ABC26GN0276 · Number System

Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:

If the seven-figure number 30X0103 is a multiple of 13, then $X$ is
(a)1
(b)6
(c)7
(d)8
Answer
Answer (as printed): D
Explanation
We first divide the number into groups of 3 digits from the right → 3 0X0 103 Difference of sum of numbers at odd and even places = (103 + 3) – 0X0 = 106 – 0X0, which must be divisible by 13. 106 – 0X0 is divisible by 13 only for X = 8.

Explanation as extracted from the printed page; notation may be imperfect.

Open in whiteboard · Browse this chapter in the app