find the proper divisors of the number 8400. Also find their sum
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Answer:
8400=4×100×21=18×25×3×7
=3
1
.2
4
.5
2
.7
1
=3
a
2
b
5
c
7
d
a varies from 0 to 1, b from 0 to 4, c from 0 to 2 and d again from 0 to 1.
Thus the total number of divisors is
= 2×5×3×2=60
We have to exclude the divisor 1 and the divisor 8400. i.e. the number itself. Hence required number is 60−2=58.
Sum of the divisors
Any divisor is of the form 3
a
2
b
5
c
7
d
Sum = ∑
0
1
3
a
∑
0
4
2
b
∑
0
2
5
c
∑
0
1
7
d
=(1+3)(1+2+2
2
+2
3
+2
4
)(1+5+5
2
)(1+7)
=4×31×31×8
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