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find a,b,c

Albert

Well-known member
Jan 25, 2013
1,225
a,b,c are positive integers
(a-b)(b-c)(c+a)=- 90
(a-b)(b+c)(c-a)=42
(a+b)(b-c)(c-a)=- 60
find a,b,c
 

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
a,b,c are positive integers
(a-b)(b-c)(c+a)=- 90
(a-b)(b+c)(c-a)=42
(a+b)(b-c)(c-a)=- 60
find a,b,c
My first solution:
Consider $(a-b)(b+c)(c-a)=42=2\cdot7\cdot3$, we know we must have $c>a>b>0$ and by applying AM-GM inequality to the terms $(a-b),\,(b+c),\,(c-a)$ gives $\dfrac{(a-b)+(b+c)+(c-a)}{3}\ge \sqrt[3]{42}$ or simply $c \ge 5.21$.

An educated guess by observing again the equality of $(a-b)(b+c)(c-a)=42$ reveals that $c=6,\,b=1$ and hence $a=3$ work!:eek:


Second solution:
If we are to add the last two given equations, we have

$(c-a)((a-b)(b+c)+(a+b)(b-c))=42-60$

Expand and simplifying we get

$b(c-a)^2=9=1\cdot3^2$

This tells us $b=1$, $c-a=3$ and replacing $c-a=3$, $b=1$ into $(a-b)(b+c)(c-a)=42$ we have

$(a-1)(1+c)(3)=42$

$(a-1)(1+3+a)=14$

$a=-6$ or $a=3$

Since $a>0$, $\therefore a=3$ and hence $c=6$.

Note that if we take $b=9$ and $c-a=1$, we ended up with $a=11,\,b=9,\,c=12$ but this set of solution doesn't satisfy the condition of the given equation.
 
Last edited:

kaliprasad

Well-known member
Mar 31, 2013
1,322
a,b,c are positive integers
(a-b)(b-c)(c+a)=- 90
(a-b)(b+c)(c-a)=42
(a+b)(b-c)(c-a)=- 60
find a,b,c
Asked by you July 19 2013 and answered by me
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,755
Asked by you July 19 2013 and answered by me
Here is the link:

http://mathhelpboards.com/challenge-questions-puzzles-28/find-b-c-5637.html

Given all the problem generously posted by some of our members, this is bound to happen from time to time. :D
Oh My...I was totally unaware of that challenge problem been posted before at our site, and even if I did see it before, I didn't remember it at all.

Fortunately my solutions don't match with either kaliprasad's or Opalg's method, or else I may risk myself for being viewed as the "copycat" at our site. :eek:
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
My first solution:
Consider $(a-b)(b+c)(c-a)=42=2\cdot7\cdot3$, we know we must have $c>a>b>0$ and by applying AM-GM inequality to the terms $(a-b),\,(b+c),\,(c-a)$ gives $\dfrac{(a-b)+(b+c)+(c-a)}{3}\ge \sqrt[3]{42}$ or simply $c \ge 5.21$.

An educated guess by observing again the equality of $(a-b)(b+c)(c-a)=42$ reveals that $c=6,\,b=1$ and hence $a=3$ work!:eek:


Second solution:
If we are to add the last two given equations, we have

$(c-a)((a-b)(b+c)+(a+b)(b-c))=42-60$

Expand and simplifying we get

$b(c-a)^2=9=1\cdot3^2$

This tells us $b=1$, $c-a=3$ and replacing $c-a=3$, $b=1$ into $(a-b)(b+c)(c-a)=42$ we have

$(a-1)(1+c)(3)=42$

$(a-1)(1+3+a)=14$

$a=-6$ or $a=3$

Since $a>0$, $\therefore a=3$ and hence $c=6$.

Note that if we take $b=9$ and $c-a=1$, we ended up with $a=11,\,b=9,\,c=12$ but this set of solution doesn't satisfy the condition of the given equation.
we know we must have c>a>b>0

I do not know the basis for the same.