Question 1199769
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<pre>

    +--------------------------------------------------------+
    |     Looking at the last digits of these numbers,       |
    |  you notice that neither 2 nor 3 nor 4 is the base.    |
    +--------------------------------------------------------+


Let the base be "b".  Then we can write this equation

    (b+3) + (2b+4) + (4b+3) + (4b+4) = {{{b^2 +2}}}.


Simplify and write in standard form of quadratic equation

    11b + 14 = {{{b^2}}} + 2

    {{{b^2}}} - 11b - 12 = 0


Solve by factoring

    (b-12)*(b+1) = 0


The appropriate solution/answer is  b= 12.
</pre>

Solved.


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It is nice entertainment problem with non-standard setup.


And making right setup is the most engaging/enlightening part of its solution.