Help with maths homework?

[Jelly]

Is there anybody who would be willing to help me with my homework? If anybody is, I'll post the bit I'm struggling with. And no, I'm not asking for someone to do it for me.  ;)

I have until tommorow to do it.

[loki redfern]

if you post , they will come :D ..well thye usually do.. helping people do maths is fun ;D (yes some of us are strange ;) )

[krelianzg]

Go for it. I love math. It's a hell of a lot better than what I'm doing at work at the moment :-\

[Jelly]

The homework

 

2. The prime factorisation of 560 is 2x2x2x2x5x7

a) without doing any calculation, decide which of these are factors of 560.

    Explain how you decided.

    2,8,6,10,16,9,20,35,70,60,15,14,140.

b) Check your answers by calculating

 

There's more, but that's basically what it's about. I just don't understand how to do it.

Thankyou!

[krelianzg]

The homework

 

2. The prime factorisation of 560 is 2x2x2x2x5x7

a) without doing any calculation, decide which of these are factors of 560.

    Explain how you decided.

    2,8,6,10,16,9,20,35,70,60,15,14,140.

b) Check your answers by calculating

 

There's more, but that's basically what it's about. I just don't understand how to do it.

Thankyou!

 

Ok, prime factors are all the factors of a number that only have themselves and 1 as factors. So, any other factors of the number in question (560), can be made by multiplying together the prime factors.

 

To decide which numbers from that list are factors, just check to see if they can be gotten by multiplying two or more of the prime factors.

 

E.G. 2x2x5 = 20, so 20 is a factor of 560.

 

For part B, simply divide 560 by each of the factors you decided on in part A. If they're correct, you'll get a whole number.

 

Feel free to ask for further explanation if you need it.

[Jelly]

Ah, I see. Thankyou so much!!!!!!!!!  ;D

[krelianzg]

You're quite welcome 8)

[Cadsuane]

Exactly. You could start by making a list of all of the numbers that can be made by multiplying different combinations of 2, 2, 2, 2, 5, and 7.

 

ex.

 

2x2=

2x5=

2x7=

 

2x2x2=

2x2x5=

2x2x7=

 

2x2x2x2=

...

 

and so on, adding in one more of the remaining available numbers are all added in.

[loki redfern]

..im not sure if this counts as calculation but if you work out the 'trees' that you do to get the prime factorisation

 

(e.g of the bottom part of the trees) there should be two different ones (hint 2x5, 2x7..ect

 

...      ...    ...

  |          |        |

  4    x    4    x  35

  |            |          |

/ \          / \        /  \

|    |      |    |    |    |

2 x 2    2 x 2  5 x 7 

 

 

er if you just multiply them ..you get other numbers too dont you.. 2x2x2=6 and 6 dont go? ..or did i do something wrong there ..just poke me you dont understand

[Cadsuane]

Loki, 2x2x2=8, not 6.  ;)

[Jelly]

Thankyou for all the help so far  :)

 

It gets more difficult though...

 

Question 4

The prime factorisation of 924 is 2²x3x7x11

a) Without doing any calculation, decide which of these are factors of 924.

  2  2³x3  3²x7  2²x7  3x11  7²x11  2x11

 

b) Check your answers by calculating.

 

Does anyone understand this?

[krelianzg]

Thankyou for all the help so far  :)

 

It gets more difficult though...

 

Question 4

The prime factorisation of 924 is 2²x3x7x11

a) Without doing any calculation, decide which of these are factors of 924.

   2  2³x3  3²x7  2²x7  3x11  7²x11  2x11

 

b) Check your answers by calculating.

 

Does anyone understand this?

 

Again, what you want to do is look at the list of prime factors, and see if any of the choices presented are created by multiplying together the prime factors. So, 2²x3 is a factor, 2²x11 is a factor, etc.

 

I can't get much clearer without pointing out the answer, but if you don't mind and would like a fuller explanation, let me know.

 

For part B, you should just do out the multiplication for the term you selected, then test it as a factor (divide 924 by it).

[loki redfern]

oh hel, :-[ ignore that though there was something wrong

[Cadsuane]

Thankyou for all the help so far  :)

 

It gets more difficult though...

 

Question 4

The prime factorisation of 924 is 2²x3x7x11

a) Without doing any calculation, decide which of these are factors of 924.

   2  2³x3  3²x7  2²x7  3x11  7²x11  2x11

 

b) Check your answers by calculating.

 

Does anyone understand this?

If you remember that 2² = 2x2, 2³ = 2x2x2, 3²  = 3x3, and that 7²  = 7x7, then you know that this is the same kind of problem as the first one, it's just written a little differently.

 

The basic idea is that the final number (924, in this case), in it's most basic form, is made up of the given prime factors (multiplied together). Therefore, any other number that is made up of a selection of those same prime factors will be a factor of the final number. Like ingredients. Pumpkin pie (924) is made with pumpkin (11), flour (7), butter (3), cinnamon (2), nutmeg(2). Anything else that is made with those numbers is a factor of pumpkin pie.

 

More explanation related to pies (this is just a simple picture for reference, it has nothing to do with pie charts or percentages):

[Jelly]

Thankyou so much everyone, you've all been a great help, and I've finished my homework.  :)

Thankyou again.

 

Goodnight.

 

:)