Given the lowest common multiple and highest common factor (greatest common divisor) of (continue)?

Question

positive integer x and y, find all pairs of x and y.
(what method is to be used to solve this problem?)
Name the method, give me (an/some) example(s) and show me the working, please please please.
I know nothing about number theory.

Referring to Finndo's answer, what does the capital and small letter mean?

Answer 1

1) find the divisors of the lcm. reduce it to a formula A^a*B^b*C^c... etc
2) do the same thing for the gcd
3) you have just tranformed a complicated maths problem in a simple algorithmic one. Find the all the possible combinations of exponents.

simple does the trick


PS: capital letter = a divisor of the lcm
small leter: number of times the corresponding divisor is in the lcm. obviously.

Answer 2

LCM or lowest common multiple is the smallest number that can divide a set of numbers.

For example:
For numbers 144, 48, 36, the LCM is 2, because 144 can be
divided by 2; and so with 48 and 36. The given numbers can
also be divided by 3, but 2 is smaller than 3, therefore, the
answer is 2!

Another example:
63x - 49y = 7(9x - 7y) > We factor out 7 because it's the
least common multiple or LCM,
therefore simplifying the equation...
Another example:
xyz + xy - yz = xy(z + 1 - z/x)! x cancels out! So, it's not just
in numbers....

HCF or highest common factor, is similar to LCM, but this time, it is the biggest unit that is considered.

For example:
14 abc + 210 bcd - 56 abd

14' factors are 2 x 7....
210' factors are 7 x 3 x 2 x 5...
56' factors are 7 x 4 x 2....

We can see that 2 is a common factor and so is 7, which is
higher than 2, therefore we choose 7, which becomes the
common divisor of 14, 210 and 56!
We should not forget, too, that b is common in the equation!

Answer 3

Use the theorem: LCM(a,b) * HCF(a,b) = ab for all natural numbers a and b.

Suppose there are two distinct of pairs of natural numbers (a,b) and (c,d) such that LCM(a,b)=LCM(c,d) and HCF(a,b)=HCF(c,d). Then by the theorem ab=cd.

Let a' = a/HCF(a,b), b' = b/HCF(a,b), c' = c/HCF(c,d) = c/HCF(a,b) and d' = d/HCF(c,d) = d/HCF(a,b). Then a' * b' = c' * d', and a' and b' must be coprime, i.e. HCF(a',b') = 1, similarly HCF(c',d') = 1, hence LCM(a',b') = a' * b' = c' * d' = LCM(c',d'), and the conditions are satisfied, using the theorem again.

Therefore LCM(a,b)=LCM(c,d) and HCF(a,b)=HCF(c,d) whenever c = c' * LCM(a,b), d = d' * LCM(a,b) for some coprime natural numbers c' and d', such that a' * b' = c' * d'.

That's a lot of mathematics to comprehend for someone who doesn't know number theory, so here's an example.

Take a = 60 = 2^2 * 3 * 5 and b = 35 = 5 * 7. Now HCF(a,b)=5 and LCM(a,b)=420, and a' = a/HCF(a,b) = 2^2 * 3 = 12, and b' = b/HCF(a,b) = 7. Every coprime pair, c',d', of natural numbers such that c' * d' = a' * b' = 12 * 7 = 84, will give a solution to your problem. In this case the solutions are (84,1), (12,7), (28,3), (4,21). (note that (6,14) is not a solution since this pair is not coprime)

Multiply these pairs by HCF(a,b)=5 and you will get your solutions, (420,5), (60,35), (140,14) and (20,105). These will be the only pairs of natural numbers with HCF of 5 and LCM of 420.

Answer 4

It's 24 because 24*2 = 48 and 24*3 = 72

Answer 5

To find the solution for the Highest Common Factor and Lowest comon Multiplier.

Click on the URL below for additional information

www.jamit.com.au/htmlFolder/FRAC1004.html

Answer 6

Pythagorean equation by changing the powers to include any positive integer, x n + y n = z n for n = 3, 4, 5, . . . And according to Fermat there appeared to be no three numbers that would perfectly fit this equation.

Answer 7

if the HCF is A the LCM is B ......u can be sure that
{A* (x/A)*(y/A)} = B
=>(x*y)/A = B
as the LCM would be equal to the product of the HCF of the no.s and the LCM of the two no.s left after dividing x and y by their HCF .Also as the these two no. wont have any more factors as otherwise they would have got incorparated in the HCF so their LCM would be their product...(that is (x/A)*(Y/A) )
hence the LCM would be (A)*(x/A)*(y/A)

Answer 8

If you want an answer, you could send me an example of the question through email, and I might be able to solve it or get some help. Good luck!

Answer 9

no idea ... but tt's 1 FREAKY question!!
lol ... cool pic btw...