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The smallest 5 digit number exactly divisible by 41 is:

Question:
The smallest 5 digit number exactly divisible by 41 is:

[A].

1004

[B].

10004

[C].

10045

[D].

10025

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.