The worlds first riddle!

triangular-based pyramid

I've taken this as far as solving for A-E, but I end up with DE and CDE as primes.

A must be even, and B & C are 1, 3, 7, or 9. All but
A=4, B=3, C=1
A=6, B=1, C=7
violate the prime requirements.

Now, letting D & E be 1, 3, 7, or 9, all but
A=4, B=3, C=1, D=3, E=1
violate the prime requirements. However, DE=31 and CDE=131 violate the non-prime requirements.

I don't want the solution, but am I on the wrong track. Does "except" mean the numbers aren't prime or that they don't have to be prime?

I've assumed that leading zeros (A=0) aren't allowed. Is that a false assumption?

Let me state right off the bat; IMHO this is the most fiendishly difficult problem ever poised. The fact anyone has even attempted it is truly amazing! You have cracked the first part; except for E which I have =3.

Not only are you on the right track - you are cruising the freeway. Congratulations for stepping up to the plate.


For all of you who wish to follow Mark's progress in finding a solution to the seemingly impossible -

The following are all the possible prime numbers available to use:

2 digit primes
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

3 digit primes
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997

A cannot equal 0 and it is for this reason. DE must not be prime, but ADE must be prime. If A was 0, this would be impossible to satisfy. This is a good find because it now shortens our possible scenarios to just 3.

We now have ABC as either 419, 431, or 617


Scenario 1: ABC as 419
Where to start? How about with the 9 and looking at CH and CI which both must be prime. And the only two digit prime beginning with 9 is 97. Therefore, H and I are both 7. Making HI = 77, which is not a prime, and so this scenario will never work.

Scenario 2: ABC as 617
For this one I started on a point that produced the least number of possibilities, so I went for AD which could only be 61 or 67. And this meant that BCD must be 171 or 177. Since neither of these numbers is prime, this scenario will never work.

Scenario 3: ABC as 431
BCD can be 311, 313, 317. Therefore, D is 1, 3 or 7
AE can be 41, 43, 47. Therefore, E is 1, 3 or 7

So our choices for ADE are now 411, 413, 417, 431, 433, 437, 471, 473, 477. 431 and 433 are the only primes. This means that D = 3 and E can now only equal 1 or 3. So DE can be 31 or 33.
31 is a prime number, and DE must not be prime, therefore E = 3

We now have definite values for ABCDE which are 43133


Good luck, we are all pulling for ya.




NOVERCEDINCTT


OUFDT

Is this one finding out?
I know I am easily amused

NOVERCEDINCTT

innocent verdict? (Maybe?) :wink:

Hi mismi40
Gee, I must of been really tired this morning since I lost my first post that I posted here
Good thing it wasn't something really important

PYRAMID
[size=7]My problem was that my Excel table was missing half the possibilities for ABCDE after figuring out the possibilities for ABC.

Q is the position.
Thoh had a 1.
Mark had a 7.

Good problem![/size]

The chief function of your body is to carry your brain around. Which is why Noah didn't wait for his ship to come in...he built one! After all; success consists of getting up just one more time than you've fallen down.


Mark:

PYRAMID
"My problem was that my Excel table was missing half the possibilities for ABCDE after figuring out the possibilities for ABC."

Q is the position.
Thoh had a 1.
Mark had a 7.

Good problem!


Fantastic result!

How he does it using Excel is a wonder to me. I wonder if Microsoft knows how useful it is. However, for the rest of us who have never even looked at that programme….

ADJ can be 431, 433, 439, meaning J can be 1, 3 or 9
DJ can then be either 31, 33 or 39. 33 and 39 are not prime, therefore DJ is 31 and so J = 1

AEL can be 431, 433, 439 making L be 1, 3 or 9
EL = 31, 33 or 39. As we know from above 31 is the only prime, therefore EL = 31 and so L = 1

JK can be 11, 13, 17 or 19, so JKL can be 111, 131, 171 or 191. 111 and 171 are not prime, therefore K can be 3 or 9. So KL can be 31 or 91. 91 is not prime, so KL = 31 and K = 3

BF can be 31 or 37, so F is 1 or 7
BG can be 31 or 37, so G is 1 or 7
BFG is either 311, 317, 371 or 377. 371 and 377 are not prime, therefore F = 1

BG can be either 31 or 37, so G is either a 1 or 7. Seeing as the last 6 letters we have found have been a 1 or a 3, I'm going to first try with G as 1.

The following is with G = 1
FGH is 113, therefore H = 3
HI can be 31 or 37, so I can be 1 or 7. HIJ can be 311 or 371, and 371 is not a prime, therefore I = 1

CIR can only be 113, so R = 3
QR can be 13, 53 or 73, so Q can be 1, 5 or 7
CHP can be 131, 137 or 139, so P can be 1, 7 or 9.

From this we can write down all possible PQR values which are: 113, 153, 173, 713, 753, 773, 913, 953, 973. The only possible primes from this are 113, 173, 773 and 953. Looking at the first two digits of each, we know they must also be prime, and 77 and 95 are not prime. Therefore, PQR can only equal 113 or 173 and so P = 1

Taking a look at what we have now:

A B C
4 3 1
D E F G H I
3 3 1 1 3 1
J K L M N O P Q R
1 3 1 - - - 1 - 3

BGO can be 311, 313 or 317, so O can be 1, 3 or 7
If O is 1, then OPQ must be 113, which isn't possible as we know from above that Q can not equal 3. Therefore O is 3 or 7. If O is 7, then OPQ can be 711 or 715 or 717, of which none are prime. Therefore O = 3

NO can be 13, 53 or 73, so N can be 1, 5 or 7
BFM canbe 311, 313 or 317, so M can be 1, 3 or 7
MN can be 11, 15, 17, 31, 35, 37, 71, 75 or 77. Out of these only 11, 17, 31, 37 and 71 are primes.
So LMN could be 111, 117, 131, 137 or 171. 131 and 137 are the only primes here, therefore M = 3

MN can only be 31 or 37, so N can be 1 or 7. Therefore, NOP can be 131 or 731. 731 is not prime, therefore N = 1

So we are left with Q which should hopefully take on the value of two different numbers. PQ can be 11, 13, 17 or 19, so Q can be 1, 3, 7 or 9. This makes PQR equal either 113, 133, 173 or 193. 133 is not prime. This leaves us with these values for QR: 13, 73 or 93. 93 is not a prime, therefore the only values possible for Q are 1 or 7, so Q = 1 or Q = 7

The finished article looks like this:

A B C
4 3 1
D E F G H I
3 3 1 1 3 1
J K L M N O P Q R
1 3 1 3 1 3 1 (1 or 7) 3

Just as Mark foretold!




TTH:

OUFDT
Is this one finding out?

Yup! F(IN)D OUT - particularly difficult I thought.


Missmi:

NOVERCEDINCTT = innocent verdict

Certainly in her case; and do get better soon.



KSMILOKLGS


ROREIGCEAIPLT

smoking kills (that is why I quit) Need more time on this one

ROREIGCEAIPLT
[size=7]original receipt[/size]

ROREIGCEAIPLT = original receipt (I might have looked at the fine print mark :wink:

I have strep for sure - dr. verified - happy thing about that is in 24 hours I will be feeling better now that I have a z-pak AND he has said more rest - too busy apparently. My defenses are down and it is the cold season and I am up at the school around all the cute little snotty children a lot! Thanks Try

You all have a great weekend! Hope to see you all soon!

Thanks mismi40, take care of yourself & see ya soon

TTH:

KSMILOKLGS = smoking kills

It said so on the packet!


Mark & Missmi:

ROREIGCEAIPLT = original receipt

You will need it for returns!




Four A2K'ers Andy, Bandy, Candy and Dandy were talking about where they lived. They each wrote down their two-figure house number and then showed me the four different numbers without telling me which was which.


It turned out that they were all perfect squares. Andy said, "Mine uses the digit 1." Bandy said, "Mine doesn't use the digit 4." Candy said, "Mine's even." Dandy said, "The first digit of mine is higher than the second."

It was still impossible for me to work out which was which. Even I had known the house number of the fair-haired poster I would still not have been able to work out which was which of the other numbers.

Alternatively, even if I had known the house number of the brown-haired member I would still not have been able to work out which was which of the other numbers. However, the ginger-headed one told me his house number and I was then able to work out all their numbers.

And now if I were tell you which of the four people was ginger-headed then it would be possible for you too to work out their house numbers.

What (in the order A, B, C, D) are their house numbers

Tryagain...

I'll tried it!

If the ginger is Bandy and she (he?) tells you that his/her house is #16, you can figure out that

Andy = #81
Bandy = #16
Candy = #36
Dandy = #64

EDIT:
Alternatively, if Bandy told you that his/her house was #81, you could figure out that

Andy = #16
Bandy = #81
Candy = #36
Dandy = #64

Best I could come up with, I'm still trying other scenarios now!

I was laughing too hard to read further

I found your Christmas tree

Try's tree is really pretty TTH...thanks for the well wishes too...hope you have a great weekend...
mis

Hi mismi40
Hope you are feeling better. The tree looks kinda lonely though
I don't know how to add a present with it.

TehMeh:

I'll tried it!

If the ginger is Bandy and she (he?) tells you that his/her house is #16, you can figure out that

Andy = #81
Bandy = #16
Candy = #36
Dandy = #64

"That's one small step for man, one giant leap for TehMeh"


EDIT:
Alternatively, if Bandy told you that his/her house was #81, you could figure out that

Andy = #16
Bandy = #81
Candy = #36
Dandy = #64

True!



First off, write down the possible two digit squares:
16, 25, 36, 49, 64, 81

Now write down the possible squares for each member:
Andy = 16, 81
Bandy = 16, 25, 36, 81
Candy = 16, 36, 64
Dandy = 64, 81

Now just work your way through the numbers, until you find a number to start on that allows you to work it all out.

So start with saying Andy is 16, from this it is not possible to get any more house numbers. Try with Andy is 81. This makes Dandy 64. But that's it. No more house numbers possible.

Next try Bandy is 16. This makes Andy 81, which then makes Dandy 64. And this then makes Candy 36.

Answer is:
A = 81, B = 16, C = 36, D = 64



I love the tree TTH, so much so that I am gonna buy a set of blue lights for this year.

"I don't know how to add a present with it."

It is better to give than to receive; so I'm gonna give the receiver one!



PWADONWE

TPREEE

window pane
pine tree

Haha, "I'll tried it!"
Stupid typing mistake that I often seem to make!

Can't seem to figure out your word riddles though...

TTH:

PWADONWE = window pane
TPREEE = pine tree

With nice blue lights!


TehMeh writes, "Can't seem to figure out your word riddles though..."

That makes two of us; unless I keep a careful note, I can seldom work them out.



I was a sitting on the front porch whittling me a grid, then I put one digit in each square of the grid, so that I can read eight different four-figure numbers across the rows or down the columns, four of them being odd and four even.

One of the across numbers is a cube and the other a fourth power, and the same is true of the numbers down.

Two of the other numbers are squares.

Which two numbers in the grid are not perfect powers


Just draw a 4 x 4 grid to save time.