0
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Solve this Alphametic puzzle

   T H I S
+ S I Z E

=========
S H O R T

All those who know and remember me. I am posting on this website after a long time. I am extremely sorry for not being on this website.

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  • $\begingroup$ Can this be attributed to some news paper edition? If so, please do. As, the same appeared as a mind bender in 17th July 2019 supplementary edition of Times of India. $\endgroup$ – Mea Culpa Nay Jul 16 at 17:04
  • $\begingroup$ When ???? 17 July ??? ; ) $\endgroup$ – Ak19 Jul 16 at 17:07
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A nice easy puzzle.

First. We see that S=1 by obvious arithmetic.
From there, we can conclude that H=0 as T+1 can only be 10 or 11. As H!=S, H=0 And T=8 or 9.
As H=0, we see I+0=O. This means, there's a carry from the previous addition. Again as O!=H, it follows that, there's no carry from that specific addition of I+0=O. Hence, we conclude that T=9. From E+1=T, we conclude that E=8.
From maths, we can see that I+1=O. This also means Z and I are not consecutive integers. Also, the maximum value that I or Z can take is 7. From this, we can see that R can be 2 or 3 (12 or 13, that is).
R=3 is eliminated because I and Z will have to be 7 and 6 in some order and that contradicts with some of our other findings. Hence, R=2.
From R=2, we see that Z has to be 7 and I has to be 5 with O=I+1, implying O=6.

That means our final findings are:

T=9.
S=1.
E=8
H=0.
I=5.
Z=7.
R=2
O=6.

Final answer being:

9051
+ 1578
= 10629

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  • $\begingroup$ Great explanation but the answer is already given $\endgroup$ – Quark-epoch Jul 15 at 20:02
  • $\begingroup$ @Quark-epoch You need not accept the first correct answer. Our guidance is and always has been that which answer to accept is entirely at the asker's discretion; usually they pick the "best" answer, using whatever criteria they feel to be best. See in particular How do we choose which answer to accept from a set of high quality answers and its answers. For two equally good answers it seems most fair to award the first, but quality should trump speed. If you think this answer is more complete or better explained, feel free to Accept it. $\endgroup$ – Rubio Jul 18 at 0:31
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There are actually 11 solutions to this problem, to my knowledge. Duck's solution appears to be the only one where all the digits are unique, but there are also the solutions of:

THIS = 9191, SIZE = 19Z8, where Z is any digit 0-9, and SHORT = 11OR9, where OR is a concatenated number equal to Z+9.

In fact, there are hundreds of solutions to this problem if you account for the possibility of S = 0.

I just ran a script and found 991 different solutions to this equation in addition to the existing 11 above, for a total of 1002 solutions.

However, this results in no other solutions with unique digits as S+E=T and since S=0, E must be equal to T in all these cases.

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  • $\begingroup$ Of Course but Duck's answer is the one I expected $\endgroup$ – Quark-epoch Jul 15 at 20:02
  • $\begingroup$ (I believe the convention with alphametics is that leading 0s don't happen, and every letter stands for a unique digit.) $\endgroup$ – Rubio Jul 18 at 0:26
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Is it:

THIS=9051 SIZE=1578 SHORT=10629

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  • $\begingroup$ Good ! I didn't expectit to come so fast. $\endgroup$ – Quark-epoch Jul 15 at 19:05

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