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The smallest 5-digit number exactly divisible by 41 is _______.

This question was previously asked in
NPCIL Assistant Grade-1 (Advance) Official Paper-II (Held On: 08 Sept, 2018)
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  1. 10041
  2. 10004
  3. 10045
  4. 10025

Answer (Detailed Solution Below)

Option 2 : 10004
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Detailed Solution

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Calculation:

Smallest 5-digit number = 10000

When we divide 10000 by 41 we get the remainder 37

So, the smallest 5-digit number that is a multiple of 41 is

10000 + (41 - 37) = 10000 + 4 = 10004

∴ The correct answer is Option 2).

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