Question:
The smallest 5 digit number exactly divisible by 41 is:
[A].1004
10004
10045
10025
The smallest 5 digit number exactly divisible by 41 is:
[A].
[B].
[C].
[D].
Answer: Option B
Explanation:
The smallest 5-digit number = 10000.
41) 10000 (243
82
—
180
164
—-
160
123
—
37
—
Required number = 10000 + (41 – 37)
= 10004.