Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
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)$r_{1}+r_{2}-r_{3}$
(e)Cannot be determined
Answer
Answer (as printed): D
Explanation
Let P = x + r1 and Q = y + r2, where each of x and y are divisible by the common divisor. Then, P + Q = (x + r1) + (y + r2) = (x + y) + (r1 + r2). (P + Q) leaves remainder r3 when divided by the common divisor. ⇒ [(x + y) + (r1 + r2) – r3] is divisible by the common divisor. Since (x + y) is divisible by the common divisor, so divisor = r1 + r2 – r3.
Explanation as extracted from the printed page; notation may be imperfect.