When a certain positive integer P is divided by another positive integer, the remainder is $${r_{1}}$$ . When a second positive integer Q is divided by the same integer, the remainder is $${r_{2}}$$ and when (P + Q) is divided by the same divisor, the remainder is $${r_{3}}$$ . Then the divisor may be :
A. $${r_{1}}$$$${r_{2}}$$$${r_{3}}$$
B. $${r_{1}}$$ + $${r_{2}}$$ - $${r_{3}}$$
C. $${r_{1}}$$ - $${r_{2}}$$ + $${r_{3}}$$
D. $${r_{1}}$$ + $${r_{2}}$$ - $${r_{3}}$$
Answer: Option D
Solution(By Examveda Team)
Let P = x + $${r_{1}}$$ and Q = y + $${r_{2}}$$, where each of x and y are divisible by the common divisor.Then, P + Q = (x + $${r_{1}}$$) + (y + $${r_{2}}$$)
= (x + y) + ($${r_{1}}$$ + $${r_{2}}$$)
(P + Q) leaves remainder $${r_{3}}$$ when divided by the common divisor.
⇒ [(x + y) + ($${r_{1}}$$ + $${r_{2}}$$) - $${r_{3}}$$] is divisible by the common divisor.
Since (x + y) is divisible by the common divisor, so divisor = $${r_{1}}$$ + $${r_{2}}$$ - $${r_{3}}$$
This Question Belongs to Arithmetic Ability >> Number System
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I think there is a possibility that r1 + r2 - r3 is multiple of the common divisor, am i correct if not correct me with the reason.... please
Why is it -r3 and not +r3
Why cannot r1+r2-r3 be a multiple of the divisor?