Pink Poogle Toy Forum

The current prize for the first 250 correct entries is a Remnok the Nomad. You will also get a cool trophy that people can see when they do a user lookup on you! And all correct entries will split the 2,000,000 Neopoint prize pool.

I just figured it out. All on my own too. And, I don't mean to gloat. It's just as soon as I saw it, it put a bug up my behind. And so I knew I wouldn't rest until I solved it. I doubt if I am in the first 250 but still and all, it felt good to figure it out without any clues or help from anyone.

Same here, seemed something that I just WANTED to solve.... And I did without even computer help (except a calculator like once) If I win this will be my second time! =)

YAY -I solved this one for once

Can someone plz post ther answer here since you already know it!

We do not post answers here as that is unfair to all of the players who have solved the puzzle on their own. People can provide hints to help others, as long as those hints are not too revealing.

I actually figured this one out. Yay!

Here is a tip to help you:

Since you're multiplying your original set of numbers by 4, you know the digit in the ten thousand spot has to be a 2 or less. If you have a 3 in there... answer would be too large.

These kinds of puzzles are possible to figure out. It just takes some thinking. If you guess and check, that might work; although it is time consuming.

Yay, I think I got it right!

Awesome. This was actually pretty easy:

NOMAD x4 = DAMON

We know N has to be an even number, because DAMON is a multiple of 4. Because a greater number would cause a six-digit result, the only things N can be are 1 and 2, so N = 2.

From there, we know that 3x4 and 8x4 are the only ways to get a number ending in 2. Since 2x4 is 8, D = 8.

We also know O must be less than 2, or else it would carry over to the next place, making Nx4=9. Since 2 is already taken, O must be 1 or 0. However, since Dx4 = 32, we know that Ax4+3 = 0. Since Ax4 must be even, Ax4+3 must be odd, so the only possible value for O = 1.

Back to A. Ax4+3 = something ending in 8, because 8+3 = 11. The only two choices are 2 and 7. However, Mx4 cannot equal M unless it is added to by carrying. That means A would have to be greater than two, so A = 7.

So far, we have this: 21M78 x4 = 87M12. M is a place value of 100, so we can solve it with linear algebra.
(21078 + 100M) x4 = 87012 + 100M
84312 + 400M = 87012 + 100M
300M = 2700
M = 9

Put it together, and you get DAMON = 87912.