Question

Let N be the number of positive
integral divisors of 1998, the sum
of Nis
1) 16
2)5
3) 7
4) 8​

Answer 1

$Given \: number \: 1998$

2 | 1998

________

3| 999

________

3| 333

________

3| 111

________

*** 37

1998 = 2¹ × 3³ × 37¹

_____________________

$We \:know \:that ,$

$If \: N = a^{p} \times b^{q} \times c^{r} \times \ldots$

$Where \: a, b , c \: are \: prime \: numbers$

$Then \: Number \:of \:divisors = (p+1)(q+1)(r+1)$

_____________________

i) Number of divisors(N) = (1+1)(3+1)(1+1)

= 2 × 4 × 2

= 16

ii ) Sum of divisors = 2 + 4 + 2 = 8

$Option \: \pink{ ( 4) } \: is \: correct.$

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