Question

a>b>c>0, what could be the possible values of a,b,c if abc=ab+bc+ca

Answer 1

welcome to Open Study.

Answer 2

Thank you

Answer 3

one set of values are 3,6,10

Answer 4

3,6,10 dont satisfy the given equation right ?

Answer 5

And, any restrictions given on a,b,c ? are they integers ?

Answer 6

no more info given....sorry the given set is wrong, 3,6.10 :(

Answer 7

i think there wil be finite # of solutions only if a,b,c are integers

Answer 8

a,b,c cant be fractions, 'cos then i divided by the number exceeds 1 in that case

Answer 9

can you think of a way to get the solutions?

Answer 10

\(abc=ab+bc+ca\)

divide abc thru out

\(1=\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\)

only one solution : \((2, 3, 6)\)

Answer 11

no more solutions possible in integers

Answer 12

how did you make sure?

Answer 13

Notice that \(\dfrac{1}{3} +\dfrac{1}{3}+ \dfrac{1}{3} = 1\),

mess with it - fix one value and see what values other two can take

Answer 14

here is a nice explanation :
http://krismanohar.com/blog/?p=157

Answer 15

Thank you for all the help :)

Answer 16

you're welcome :)

Answer 17

i wish there were a generalized solution for solving for a1, a2, a3...an...with the condition 1/a1+1/a2 ....=1 like mentioned in the blog you posted

Answer 18

yeah should be possible... definitely worth trying xD

Answer 19

solve in distinct positive integers :

\(\dfrac{1}{a_1} + \dfrac{1}{a_2} + ... + \dfrac{1}{a_n} = 1\)

Answer 20

trying with 4 integers...

Answer 21

try (2,4,6,12)

Answer 22

it looks extremely difficult to cookup a formula

Answer 23

wow....u got dat

Answer 24

it feels frightening to just plug and play with numbers

Answer 25

Hey Ganeshie8, found a nice blog on how to solve such ones, http://mathforum.org/library/drmath/view/56821.html

Answer 26

Ahh nice :) so basically they're doing :

\[ 1 = \dfrac{1}{2} + \left(1 - \dfrac{1}{2}\right)\]

Answer 27

\[ 1 = \dfrac{1}{2} + \dfrac{1}{3} + \left(1 - \left[\dfrac{1}{2} + \dfrac{1}{3}\right]\right)\]

Answer 28

\[ 1 = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \left(1 - \left[\dfrac{1}{2} + \dfrac{1}{3}+ \dfrac{1}{7}\right]\right)\]

Answer 29

^gives the solution : \((2,3,7, 42)\)

Answer 30

yes!!! Thank God i found that site...a blessing!!! Thank you also for being there, supporting!!!

Answer 31

np :) lol i was almost giving up... and u gave me that site xD thank you too :))

Answer 32

for 5 ints you have: (2,3,7,43,1806)

Answer 33

yeah and a plenty more
http://www.wolframalpha.com/input/?i=1+-%281%2F2%2B1%2F3%2B1%2F7%2B1%2F43%29

Answer 34

i think wolfram is struggling
http://www.wolframalpha.com/input/?i=solve+in+integers+1%2Fa%2B1%2Fb%2B1%2Fc%2B1%2Fd%2B1%2Fe+%3D+1

Answer 35

That's nice site, thank you again!

Answer 36

i think 1806 is max value possible for a/b/c/d/e, so it looks like there will not be infinite combinations for 5 ints

Answer 37

for 4 ints, also i think 42 is the max possible value for a/b/c/d, so these solutions are also finite

Answer 38

but not fully sure... il need to think bit more thoroughly

Answer 39

by the way, according to the method given on Dr.Math for 5 ints you have: (2,3,7,43,1806)
should work...not sure why the difference ends up in 1/1806...oh...that's b'cos it takes the decimals and computes the answer so there's a difference coming of 1/1806

Answer 40

the above post was wrong...i messed up thinking 1/1806 is got after unit factors of 5 ints are subtracted from 1..

Answer 41

but Dr.Math said that there are infinite solutions possible?

Answer 42

yeah need to think a bit more - consider 4ints :
can there be solutions such that any one of a,b,c,d is greater than 42 ?

Answer 43

im thinking the value 42 is the upper limit for 4ints, il think more on this later... for now i need to go work on something else...

Answer 44

sure, TC

Answer 45

solution : \((a, b, c, d)\) all \(\le 42\) is what my best guess is. see if u can give a counter example

Answer 46

have fun :)