Question

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

Answer 1

The smallest five-digit number is 10000.

Solution

We will use Division algorithm

We know that

Dividend = divisor × quotient + remainder

Dividend = 10000

Divisor = 41

10000 = 41 × 243 + 37

To find required number we have to

add number = divisor – remainder

= 41 – 37

= 4 to 10000

Therefore, Smallest 5 digit number exactly divisible by 41 is 10004.

Answer 2

Answer:

$\huge{ \underline{ \boxed{ \bold{ \star{ \blue{ \green{ \orange{ \pink{ \purple{ \red{ \mathfrak{ \pink{ \red{ \orange{ \green{a}{n} }{s} }{w} }{} }{} }{e} }{r} }{} }{} }{} }{} }{} }{} }{} }{} }{}$

10004 is the answer