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Find-all-three-digit-numbers-abc-with-a-0-such-that-a-2-b-2-c-2-is-divisible-by-26-

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Question Number 19182 by Tinkutara last updated on 06/Aug/17
Find all three digit numbers abc (with a ≠ 0) such that a^2 + b^2 + c^2 , is divisible by 26.
$$\mathrm{Find}\:\mathrm{all}\:\mathrm{three}\:\mathrm{digit}\:\mathrm{numbers}\:{abc}\:\left(\mathrm{with}\right. \\ $$$$\left.\mathrm{a}\:\neq\:\mathrm{0}\right)\:\mathrm{such}\:\mathrm{that}\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} ,\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by}\:\mathrm{26}. \\ $$
Answered by RasheedSindhi last updated on 08/Aug/17
a∈{1,2,∙∙∙,9}; b,c∈{0,1,...,9} a^2 +b^2 +c^2 is divisible by 26⇒ a^2 +b^2 +c^2 ≡0(mod 26) Give any values to any of two from their domains and find the third : c^2 ≡−(a^2 +b^2 )(mod 26) ^• Let a=3 and b=4 c^2 ≡−(3^2 +4^2 )(mod 26) ≡−25≡1(mod 26 c≡1(mod 26) The required numbers: 143,413,314,... ^• Let a=6 , b=3 c^2 ≡−(6^2 +3^2 )(mod 26) ≡−45≡7≡33≡59≡85(mod 26) There is no value in the domain of c for a=6 and b=3 ^(•Let a=7,b=5) c^2 ≡−(7^2 +5^2 )(mod 26) ≡−74≡4(mod 26) c=2
$$\mathrm{a}\in\left\{\mathrm{1},\mathrm{2},\centerdot\centerdot\centerdot,\mathrm{9}\right\};\:\mathrm{b},\mathrm{c}\in\left\{\mathrm{0},\mathrm{1},…,\mathrm{9}\right\} \\ $$$$\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\mathrm{26}\Rightarrow \\ $$$$\:\:\:\:\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{26}\right)\: \\ $$$$\mathrm{Give}\:\mathrm{any}\:\mathrm{values}\:\mathrm{to}\:\mathrm{any}\:\mathrm{of}\:\mathrm{two}\: \\ $$$$\mathrm{from}\:\mathrm{their}\:\mathrm{domains}\:\mathrm{and}\:\mathrm{find} \\ $$$$\mathrm{the}\:\mathrm{third}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} \equiv−\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right)\left(\mathrm{mod}\:\mathrm{26}\right) \\ $$$$\:^{\bullet} \mathrm{Let}\:\mathrm{a}=\mathrm{3}\:\mathrm{and}\:\mathrm{b}=\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} \equiv−\left(\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \right)\left(\mathrm{mod}\:\mathrm{26}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\equiv−\mathrm{25}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{26}\right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{c}\equiv\mathrm{1}\left(\mathrm{mod}\:\mathrm{26}\right) \\ $$$$\mathrm{The}\:\mathrm{required}\:\mathrm{numbers}: \\ $$$$\:\:\:\mathrm{143},\mathrm{413},\mathrm{314},… \\ $$$$\:^{\bullet} \mathrm{Let}\:\:\mathrm{a}=\mathrm{6}\:,\:\mathrm{b}=\mathrm{3} \\ $$$$\:\:\:\mathrm{c}^{\mathrm{2}} \equiv−\left(\mathrm{6}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\left(\mathrm{mod}\:\mathrm{26}\right) \\ $$$$\:\:\:\:\:\:\:\equiv−\mathrm{45}\equiv\mathrm{7}\equiv\mathrm{33}\equiv\mathrm{59}\equiv\mathrm{85}\left(\mathrm{mod}\:\mathrm{26}\right) \\ $$$$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{value}\:\mathrm{in}\:\mathrm{the}\:\mathrm{domain} \\ $$$$\mathrm{of}\:\mathrm{c}\:\mathrm{for}\:\mathrm{a}=\mathrm{6}\:\mathrm{and}\:\mathrm{b}=\mathrm{3} \\ $$$$\:^{\bullet\mathrm{Let}\:\mathrm{a}=\mathrm{7},\mathrm{b}=\mathrm{5}} \\ $$$$\:\:\:\:\mathrm{c}^{\mathrm{2}} \equiv−\left(\mathrm{7}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} \right)\left(\mathrm{mod}\:\mathrm{26}\right) \\ $$$$\:\:\:\:\:\:\:\:\equiv−\mathrm{74}\equiv\mathrm{4}\left(\mathrm{mod}\:\mathrm{26}\right) \\ $$$$\:\:\:\mathrm{c}=\mathrm{2} \\ $$$$ \\ $$
Answered by Tinkutara last updated on 09/Aug/17

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