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An instance of this puzzle is here: https://www.theguardian.com/science/2020/nov/16/can-you-solve-it-the-srcmalbed-nmebur-plzuze

a) It has been swhon taht to raed a txet the oedrr in wihch the lrtetes of ecah idniadiuvl wrod aepapr is not ipmotanrt, so lnog as the fsrit and lsat ltetres are the crorect oens. Tihs is not the csae wtih nmuebrs baecsue if one slcarbmes the ditgis of a nmbeur it is not psisolbe to wrok out waht the ogirianl nemubr was.

Tehre are, hevoewr, cirtaen cesas in wchih tehre is sifuficnet inmoartfion to fnid out the onriiagl neumbrs if olny the interior diigts of ecah of them wree mxied wlihe the fsirt and lstt digtis wree lfet untaelred. Scuh is the csee in the fonlwloig aditiodn:



Can you rseotre the oigrianl atdiiodn and its sum?

(Hint: That lonely red 2 [marked by brackets] is where it should be.)

b) Without the hint of the lonely red 2, there are 13 solutions to the above puzzle. Can a similar puzzle, with a unique solution, be devised, with four five-digit summands, one six-digit sum, and preferably no number with interior digits repeated?


2 Answers 2


Partial Answer

Solution for a):


Process for a):

First started out by noting the sum of the digits in the ones column to be 11, meaning 1 will be added to whatever digits are placed in the 10s column. I then looked to the sum of the digits in the 10,000s column and on its own it equals 18. Since we know the overall sum has a 2 in the 100,000s column 2 or more must be carried over from the 1,000s column into the 10,000s column.
From there I looked at possible combinations of numbers that could go in each column and sum to a value that is a part of the final answer (0, 1, 2, 9). Starting from the 10s column since we know 1 will be carried over from the 1s column and then slowly working up the columns listing possible values. After I had all possible values listed I scanned for further possible ways to cancel out options. Finally, I looked at the remaining column options I had and looked for a combination from each column that could be validly combined without using the same numbers twice or missing any required numbers. This led me to my final solution.


Solution for b)

The answer is

It is possible to construct a puzzle satisfying all the conditions (including the "preferably" part).


$$\begin{array}{cccccc} &5&9&1&2&3 \\ &5&9&1&2&3 \\ &5&9&1&2&3 \\ +&5&9&1&2&3 \\\hline 2&3&6&4&9&2\end{array}$$

The shuffled digits in the sum are 3, 4, 6, 9. Since the smallest digit combination for 10,000's and 1000's digits is 34 (which cannot be reached with three 9's and a 2 or lower, even with a carry), all the 1000's digits of summands must be 9. The lower places cannot create a carry to the 1000's, so 10,000's and 1000's digits of the sum are 36. Then the maximum sum for 100's and lower is 852, so 9 in the sum can only be at 10's position. The only way to get a 9 there is to have all 2's in the summands at 10's positions. Therefore, the answer is unique.


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