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

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1004

10004

10045

10025

answer is B.

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Detailed Solution

Concept- To answer this question, we must identify the largest four-digit number that is divisible by 41. Therefore, by adding 41 to it, the smallest five-digit number divisible by 41 will be obtained.
The largest four-digit number that is divisible

by 41

will be sought after first. We shall apply the remainder formula, which goes as follows:

Dividend = Quotient \times Divisor + Remainder .

The remainder in this case is lower than the divisor. The number that is being divided is called the dividend. The number being divided is known as the divisor, and the largest integer that can be multiplied by the divisor while still resulting in a result that is less than the dividend is known as the quotient.
Dividend and the sum of the products of the quotient and the divisor are equal to the remainder.
In fact, because the quotient is an integer, this also indicates that the result would be exactly divisible by the divisor if the remainder were subtracted from the dividend, i.e.10000 divided by 41 yields a quotient of 243 and a leftover of 37.This indicates that the largest four digit number divisible by