what is the remainder when 41²*36³*31² is divided by 15 ?
Answers
Answer:
6 is the answer
Step-by-step explanation:
Let us solve it by applying Modular Airthmetic.
41^2×36 ^3×31^2=REM? (mod 15 )
Prime Factorisation of 15 =3×5
So the above congruence can be written into two simultaneous equations to ease computations.
41^3×36^3×31^2=REM (mod 5)
41^2×36^3×31 ^2=REM (mod 3) ______ (i)
Please observe (i) above wherein a term 36^3 is divisible by Modulo 3 . So this congruence is inconsistent and can't be made consistent by changing the Modulo without loosing the generality of the congruence. So we can't arrive at a solution. The two congruences can't be construed as simultaneous equation.
So the only solution, we have to calculate the Remainder(s) directly to arrive at the result.
41^2×36 ^3×31^2=(41×41)(36×36×36 )(31×31) =11 ×11×6×6×6×1×1=(11×6)(11×6)×(6)=( 6×6)(6) = 6×6=6 (mod 15)
Therefore the REMAINDER is 6. ■ANSWER
Answer:
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