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Question

If A=102020213, prove that A36A2+7A+2I=0

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Solution

A2=AA=102020213102020213

=1+0+40+0+22+0+60+0+00+4+00+0+02+0+60+2+34+0+9=5280408513
Now, A3=A2.A
=5280408513102020213
=5+0+162+0+108+0+260+0+80+8+00+0+010+0+244+4+1516+0+39
=211234080342355
A36A2+7A+2I
=21123408034235565280408513+7102020213+2100010001
=2112340803423553012480240483078+7014014014721+200020002
=21+7+212+0+034+14+00+0+08+14+20+0+034+14+023+7+055+21+23012480240483078
=30124802404830783012480240483078
=000000000=0
A36A2+7A+2I=0


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