Question 1204004
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Let the last 2 digits of k be x and y, so that the number is 10x+y.  Then k^2 is 100x^2+20xy+y^2.<br>
We want to know the condition(s) that make the last two digits of k^2 = 100x^2+20xy+y^2 the same.<br>
The last 2 digits of 100x^2 are always "00", so the last 2 digits of k^2 are determined by 20xy+y^2.<br>
So, for each last digit y, we want to find the values of next-to-last digit x for which the last 2 digits of 20xy+y^2 are the same.<br><pre>

  last digit y  20xy+y^2  value(s) of x that make last 2 digits of 20xy+y^2 the same
 ---------------------------------------------------------
      0            0        any (trivial case -- the square of any integer with last digit 0 has last 2 digits 00)
      1         20x+1       none
      2         40x+4       1 or 6 (last 2 digits "44")
      3         60x+9       none
      4         80x+16      none
      5        100x+25      none
      6        120x+36      none
      7        140x+49      none
      8        160x+64      3 or 8 (last 2 digits "44")
      9        180x+81      none</pre>
Summary....<br>
The last 2 digits of k^2 are the same if, and only if...
(1) the last digit of k is 0 (last 2 digits of k^2 "00"); or
(2) the last 2 digits of k are 12 or 62, or 38 or 88 (last 2 digits of k^2 "44")<br>
It's not clear what kind of answers are wanted for the questions that are asked, so I leave it to you to use those results to answer them.<br>