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

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 \in {1, 2, \cdot \cdot \cdot, 9}; b, c \in {0, 1, \dots, 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, \dots

^{∙} 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