Math questions on prime numbers

[LanaTron]

Ds started Singapore's Discovering Mathematics yesterday. The book states that "to determine whether a number is prime, we can check if it is divisible by prime numbers less than the given number."

 

Then one of the problems is:

 

Determine whether the following numbers are prime numbers. (a) 103, (b) 229, © 817.

 

I did not have my ds do this, because it seems it would take forever to figure that out if you have to try to divide the number by all primes less than that number. It seems like it would take forever just to find the primes less than those numbers. I read in the teacher manual something about that you really only need to divide by primes less than the square root of the number, which makes sense, but then you are stuck figuring out the square root of 817 (and we haven't done those yet, this is the first lesson in the book).

 

The rest of the lesson is on prime factorisation and index notation, and that all went really well.

 

Is it a crucial skill to be able to determine if a larger number is prime?

 

Thank you!

[nmoira]

I read in the teacher manual something about that you really only need to divide by primes less than the square root of the number, which makes sense, but then you are stuck figuring out the square root of 817 (and we haven't done those yet, this is the first lesson in the book).

Or you could just use 30^2=900 as a starting point. 29^2 is too big, 28^2 is too small.... so check all primes less than 28.

 

I think it's a useful skill, especially if you don't have a calculator that can factor numbers.

[LanaTron]

Okay, that makes the task seem more manageable; I don't know why I didn't figure that out. Thanks.

[8filltheheart]

There is a faster way. There are very simple ways to determine of the numbers are divisible by 2,3,5,7, and 11. (There are rules for all the prime numbers under 50, but if you know the easiest ones, then I think that trial and error isn't that big of a deal with the larger #s.)

 

A number is divisible by 2 if its last digit is also (i.e. 0,2,4,6 or 8).

 

A number is divisible by 3 if the sum of its digits is also. Example: 534: 5+3+4=12 and 1+2=3 so 534 is divisible by 3.

 

A number is divisible by 5 if the last digit is 5 or 0.

 

Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.

 

Test for divisibility by 11. Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.

Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.

 

 

I simply pasted these from http://home.egge.net/~savory/maths1.htm

[Jackie in AR]

I simply pasted these from http://home.egge.net/~savory/maths1.htm

 

Thanks for the link!

[LanaTron]

Thanks! We knew the 2 & 5, and ds knew the 3 rule from LoF...thanks for the others, that simplifies even more.