the smallest number which when divided by 39 52 65 leaves a remainder 7 in each case
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Answered by
1
So, the smallest number is \( 785\), which when divided by \( 39\), \( 52\) and \( 65\) leaves a remainder of \( 5\) in each case.
Answered by
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""" ❤️ Answer ❤️ """
For this first, we have to compute the
LCM of
39,
52 and
65
Prime factorization of
39
=
3
×
13
Prime factorization of
52
=
2
×
2
×
13
Prime factorization of
65
=
5
×
13
LCM
=
2
×
2
×
3
×
5
×
13
LCM of 39 , 52 and
65
=
780
The required smallest number
=
780
+
5
=
785
So, the smallest number is
785, which when divided by
39,
52 and
65 leaves a remainder of
5 in each case.
Let us check our answer
785 ÷ 39
Quotient
=
20
Remainder
=
5
785 ÷ 52
Quotient
=
15
Remainder
=
5
785 ÷ 65
Quotient
=
12
Remainder
=
5
So, when
785 is divided by
39,
52 and
65, the remainder is
5 in each case.
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