Math Is Fun Forum

jozou

Member

Registered: 2011-07-31

Posts: 8

divisibility of numbers

Find all integers of the form 1 ≤ a < b < c, s.t.:
1* ∃k a+b = kc
2* ∃l a+c = lb
3* ∃m b+c = ma

Attempt: Well, I have answer that solution is of the form a, 2a, 3a. I can prove that if b = 2a, then c = 3a:
From 2* we have c = a(2l - 1)
Then from 1* and previous we have 3a = ka(2l - 1). So k = 1 or k = 3:
If k = 3, then
1 = 2l - 1, l = 1, a +c = 1.b = 2a what contradicts assumption that a < c.
So k = 1. Then a + b = 3a = 1.c

But I can't prove that b=2a. Any hints? Thanks!

Last edited by jozou (2011-08-15 07:33:57)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: divisibility of numbers

hi jozou,

This is really just one question, not three?

And must k, L and m be integers?

So far I've managed to show there are no solutions ???

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: divisibility of numbers

hi jozou

i have found an equation which links k,l and m,but i am not sure if it is true:

if you can find any numbers that satisfy this equation than you can probably find a,b and c by solving a system of equations given by the first three equations you posted.

Last edited by anonimnystefy (2011-08-15 02:51:44)

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“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: divisibility of numbers

hi anonimnystefy

I think it's right.  I got that too.

Then the only integer solutions I could find for k, L and m are 1,3,3 and 2,2,2.

Neither seemed to lead to a set of values for a, b and c.

Bob

Last edited by Bob (2011-08-15 03:02:15)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: divisibility of numbers

well did you try solving:
a+b=c
a+c=3b
b+c=3a
?
or:
a+b=2c
a+c=2b
b+c=2a
?

---

“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: divisibility of numbers

hi anonimnystefy

Both.

former leads to a = b

latter leads to b - c  = 2(c - b) =><=

??

later edit:

Oh silly me.  I should have read your first post more carefully.

k =1, L = 2; m = 5 works as well and leads to the a; b=2a; c=3a solution you have found.

So remains to show:

that no other solutions to k + L + m + 2 = kLm exist.

working on that now BST 20.12

Bob

Last edited by Bob (2011-08-15 07:12:16)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

jozou

Member

Registered: 2011-07-31

Posts: 8

Re: divisibility of numbers

hi jozou

i have found an equation which links k,l and m,but i am not sure if it is true:

if you can find any numbers that satisfy this equation than you can probably find a,b and c by solving a system of equations given by the first three equations you posted.

Can you please expand, how you got

?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,626

Re: divisibility of numbers

hi jozou

What I did was eliminate a,b and c from the original equations.  It's 3 simultaneous equations with three unknowns (a,b,c).  This leaves a single equation in k, L and m.

I have worked through all other possibilites for k, L and m now.  1,3,3;   2,2,2; and 1;2;5 are the only solutions for anonimnystefy's equation.  They do not all lead to solutions however as (i) you have the a < b < c restriction plus they must be integers.

Can you do the elimination yourself or do you want me to post my working?

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

anonimnystefy

Real Member

From: Harlan's World

Registered: 2011-05-23

Posts: 16,049

Re: divisibility of numbers

hi jozou

here it is:

---

“Here lies the reader who will never open this book. He is forever dead.
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.

jozou

Member

Registered: 2011-07-31

Posts: 8

Re: divisibility of numbers

Thanks guys. I have nice proof now (Im not an author):
From 1* we have kc = a+b < c+c. So k < 2 and k must be 1.
Next from 2* we have that l > 1 (otherwise a = 0). 2a + b = lb < 3b. So l < 3 and l must be 2.
From 2* now we have b = 2a. Finally from a+b = c we have a + 2a = c.