n is a positive integer. The next-to-last digit in the decimal expression
of n^2 is 7.
What’s the last digit?

Please provide your reasoning, not only the digit.

n = {…xy}   where {} means concatenates digits 

n^2 = {…7m}

where m is one of (0,…9)  

sup. y^2 = {b1 b2}

        xy   = {c1 c2}

set up the n^2 multiplication in the grade school way:

    {     …x       y}

 X {    …x       y}

_______________ notes:

 (c2)                    [carries for x (xy)]

 (c1)  (b1)           [carries for y (xy)]

         c2      b2    [y(xy)]

+       c2              [x(xy)]

___________________________

   …    7       m

{… 7} = b1 + 2 c2, so b1 must be odd

the squares from 1 digit integers that have an odd ten’s place are (16 and 36) = y^2 = {b1 b2}

b2 is the ones place of y^2, so, b2 = m = 6. n^2 ends in 76

Edited on November 12, 2018, 5:48 pm

Posted by Steven Lord on 2018-11-12 17:46:50