NeoPets Riddles (Lenny Conundrums) and Answers Here

I'm under the impression it's not base 10

what do you mean, it is not base 10?
You mean, each letter (E & O) does not represent a number below 10?
Like E could be 12 or 14 etc, and O could be 11 or 13 etc?

But doesn't that mean that there are even more possible solutions?

I'll try it out to see if that works.

Based on the odd and even indications, you can tell whether each column of the sum requires carrying a 1 to the next column. This quickly leads to the position of the 1 and 0 in the problem. Then work the right-most column to figure out the bottom value and the possible values for the top two numbers. After that, you can do similarly with the other two columns and then the sum to 9 to figure out which order the values belong in the columns.

"you can tell whether each column of the sum requires carrying a 1 to the next column. "
^
I manage to do this, but for some reason I'm not able to figure out the 0. Thanks for the hint/explanations, I'll look though my workings again, I might have made a mistake somewhere because one of my Es is an odd number.

Also, there's no way for all Es to be the same number and all Os be the same number.
I quickly realized there were multiple possibilities then (603+433=1047, 441+611=1052)
So I looked at the dates and decided maybe we can't use any of the numbers given, but that won't work because you know the OEOE must be 1EOE, so I was thinking maybe we just can't use 8 (because the 12 and 16 are not digits) But there's still multiple answers.

So I'm trying to find some clue with the foreman or Y8 or something :/

@lenny fan
"This quickly leads to the position of the 1 and 0 in the problem."
I'm not sure what you mean by that.

@lennyfan,
"you can tell whether each column of the sum requires carrying a 1 to the next column. "

Nevermind, I figured it out, answer submitted, good luck everyone.

Oh sorry, the normal current system used is base 10, that is, 0-9 before moving to the next digit, but don't worry, I was incorrect in assuming it wasn't.

Here is the only Answer I can come up with.
Dates are 12, and 16...
E + E + O = 9 (6+2+1=9)
E + O + O (6+1+1 = 8)
__________
(621+)+(611)=1O2E3O2E

Answer was given using ONLY answers from the dates (12, 16 and 8)

i noticed that there are 5 O's and 5 E's, does this mean that each digit must be different, therefore including every number from 0-9 in the equation?

because if so, im 99.9% sure i just submitted the correct answer. i wonder if ill get a gold trophy!
good luck everyone

If so (although the problem doesn't indicate that), the answer is 1098.

423+675=1098

So... I know this might be far fetched, but I think I found the right answer by total accident.

1612. I added 801+811 (EEO+EOO)=OEOE - 1612. And I looked at the dates and years and realized... both year 8 (so 801/811) and dates 16 and 12 literally. It was just my first guess. But doesn't it seem very possible? Anyway... hoping I get a gold!

Anybody think this could be right?

The current prize for the first 250 correct entries is a Chocolate Doughnut Hole and a trophy.
All correct entries will split the 4,000,000 Neopoint prize pool.
THIS WEEKS PUZZLE - Ends: probably on Wednesday afternoon

As I was going to Meridell
I met a Draik with five Spardels
Every Spardel had five Pinchits
Every Pinchit had five meatballs
Every meatball had five spices
Spices, meatballs, Pinchits, Spardels.
How many were going to Meridell?

Please submit the answer in standard word form.

This one is a classic:

As I was going to St. Ives
I met a man with seven wives
Every wife had seven sacks
Every sack had seven cats
Every cat had seven kits
Kits, cats, sacks, and wives
How many were going to St. Ives?

he current prize for the first 250 correct entries is a 101 Recipes For Courgettes and a trophy.
All correct entries will split the 4,000,000 Neopoint prize pool.
THIS WEEKS PUZZLE - Ends: probably on Wednesday afternoon

BN: A, B, C, D, E

A = 0100010101
B = UNKNOWN
C = UNKNOWN
D = UNKNOWN
E = UNKNOWN
A XOR C XOR D = 1000011010;
A XOR C XOR E = 0000101010;
A XOR B XOR C XOR D = 0011101011

B XOR C XOR E = ???
Please fill in the correct answer for "???".

1111001110

I get that it's about exclusive or gates, but what did you guys do about the tenth term? If A XOR C XOR D is 0, that means at least one of either C or D is 1. This contradicts A XOR B XOR C XOR D being 1, since there are at least two 1s, which would make it a 0.

Or am I wrong in assuming that TNT uses correct digital logic notation, and instead one or more of the ports is added at a different time? Should it really be written (A XOR B) XOR (C XOR D)?

With regard to the tenth bit (term):
A = 1 (given)
Since A^C^D = 0, C or D (but not both) is 1.
Since A^B^C^D = 1, B must be 1. That results in three 1s and one 0.

"...since there are at least two 1s, which would make it a 0."
The result is 0 when there are an even number of 1s. Three ones would make it a 1.

xor is commutative and associative, so you can move or remove your parentheses as you like. You can also rearrange the terms any way you like.

It's binary/ You already know that.
two (0)s make a (0), two (1)s make a (0), but one (1) and one (0) make a (1)
so...
ABCD-ACD=B
ABCD (0011101011) - ACD (1000011010) = B (1011110001)
ACE-A= CE
ACE(0000101010) - A(0100010101) = CE (0100111111)
now we have B and CE. Add them together
B+CE=BCE
B(1011110001) + CE(0100111111) = BCE(1111001110)
Answer is (1111001110)