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I saw a quite vague math riddle like this - given two numbers a = 78404343 and b = 57, derive the third number c.

I think the solution to that might be this:

78404343 + 57 = 78404400
reverse
00440487

440487 + 13 = 440500
reverse
005044

5044 + 56 = 5100
reverse
0015

15 + 85 = 100

c = 57135685

I think it's a quite neat trick because when looking at a and c it is not obvious (not to me at least ) how these numbers are related.

My question is - is there a name for this trick with numbers? I wanted to read more about it, but don't know how to google it .

Also I could have avoided reversing and do dividing by hundred instead or perhaps increate the digits of added numbers each time (not sure which is the right way to do it).

Andris Zalitis
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  • This is not clear. What has $c$ got to do with $a,b$? If all you want is some operation on pairs of natural numbers, there are a lot of them. – lulu Feb 03 '20 at 18:40
  • @lulu: I am guessing that this is a stage magic thing, so you are given the first number and are supposed to predict the last one, which is the concatenation of all the two digit numbers that came before. I am not sure my approach is any easier to carry out mentally. – Ross Millikan Feb 03 '20 at 18:43
  • lychrel numbers ? –  Feb 03 '20 at 18:57
  • @lulu yes, it's not clear. However, the fact that summing a and b gives a number with two trailing zeroes gave me this idea. I think it's a pointer to a procedure like this instead of for example multiplying or dividing. – Andris Zalitis Feb 03 '20 at 18:57
  • @RossMillikan I must be missing something simple. Given $78404343, 57$, in what sense is "$57135685$" the "concatenation of all the two digit numbers that came before"? – lulu Feb 03 '20 at 19:00
  • the 57 13 56 and 85 from above concatenate to it @lulu to be fair I'm not sure I get the point ... –  Feb 03 '20 at 19:03
  • @lulu: it is the concatenation of the numbers you added, in order. The first number added was $57$, then $13$, then $56$, and finally $85$. – Ross Millikan Feb 03 '20 at 19:03

  • lychrel numbers ?

    @RoddyMacPhee interesting, somewhat related topic, thanks!

     – Andris Zalitis Feb 03 '20 at 19:06
  • @RossMillikan Ah, and at each point the number you add makes the last two digits $0$. Got it. Seems a bit arbitrary, but it does make sense. Thanks. – lulu Feb 03 '20 at 19:06

1 Answers1

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I don't have a name for it, but can explain how it works. I think it is stage magic where you describe the computation in your question, then are given $a$ and are asked to predict the final number mentally. You should only be given $a$, not $b$ as well. I do a lot of mental calculation, but this seems like it would be difficult to me because of the reordering that takes place. $a$ should have a number of digits divisible by $4$ or we have to rethink what happens at the end.

Break $a$ into two digit pieces $$78\ 40\ 43\ 43$$ Now reorder them taking the last, the first, the next to last, the next to first, and so on (if there were more digits). When you take from the front of the number, reverse the digits $$43\ 87\ 43\ 04$$ Each one after the second gets a carry in, so increase it by $1$. The last one, which was just in front of the midpoint, gets a carry in from both sides so increase it by $11$ $$43\ 87\ 44\ 15$$ Finally subtract each pair from $100$ and report the result $$57135685$$

As a longer example, take $$12\ 34\ 56\ 78\ 75\ 31\\ 31\ 21\ 75\ 43\ 78\ 65\\ 31\ 21\ 76\ 44\ 79\ 76\\ 697924562124$$ And following your procedure $$123456787531+69=123456787600\\ 6787654321+79=6787654400\\ 44567876+24=44567900\\ 976544+56=976600\\ 6679+21=6700\\ 76+24=100$$

Ross Millikan
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  • Nice, I can just comment that if b was not given, it wouldn't be possible to guess that procedure like this should be applied. I.e. b is a clue IMO. – Andris Zalitis Feb 03 '20 at 19:00
  • @AndrisZalitis: I am guessing that you have told people that the number to be added is what will bring it up to the next multiple of $100$, so we can derive $b$ from $a$. – Ross Millikan Feb 03 '20 at 19:01