then subtract, etc.)

colorcarnival

Olympus 45mm f/1.8 M.Zuiko Digital

03/30/2007 10:06:13 AM · #38

anything at 932 or 931?

edit to add: probably not since i just realized i did something wrong lol

Message edited by author 2007-03-30 10:08:45.

03/30/2007 10:10:06 AM · #39

Originally posted by thelobster:

I think I can justify it - but with one small guess.

So, long story short, you're saying 750? :) If you're confident of that, I'll run out and take a look to see if I find anything there...

Nikon AF Nikkor 50mm f/1.8D

03/30/2007 10:11:11 AM · #40

Originally posted by purpleflutterby13:

You subtract the top from the bottom, and end up with:
205+95-91 = 209.
Which would be my best guess.

It's going to have to be a number higher than 500, I'm afraid. Anything lower than 500-ish put me into a residential area next to the park :)

Message edited by author 2007-03-30 10:11:24.

David.C

03/30/2007 10:41:25 AM · #41

Originally posted by alanfreed:

That is a definite possibility. My wife had mentioned something like that at one point. It seems out of place that there would be two non-6 digits in that last string of four, though.

Originally posted by Southern Gentleman:

Since I have nothing to go on I guess. If you put each number together (top to bottom) all equal 6 except for the following:

3 5 1 5 3 / 1 4 1 5 / 5 5 2 4

4 1 5 1 3 / 2 2 5 1 / 5 1 4 3
-----------------------------
7 6 6 6 6 / 3 6 6 6 / 0 6 6 7

So my guess would be 7/3/7 or 7/3/0

does that make sense?

The number of non-6 numbers in the last set throws that off -- but not if the number of 6s is the result, but you will have to be the judge if 432 is close enough to 500ish for your purposes. But it does have the simplicity of solution these types of things usually have.

David

Canon EF-S 18-55mm f/3.5-5.6 USM

03/30/2007 11:38:11 AM · #42

I do really think it's 750 - I'm sure enough that if it was me I'd go and look - but I have been known to be wrong so I won't stake my life on it

rox_rox

Canon EOS-20D

03/30/2007 12:38:54 PM · #43

What are the chances of it being 734?

Gordon

RIM BlackBerry Curve

Canon EF 85mm f/1.8 USM

03/30/2007 01:42:36 PM · #44

Originally posted by alanfreed:

Originally posted by thelobster:
I hope this right

Honestly, that still sounds a little too "out-of-the-box" to me. I dunno...

The rest of the puzzles were such that once you recognized what they were, you pretty much KNEW they were right. We had solved everything else, and I went out and found one of the hidden caches (with yet more clues in it) yesterday.

Perhaps the methods used to solve the other clues would be useful in understanding the remaining problems ?

colorcarnival

Olympus OM-D E-M1 Mark III

Olympus 45mm f/1.8 M.Zuiko Digital

03/30/2007 01:45:07 PM · #45

Originally posted by thelobster:

I do really think it's 750 - I'm sure enough that if it was me I'd go and look - but I have been known to be wrong so I won't stake my life on it

I used a different way out there method and got 745 lol. Kind of close but probably not close enough.

DrAchoo

Panasonic Lumix DMC-LX5

Canon EF 24-105mm f/4.0L IS

03/30/2007 02:00:58 PM · #46

I think lobster is onto the answer...

chesire

Canon EOS-60D

Canon EF 24-70mm f/2.8L USM

03/30/2007 02:09:13 PM · #47

Originally posted by purpleflutterby13:

Oh man I'm gonna have to go make my caches more difficult to find!!!!!

This would work :)

Sigma 24-70mm f/2.8 EX Aspherical DF for Nikon

03/30/2007 02:20:50 PM · #48

As soon as I get my daughter from school, we'll head out to check it using the 750 (and 745) coordinates. The difference between 750 and 745 is really only a matter of feet, by the way... so we'll be checking that area :)

Originally posted by Gordon:

Perhaps the methods used to solve the other clues would be useful in understanding the remaining problems ?

Nope, not really. Each puzzle had its own different methodology.

Apple iPhone

03/30/2007 03:06:01 PM · #49

..

Message edited by author 2007-03-30 16:09:53.