Let digits are a,b,c Let b=a+c because second digit is sum of last and first So number=a*100+b*10+c*1 =a*100+(a+c)*10+c*1 =a*100+a*10+c*10+c*1 =110*a+11*c =11*10*a+11*c =11*(10*a+c) That is the number is multiple of 11 so it is divisible by 11

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## proof 3digit

Let digits are a,b,c

Let b=a+c because second digit is sum of last and first

So number=a*100+b*10+c*1

=a*100+(a+c)*10+c*1

=a*100+a*10+c*10+c*1

=110*a+11*c

=11*10*a+11*c

=11*(10*a+c)

That is the number is multiple of 11 so it is divisible by 11

opksalu@gmail.com