MATHS HELP DESPERATE!?!!?!

Q1. Find all positive intergers less than 50,00 that have precisely 11 divisors
Q2 Can the sum of the first 'n' consecutive positive intergers be equal to the square of a prime number?
Q3. Find all positive intergers 'n' such that the pentagonal number P'n' is prime
Q4. IN a triangle ABC , AB = BC . L is the point on BC such that AL bisects angle BAC. IF AL=AC find the sizes of the triangle ABC

For Q1 i know that one of the numbers is 1024.

Thx in advance

Q1:

Let prime factors of any such number (x) be p₁, p₂, ... pᵢ, such that
x = p₁^n₁ * p₂^n₂ * ... * pᵢ^nᵢ

Then number of divisors of x is: (n₁+1) * (n₂+1) * ... * (nᵢ+1)
For x to have 11 divisors, this product must = 11.
But 11 is prime (11 = 1 * 11), and each of these numbers is > 1
So x must have only 1 unique prime factor, raised to the power of 10,
since (10+1) = 11

x = p¹⁰
Divisors of x are: p⁰, p¹, p², p³, p⁴, p⁵, p⁶, p⁷, p⁸, p⁹, p¹⁰ (11 divisors)

2¹⁰ = 1,024
3¹⁰ = 59,049
5¹⁰ = 9,765,625

Only 1,024 is less than 50,00 (5 thousand? or 50 thousand?)

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Q2.

Assume there is a value for n such that sum of first n numbers = square of a prime

Therefore there a number n and prime number p such that
1 + 2 + ... + n = p²
n(n+1)/2 = p²

Now p² has only 2 divisors, p and p.
Therefore n(n+1)/2 has 2 divisors, p and p.

One of n and (n+1) is even

Case 1: n is even -----> n/2 and (n+1) are integers
n/2 * (n+1) = p * p
Since p is prime, there is no other way to represent this number as product of 2 numbers.
Therefore n/2 = p and (n+1) = p
n/2 = n + 1
-n/2 = 1
n = -2
p = -1 -----> not a prime number

Case 2: n+1 is even -----> n and (n+1)/2 are integers
n * (n+1)/2 = p * p
Since p is prime, there is no other way to represent this number as product of 2 numbers.
Therefore n = p and (n+1)/2 = p