What is the smallest 5 digit number exactly divisible by 41?
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3
10004
10004Therefore, Smallest 5 digit number exactly divisible by 41 is 10004.
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15
Answer:
Answer is 10004
Step-by-step explanation:
The smallest 5 digit number = 10000
= 10000 ÷ 41
Quotient = 343;
Remainder = 371
Required number = 10000 + ( 41 - 37)
= 10004
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