Hard maths questions...

1, Note that if $n\in \mathbb{N}$ then the last digit of 5n-1 is always a 4 or a 9. List all the possible last digits of multiples of 36 and it will limit the numbers you need to try so that 5n-1 is divisible by 36.
2. You're just solving $\log (x^2)=\log (2x)$.
3. $\sin x + \cos x$ can be expressed as a single sine or cosine.

1st one
n = 29
5n - 1 will end either 4 or 9
Find first multiple of 36 that ends in a 4 or 9, add 1 (to make it a multiple of 5) then divide by 5 to find n

or practically the same thing but a bit clearer.

$5n-1 = 36p$ where p and n are both positive integers

as n is a integer, p must also be an integer.
Therefore find the first multiple of 36 where you add 1 it is divisible by 5. You will then get
$144 + 1 = 145[br]145 / 5 = n[br]n = 29[br]$

for 2) would it help if i said you could rewrite it as log(2)+log(x)=2log(x)?

for 3) you can differentiate and evaluate at those points

looks like two people have already answered 1).

Thanks
The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is.
im guessing its similar to question 1) above but not sure. Its supposed to be a big number so you wouldnt just use trial and error is there a method for it?