26-R-1-37-1-R-2-3-2-R-3-5-Find-R-1-R-2-and-R-3-

Question Number 90200 by Ar Brandon last updated on 21/Apr/20

26 \equiv R_{1} [37]

1 \equiv R_{2} [3]

2 \equiv R_{3} [5]

Find R_{1}, R_{2} and R_{3}

Commented by Rio Michael last updated on 22/Apr/20

R_{1} = 11, R_{2} = 2, R_{3} = 3

Commented by Ar Brandon last updated on 22/Apr/20

Thanks, but how ?

Commented by Rio Michael last updated on 22/Apr/20

if your question is actually

26 \equiv R_{1} (\mod 37) \Rightarrow R_{1} is the reminder when 26 is divided by 37

which rightly is 11 .

or you can say

by definition a \equiv b (\mod n) \Leftrightarrow n ∣ (a - b) \Rightarrow a - b = n k, k \in Z

\Rightarrow 26 \equiv R_{1} (\mod 37) \Rightarrow 26 = 37 k + R

k = 0 R_{1} = 26

k = 1 \Rightarrow R_{1} = - 11

k = - 1 \Rightarrow R_{1} = 63

these are just the elements of the set of least residue \mod 37 .

Commented by Ar Brandon last updated on 22/Apr/20

Thank you - thank you

Commented by mr W last updated on 22/Apr/20

h o w c a n y o u s a y

26 \equiv R_{1} m o d 37 \Rightarrow R_{1} = 11 ?

t h a t m e a n s t h e n 26 = 37 k + 11 .

t h i s i s w r o n g!

c l e a r l y :

R_{1} = 26 s i n c e 26 \equiv 26 m o d 37

R_{2} = 1

R_{3} = 2

Commented by Ar Brandon last updated on 22/Apr/20

what about - 11, - 2, and - 3 ?

Commented by mr W last updated on 22/Apr/20

b u t r e m a i n d e r i s u s u a l l y 0 o r p o s i t i v e

n u m b e r .

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