Question 796205

A two digit number is 18 less than the sum of the squares of its digits. How many such numbers are there?


Let tens and units digits be T and U, respectively
Then: {{{10T + U = T^2 + U^2 - 18}}}
{{{T^2 - 10T + U^2 - U - 18 = 0}}}


Since the number is 18 less than the sum of the squares of its digits, then its units digit is greater than its tens digit, or U > T.

Therefore, with 9 being the largest digit, we:


Let U = 9
{{{T^2 - 10T + U^2 - U - 18 = 0}}}
{{{T^2 - 10T + 81 - 9 - 18 = 0}}}
{{{T^2 - 10T + 54 = 0}}} ----- T is NOT an INTEGER


Let U = 8
{{{T^2 - 10T + U^2 - U - 18 = 0}}}
{{{T^2 - 10T + 64 - 8 - 18 = 0}}}
{{{T^2 - 10T + 40 = 0}}} ----- T is NOT an INTEGER


Let U = 7
{{{T^2 - 10T + U^2 - U - 18 = 0}}}
{{{T^2 - 10T + 49 - 7 - 18 = 0}}}
{{{T^2 - 10T + 24 = 0}}}
T, or tens digit = 6 or 4


This means that the number could be {{{highlight_green(67)}}}, or {{{highlight_green(47)}}}


Let U = 6
{{{T^2 - 10T + U^2 - U – 18 = 0}}}
{{{T^2 - 10T + 36 - 6 - 18 = 0}}}
{{{T^2 - 10T + 12 = 0}}} ----- T is NOT an INTEGER


No need to go any further since T ≠ U, and U > T 


You can do the check!! 


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