Adri,

Thanks for this -- I had got the word (which, unsurprisingly, wasn't in my [fairly old] Chambers 20th Century dictionary, by the way) and my first online search didn't seem to offer up anything related to the rest of the clue!

After a break, I reconsidered the second part of the clue looking for potential anagrams that didn't involve the word I'd found, so that all the letters had to be from that phrase. In that light, the phrase quickly yielded a food item -- success at last!

It happens that although I enjoy a lot of food types from around the world, you happened to hit on one I'm not really familiar with (apart from the kebab, which isn't a favourite :-( ...) -- thanks for another learning experience :-)

Cheers - Al.

P.S. A Happy New Year to all puzzlers!

Hi Adri,

Thank you very much for the full explanations. I enjoyed the puzzle, though I did not find it easy!

Happy New Year to you and all the other puzzlers!

Cheers,
Mark

Interesting puzzles. I did not have much trouble with parts A and B, but part C is proving to be a harder nut to crack. I will let it rest for a bit and then have another go at it.

Edit: I came up with a solution for C, but it is not as elegant as I would have liked.

I have a feeling there is more than one way to legitimately solve these, but once I have one solution, I am leaving it alone.

Cheers

Unless I have made some ridiculous arithmetic error I was bedeviled by the left hand side of the the third problem for a long time. The key for me was Adri's hint about raising numbers to a power. I spent a longer time than I should have on combining the first two terms of the left had side (that leading zero maybe having some significance on the second term) to come up with a number which is a power of 2. But I eventually abandoned that line of reasoning before it drove me around the bend. I thought it would lead to an elegant solution, but what I did come up with is awkward and cumbersome, but it works. That left hand part of the third problem was the stickler for me.
It will be interesting to see the solutions Adri gives us.

Cheers

P.S. In rereading what I wrote, I inadvertently wrote "combing" instead of "combining" which would bring up a rather comical mental picture.

Dear puzzle(d) friends, here are the solutions to post 700591.
Once you understood that the intended sum was 2024 + 1 = 2025, it shouldn't have been too difficult to solve at least one (part of a) puzzle (of course that's easy for me to say).

a.
I was playing around a bit with the numbers 2024 and 2025 around New Year and noticed some sort of puzzle based on the digits 4 and 5:

4×5, 4×6 => 20, 24
4×5, 5×5 => 20, 25
So I decided to rewrite this as roman numerals:
Ⅳ Ⅴ Ⅳ Ⅵ
Ⅳ Ⅴ Ⅴ Ⅴ
And there you have the first puzzle with minimal effort!

{Ⅳ}{Ⅴ}{Ⅳ}{Ⅵ} + 1 = {Ⅳ}{Ⅴ}{Ⅴ}{Ⅴ}
   4×5   4×6           4×5  5×5
     20 24              20 25

b. {10}{3}{2}{10} + 1 = {3}{2}{5}{5}{3}{2}
For this one, you needed to use addition, multiplication and raising to powers.
My favourite fact is that 2 to the tenth is 1024. Add one thousand and you have 2024.
So 1024 + 1000 = 2024, or, for this sum, 1000 + 1024 = 2024. Rewrite this as 10^3 + 2^10 using the element numbers {10}, {3}, {2} and {10}. That's the basis of the left hand part.
Now, I don't know if you know this: 45 squared is 2025 (so 45² = 2025).
There is a neat little UNIX/Linux program called 'factor'. It prints the prime factors of a specified integer, and so it does this for 2025 like so:

$ factor 2025
2025: 3 3 3 3 5 5

These are the prime factors of 2025. Rewrite this as 3² 3² 5 5. Change the order: 3² 5 5 3². Now write this as element numbers: {3}, {2}, {5}, {5}, {3}, {2} and there you have the right hand part.
3² × 5 × 5 × 3² = (9×5)×(5×9) = 45×45 = 45² = 2025.

Puzzle c.
Another fun fact: all numbers that end in a 5, raised to the second power, end in 25, i.e.: 5²=25, 15²=225, 25²=625, 35²=1225. Not only that, if you split the number that ends in a 5 in two (e.g. 2025 will become 202 and 5) and multiply the first part (202) by its successor (203), yielding 202×203=(200+2)×(200+3)=40000+400+600+6=41006, then append the second part (5) squared (i.e. 25), we will have 4100625. Too fast? OK, let's do this: according to puzzle b, 45² = 2025 (or to 'square' 45, split "45": take the 4, multiply 4 by its successor (5), that's 20, then take the 5, its square is 25, finally put 20 and 25 together and there you have 2025).
So, again, 45² = 2025. So (45²)² = 2025² = 4100625. Still following this? Now for an easy one:
So taking the square root of 2025² will yield 2025, of course, and it's the same as taking the square root of 4100625.
But wait! Taking the square root of a number is the same as raising a number to the power of 0.5.
So 4100625 to the power of ½ equals 2025.
And there we have the right hand part element numbers: {4 100 625} and {½}.
It doesn't take much to notice that {4 096 256} and {½} stand for the square root of 4096256 and that is 2024, since all three puzzles are 'the same'.

One more thing I'd like to note is that 4096256 consists of two parts: 4096 and 256, where 4096 = 2^12 and 256 = 16^2.

Thank you very much for participating!
Adri
PS I'd like to hear your comments, while I'm taking the train to a city nearby.