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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

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