Question

P is a prime number such that p-1 has exactly 10 positive factors while p+1 has exactly 6 positive factors. Find .[p^(1/2)]​

Answer 1

Given :-

• p is a prime number .
• (p - 1) has exactly 10 positive factors.
• (p + 1) has exactly 6 positive factors.

To Find :-

• p^(1/2) = ?

Solution :-

Let us Assume that, p is Equal to (6k - 1) .

Than,

p = 6k - 1

p + 1 = 2*3*k

6 factors than, 2 * 3 (2¹ * 3² or, 2² * 3¹)

k can be = 3 or, 2

So,

if k = 3

p + 1 = 6 * 3 = 18

p = 17

if k = 2 ,

p + 1 = 12

p = 11

So, both possible

Now,

(p - 1) = (6k - 2) = 2(3k - 1)

putting both k

→ p - 1 = 2(3*3 - 1) = 16

→ p - 1 = 2(3*2 - 1) = 10

Now,

10 = 2 * 5

16 = 2⁴

none of them has 10 factors.

conclusion :- p ≠ (6k - 1) .

So,

Let us Assume that, p is Equal to (6k + 1).

Than,

→ p = 6k + 1

→ (p - 1) = 6k

→ (p - 1) = 2 * 3 * k

Now, we have given that, (p - 1) has exactly 10 positive factors .

and,

→ 10 = 2 * 5

So,

→ Numbers can be (2⁴ * 3) or , (2 * 3⁴) { as, (4+1)*(1+1) = 10 .}

Therefore,

→ k can be = 2³ or 3³ . { As, (p - 1) = 2*3*k } .

Now,

if k = 2³

→ (p - 1) = 6 * 2³

→ p - 1 = 48

→ p = 49 ≠ Prime number.

So, k ≠ 2³ .

we conclude that, value of k is 3³.

Now, given that, (p + 1) has 6 factors.

→ p = (6k + 1)

→ (p + 1) = (6k + 2)

→ (p + 1) = 2(3k + 1)

Putting k = 3³ Now ,

→ (p + 1) = 2(3*3³ + 1)

→ (p + 1) = 2(81 + 1)

→ (p + 1) = 2 * 82

→ (p + 1) = 2² * 41

Comparing RHS part with a^p * b^q, we get,

→ number of positive factors = (p + 1) * (q + 1)

→ (2 + 1) * (1 + 1) = 3 * 2 = 6 = Satisfy .

Hence,

→ p + 1 = 2² * 41 = 4 * 41 = 164

→ p = 164 - 1

→ p = 163 .

∴

→ p^(1/2)

→ √(p)

→ √(163)

→ 12.76 (Approx.) (Ans.)