Checking if a number is prime

[Underhiseye2]

When checking if a number is prime, is it enough to rule it out of being in any of the 2 to 10 times tables?

Do you need to check 11 too? If not, why not?

I assume you don't need to check 12 because that's 2 x 6. But would you need to check 13 too.. head hurting!

Hoping someone can explain the reasoning to me, thank you!

Even after i wrote that I realise that you would have to keep going.

The number we have to check is 1009.

It's not in the 2-12 times tables. How would I prove there are no factors?

This is y5 maths btw.

Thank you

Short homework answer is that you cheat and use a list like this one

www.mathsisfun.com/numbers/prime-numbers-to-10k.html

And yes it's prime

I might be wrong about this, but a number that ends in 9 will be prime if it is not divisible by 3, so that is probably the check they are looking for.

49 isn't prime. Neither is 119.

I would also look at the 3, you know it’s not an even number, so 3 would be next

You can rule out 5 as it doesn’t end in 5 or 0

That only leaves 7 and 9

Neither work

So it’s a prime number

But if it was 13x13 then that wouldn’t be caught by that or would it?

You know it’s not even
You know it’s not a multiple of 3 1+0+0+9 = 10
You know it’s not a multiple of 5
So if it’s not prime, it’s a multiple of two prime numbers. You can start checking those and I’d check some of them, but for year 5 math I don’t think they are really expecting that the student manually rule out every potential combination by hand so at that point I would check a chart.

For numbers divisible by 3, add the digits together. Repeat until you're left with a single number. If it's 3, 6 or 9, it's divisible by 3

E.g.

27 2+7=9
165 1+6+5=12 1+3=3
563451 5+6+3+4+5+1=24 2+4=6

You need to check whether it's divisible by every prime number until you get above the square root of the number itself.

E.g. If I want to know if 53 is prime I need to check all the prime numbers less than 8 (since we know the square root of 53 is more than 7 and less than 8). So I check 2,3,5,7 and if none go into 53 (which they don't) it's prime.

If I wanted to check whether 123 was prime I'd need to check all prime numbers below 12 so I'd need to check 2,3,5,7,11. I'd find that 3 goes into 123 so it isn't prime.

Remember the tricks if the digit sum is divisible by 3 it's divisible by 3. If the unit digit is even it's divisible by 2. If it ends in 5 or 0 it's divisible by 5.

There's no shortcut (other than looking up in a table), you need to check all primes below the square root of the number.

(Have they done square roots by yr 5?)

www.planetmath.org/howtofindwhetheragivennumberisprimeornot

So for your example you have the number 1009. I can see that 33*33 is 1089 so I only need to check primes below 33.

I can see instantly it's not divisible by 1,3,5, To check 7, 11, 13, 17, 19, 23, 29, 31 I would simply have to divide each one and see if it's divisible. Tedious but won't actually take too long.

If alternate digits add up to 11 or a multiple of 11 it is divisible by 11.

it's easy, just use Fermat’s Little Theorem for a number below 10^10.

for larger numbers use Primes is in P, an unconditional deterministic polynomial-time algorithm.

To check if 1009 was divisible by 11, I would think that 11x100=1100, and then work back from there. So 1100-99 is 1001, so 1009 is not a multiple of 11.

Thank you!

are there any other rules like that for other numbers? this homework is all about digital roots

sorry, my reply was for RancidOldHag

I've only just noticed the other replies.

I like yours, IMNOTSHOUTING

and thank you for the reminder about Fermat's Little theorem ifailed !