Bunuel wrote:
samagra21 wrote:
\(10a+b+10b+a=11(a+b)=n^2\\
11*11\\
11*44\\
11*99\\
\\
a+b=11,44,99\\
{29,38,47,56} & {92,83,74,65}\\
\)
8 values
11*11\\
11*44\\
11*99\\
\\
a+b=11,44,99\\
{29,38,47,56} & {92,83,74,65}\\
\)
8 values
a and b are digits, so how can a + b be 44 or 99?
edited! That was for general possibilities without constraints of digits.
Statistics : Posted by samagra21 • on 01 Nov 2023, 06:29 • Replies 5 • Views 130
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