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Byju's Answer
Standard X
Mathematics
Applications of HCF & LCM in Real-World Problems
The smallest ...
Question
The smallest number which when diminished by
7
, is divisible
12
,
16
,
18
,
21
and
28
is:
A
1008
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B
1015
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C
1022
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D
1032
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Solution
The correct option is
B
1015
First find the smallest number divisible by
12
,
16
,
18
,
21
and
28
.
It is the LCM of these numbers.
12
=
2
×
2
×
3
=
2
2
×
3
16
=
2
×
2
×
2
×
2
=
2
4
18
=
2
×
3
×
3
=
2
×
3
2
21
=
3
×
7
28
=
2
×
2
×
7
=
2
2
×
7
L
C
M
=
2
4
×
3
2
×
7
=
1008
Required number
=
(L.C.M. of
12
,
16
,
18
,
21
,
28
)
+
7
=
1008
+
7
=
1015
So the required number is
1015
.
If you diminish by
7
,
it will be divisible by all the given numbers.
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3
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