Math Is Fun Forum

ok say N is 15. The factors are 1 3 5 and 15. One way to do it is say,what two groups of numbers can have together,the factors 1,3,5 and 15.
one way is start with each second number and work out the possible first numbers.
so here I go with,

1,15
3,15
5,15
15,15
Now set the second no. to 5 but avoid 15 cos we already included that but it should be multiple of 3.
3,5.

we cant do it with 3 or 1.

So here we go! the answer is

1,15
3,5
3,15
5,15
15,15.

Lets do another much harder example.
N is 60.
The prime factorization is 2 squared*3*5
And all the factors are:
1,2,3,4,5,6,10,12,15,20,30,60.

the 60s we have all the factors with a 60:

1,60
2,60
3,60
4,60
5,60
6,60
10,60
12,60
15,60
20,60
30,60
60,60.

Now the 30s.
60/30=2 and 30 is divisible by 2.So the divisiblility no.has to have another 2 in it,so the other ones are factors divisible by 4 but not 60 since we already included 60.

4,30
12,30
20,30

Now the 20s.
60/20=3 but 20 is not divisible by 3.The first number has to be divisible by 3.60 and 30 are not there since they had 20 in them.

3,20
6,20
12,20
15,20
Now the 15s.
60/15 is 4.15 isn't divisible by 4 so... you get the idea.20 and 60 are excluded

4,15
12,15

and so on...

the 12s:

10,12.

the 10s:
wait a second.30 is also excluded in addition to 12 and 60 since it is a multiple of 10.

6,10

1,2,3,4,5 and 6 have no solutions since they are less than square root of 60.

So FINALLY we have our answer.

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