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Question Number 99584 by DGmichael last updated on 21/Jun/20

Answered by Rasheed.Sindhi last updated on 22/Jun/20

If  a≡b(mod m) and P(x) is a  polynomial with integer coficients  then       P(a)≡P(b)(mod m)    Now,      Let P(x)=a_n x^n +a_(n−1) x^(n−1) +...+a_1 x+a_0   where  0≤a_i ≤9  As  10≡1(mod 9)  Therefore,  a_n .10^n +a_(n−1) .10^(n−1) +...a_1 .10+a_0               ≡a_n .1^n +a_(n−1) .1^(n−1) +...+a_1 .1+a_0                                                                (mod 9)   a_n a_(n−1) ...a_1 a_0 ≡a_n +a_(n−1) +...+a_1 +a_0                                                                            (mod 9)  That is a number is divisible by  9 if its sum of digits is divisible  by 9.

$${If}\:\:{a}\equiv{b}\left({mod}\:{m}\right)\:{and}\:{P}\left({x}\right)\:{is}\:{a} \\ $$$${polynomial}\:{with}\:{integer}\:{coficients} \\ $$$${then} \\ $$$$\:\:\:\:\:{P}\left({a}\right)\equiv{P}\left({b}\right)\left({mod}\:{m}\right) \\ $$$$ \\ $$$${Now}, \\ $$$$\:\:\:\:{Let}\:{P}\left({x}\right)={a}_{{n}} {x}^{{n}} +{a}_{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} +...+{a}_{\mathrm{1}} {x}+{a}_{\mathrm{0}} \\ $$$${where}\:\:\mathrm{0}\leqslant{a}_{{i}} \leqslant\mathrm{9} \\ $$$${As}\:\:\mathrm{10}\equiv\mathrm{1}\left({mod}\:\mathrm{9}\right) \\ $$$${Therefore}, \\ $$$${a}_{{n}} .\mathrm{10}^{{n}} +{a}_{{n}−\mathrm{1}} .\mathrm{10}^{{n}−\mathrm{1}} +...{a}_{\mathrm{1}} .\mathrm{10}+{a}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\equiv{a}_{{n}} .\mathrm{1}^{{n}} +{a}_{{n}−\mathrm{1}} .\mathrm{1}^{{n}−\mathrm{1}} +...+{a}_{\mathrm{1}} .\mathrm{1}+{a}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({mod}\:\mathrm{9}\right) \\ $$$$\:{a}_{{n}} {a}_{{n}−\mathrm{1}} ...{a}_{\mathrm{1}} {a}_{\mathrm{0}} \equiv{a}_{{n}} +{a}_{{n}−\mathrm{1}} +...+{a}_{\mathrm{1}} +{a}_{\mathrm{0}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({mod}\:\mathrm{9}\right) \\ $$$${That}\:{is}\:{a}\:{number}\:{is}\:{divisible}\:{by} \\ $$$$\mathrm{9}\:{if}\:{its}\:{sum}\:{of}\:{digits}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{9}. \\ $$

Commented by DGmichael last updated on 23/Jun/20

thanks sir !��

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