[Guest]

This is very hard. Please don't guess cos u will get it wrong. (anyway if u don't know just guess)

[Guest]

WLOG a<=b<=c<=d. Suppose abcd is no divisible by 3 or 5, then none of a,b,c,d is divisible by 3 or 5.

Since a+b+c+d is lcm(a,b,c,d), the four number cannot be the same.

Now, d divides a+b+c+d<4d, so we have that a+b+c=2d or a+b+c=d.

If a+b+c=2d then a+b+c+d=3d, since a+b+c+d is lcm(a,b,c,d) and is divisible by 3, then one of a,b,c,d must be divisible by 3.

Therefore a+b+c=d. We then have d=a+b+c<=3c, therefore 2d=a+b+c+d=2a+2b+2c= 3c,4c,5c or 6c.

Since d is not divisible by 3 or 5, we must have d=2c.

Now, a+b+c=d=2c, therefore a+b=c and a+b+c+d=4(a+b). So a divides 4(a+b) and consequently divides 4b. Analogously b divides 4a.

Since a<=b, we must have b= a,2a or 4a.

If b=a or if b=2a then all of a,b,c,d divide 2(a+b) and a+b+c+d=4(a+b) would not be the lcm.

So b=4a and lcm(a,b,c,d)=20a. Since 5 divides the lcm, 5 must divide one of a,b,c,d. A contradiction.

[Guest]

If: a=1, b=2, c=6, d=9.

Then: 1+2+6+9=18

18 is also the least common multiple.

1x2x6x9=108, which is divisible by three.

[Guest]

hmmmm.....

Edited April 15, 2010 by Lady Chem

[Guest]

I agree with Adogg's solution, but is 'abcd' the same as 'a X b X c X d'? Example: for numbers 1,2,6,9 it would have also meant 1269 as a number instead of 1 X 2 X 6 X 9. Don't know what the OP meant... Just a thought...

[Guest]

I agree with Adogg's solution, but is 'abcd' the same as 'a X b X c X d'? Example: for numbers 1,2,6,9 it would have also meant 1269 as a number instead of 1 X 2 X 6 X 9. Don't know what the OP meant... Just a thought...

In math/algebra having abcd or ab or anything like that is always short for a X b X c X d or a X b. Always.

[random7]

WLOG a<=b<=c<=d. Suppose abcd is no divisible by 3 or 5, then none of a,b,c,d is divisible by 3 or 5.

Since a+b+c+d is lcm(a,b,c,d), the four number cannot be the same.

Now, d divides a+b+c+d<4d, so we have that a+b+c=2d or a+b+c=d.

If a+b+c=2d then a+b+c+d=3d, since a+b+c+d is lcm(a,b,c,d) and is divisible by 3, then one of a,b,c,d must be divisible by 3.

Therefore a+b+c=d. We then have d=a+b+c<=3c, therefore 2d=a+b+c+d=2a+2b+2c= 3c,4c,5c or 6c.

Since d is not divisible by 3 or 5, we must have d=2c.

Now, a+b+c=d=2c, therefore a+b=c and a+b+c+d=4(a+b). So a divides 4(a+b) and consequently divides 4b. Analogously b divides 4a.

Since a<=b, we must have b= a,2a or 4a.

If b=a or if b=2a then all of a,b,c,d divide 2(a+b) and a+b+c+d=4(a+b) would not be the lcm.

So b=4a and lcm(a,b,c,d)=20a. Since 5 divides the lcm, 5 must divide one of a,b,c,d. A contradiction.

Bravo! I don't know how you did that within a couple hours! I was not able to prove it, but I'm glad that someone did though you might want to practice writing your proofs such that people can easily understand them; it took me a while to decode a lot of your logical steps, due to the vagueness.

If: a=1, b=2, c=6, d=9.

Then: 1+2+6+9=18

18 is also the least common multiple.

1x2x6x9=108, which is divisible by three.

As an extension to this:

LCM(P, 2P, 2x3P, 3²P) = 2x3²P = 18P (where P = P₁xP₂x...xP_n and where P₁,...,P_n are prime numbers other than 2 or 3)

And; P+2P+6P+9P = 18P

And; 3 divides (Px2Px6Px9P)

Thus, {P, 2P, 6P, 9P} is a family of infinite examples for a, b, c and d.

[Guest]

In math/algebra having abcd or ab or anything like that is always short for a X b X c X d or a X b. Always.

Well, I know that in Math abcd always implies a X b X c X d, but I was trying to think out of the box... forget about it if I am wrong...

[Guest]

I got

2,3,15,10

Is that right?

their least common multiple is the same as a+b+c+d

and abcd is both divisible by 5 and 3

[Guest]

Off topic, how do you add a signature?

[Guest]

Off topic, how do you add a signature?

EDIT: NVM, I found out

[Guest]

An answer would be 1,2,3,6

The LCM of all 3 numbers is 12

1*2*3*6=36, which is divisible by 3

[random7]

An answer would be 1,2,3,6

The LCM of all 3 numbers is 12

1*2*3*6=36, which is divisible by 3

LCM(1,2,3,6)=6

[random7]

As an extension to this:

LCM(P, 2P, 2x3P, 3²P) = 2x3²P = 18P (where P = P₁xP₂x...xP_n and where P₁,...,P_n are prime numbers other than 2 or 3)

And; P+2P+6P+9P = 18P

And; 3 divides (Px2Px6Px9P)

Thus, {P, 2P, 6P, 9P} is a family of infinite examples for a, b, c and d.

Actually, P can be any positive integer (ie. any natural number). For other people that find a set of 4 numbers, you can do the same.

eg. nhan1st's answer of (2,3,15,10) could be generalised with (2n,3n,10n,15n)