Sunday, January 24, 2010

Maths Homework-prime numbers continued textbook

Textbook Pg 9 #25:
The Prime factorisation of a number is 2^4 X 3^5 X 7^2 X 11.
Write down 3 factors of the number that are greater than 100.

To find the answer, we need to simplify the index notation further so that we get something like this:16x243x49x11=2095632
Then we multiple the factors:16x11=176
we then test this number 2095632/176=11907 this means that 176 and 11907 are a factor of 2095632 and they are more than 100
since 243 is a number more than 100, lets try it 2095632/243=8624
this proves to us that 243 and 8624 are a factor of 2095632.

The answer is 176,11907 and 243

Pg 9 #25 & textbook p9 #26
The Prime factorisation of two numbers are 2 X 3^2 X 7^3 X 13 and3 X 7^2 X 13^3 X 17 .
Write down 3 common factors of the numbers.

First we extend the index notation
Next we find the common numbers:3,7,13
Then we multiply the numbers 3x7x13=273. this is the HCF
Every number,except 0,has one common factor the no. 1.
11466 and 5490303 can be divided by 3

the answer is 273,1 and 3

1 comment:

  1. Harsh

    Textbook Pg 9 #25:
    That's very good reasoning :D

    Indeed, using the prime factors expressed in the index notation, we could multiply different 'combinations' of factors to result a larger factor (as required in the question).
    Indeed, apart from using the final number (2095632) to check your factor, you have rightly pointed out that the number resulted through checking is another factor for the number, too :D

    Textbook p9 #26:
    Indeed, you already found the 3 factors to the 2 numbers (3, 7 and 13).
    Interestingly, you didn't forget "1".

    You are also right that the HCF is another factor :D
    However, if you check amongst the 2 numbers, the HCF should be 3 x 7^2 x 13.