Prime Factors

Hi
Not sure how to answer this question. Btw I know how to express something as a product of its prime factors but not sure what this question wants me to do, thanks.

Given that ${25x}$² — 1 factorises to give (5${x}$-1)(5${x}$+1), express 2499 as a product of its prime factors.

$25x^2-1=(5x-1)(5x+1)$

Let $x=10$

LHS: $25*10^2-1=2500-1=2499$

So $2499=(5*10-1)(5*10+1)=49*51=7*7*51$

Ah can I let x = 20 if I want so can it be anything?

If you let x=20 then you get 25*20^2-1=10000-1 which is not what you wanted the prime factorisation for.

Oh so you think of a number which equals 2500 when 25 is multiplied by it so in this case that would be 10 giving 2500 and when you subtract 1 you get 2499 as it says 25x^2-1 ....


Is 51 a prime number?!

Oh so you think of a number which equals 2500 when 25 is multiplied by it so in this case that would be 10 giving 2500 and when you subtract 1 you get 2499 as it says 25x^2-1 ....


Is 51 a prime number?!

Well, you know that $25x^2 - 1= (5x-1)(5x+1)$ and you've already said $x=10$, so replace the $x$ in the brackets with $10$, what do you get? You will then need to further decompose these two numbers into their prime factors.

(51 is not prime).

Oh yes I'll do that and I thought it wasn't, the dude above made a mistake I think.
He wrote 49*51=7*7*51 but 51 isn't prime so he's wrong, right?

Yes sorry forgot to cancel down 51 into 17*3

Well, not wrong. But he hasn't answered the question properly, yes. He's edited his post to give the correct version now.

Ah it's OK, I think I understand now. What about 105^2 as a product of its prime factors?

Do I have to square it to get 11025 then break that down by prime factor decomposition? That's a massive number is there a simpler way?

$105^2=(105)*(105)$

$105=5*21=3*5*7$

so $105^2=(3*5*7)*(3*5*7)=3^2*5^2*7^2$

$105^2=(105)*(105)$

$105=5*21=3*5*7$

so $105^2=(3*5*7)*(3*5*7)=3^2*5^2*7^2$

How many prime number do you recommend I memorise? I know 2,3,5,7,11,13,17 already should I learn more or id that enough

Learn them up to 100. There are only 25 primes up to 100. I doubt you will need any more for GCSE.

Is 51 a prime number?!

OK thanks...

Well, not wrong. But he hasn't answered the question properly, yes. He's edited his post to give the correct version now.

$105^2=(105)*(105)$

$105=5*21=3*5*7$

so $105^2=(3*5*7)*(3*5*7)=3^2*5^2*7^2$

Quick way to check for small factors - the first two obvious ones are
if it's an even number it's divisible by 2;
if it ends in 5 or 0 it's a multiple of 5;

But also a useful one to check for primality - if the sum of the three digits is divisible by 3, then the number is divisible by 3; in your case 51 gives 5+1 = 6, divisible by 3 so not prime