an-integer-n-has-15-divisors-n-is-divisible-by-6-but-not-divisible-by-8-determinate-n-

Question Number 124513 by mathocean1 last updated on 03/Dec/20

a n i n t e g e r n h a s 15 d i v i s o r s . n i s

d i v i s i b l e b y 6 b u t n o t d i v i s i b l e b y 8 .

d e t e r m i n a t e n .

Commented by mr W last updated on 03/Dec/20

324 ?

Commented by soumyasaha last updated on 04/Dec/20

N = 2^{p} . 3^{q} 5^{r} . .

Considering 1 as not a factor

number of factors = (p + 1) (q + 1) \dots - 1

Now 15 = 16 - 1

= 4.4 - 1 or 2 \times 8 - 1 or 2 \times 2 \times 4 - 1

p + 1 \neq 4 ∵ no . is not divisible by 8

∴ p + 1 = 2 \Rightarrow p = 1

∴ q = 7 or q = 1 and r = 3 or q = 3 and r = 1

∴ N = 2^{1} . 3^{7} or N = 2^{1} 3^{1} . 5^{3} or N = 2^{1} . 3^{3} . 5^{1}

N = 4374 or N = 750 or N = 270

Considering 1 as a factor

number of factors = (p + 1) (q + 1) \dots

∴ 15 = 3 \times 5 or 5 \times 3

p + 1 \neq 5 ∵ no . is not divisible by 8

\Rightarrow p = 2 and q = 4

∴ N = 2^{2} . 3^{4} = 324

Answered by mr W last updated on 04/Dec/20

n = p^{a} q^{b} r^{c} \dots

(a + 1) (b + 1) (c + 1) \dots = 15 = 3 \times 5

\Rightarrow a = 2, b = 4

\Rightarrow n = p^{2} q^{4}

n i s d i v i s i b l e b y 6

\Rightarrow n m u s t c o n t a i n 2^{⩾ 1} 3^{⩾ 1} \dots (1)

n i s n o t d i v i s i b l e b y 8

\Rightarrow n m a y c o n t a i n 2^{⩽ 2} \dots (2)

f r o m (1) a n d (2) w e c a n s e e

p = 2, q = 3

\Rightarrow n = 2^{2} 3^{4} = 324

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