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Byju's Answer
Standard VIII
Mathematics
Divisibility by 2
The smallest ...
Question
The smallest 5 digit number exactly divisible by
41
is:
A
1004
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B
10004
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C
10045
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D
10025
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E
None of these
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Solution
The correct option is
B
10004
The smallest 5-digit number
=
10000
10000
÷
41
(Quotient
=
243
; Remainder
=
37
)
Required number
=
10000
+
(
41
−
37
)
=
10004
.
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