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

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

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