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#1 2016-11-17 22:42:01

iamaditya
Member
From: Planet Mars
Registered: 2016-11-15
Posts: 744

a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

In my maths test there came this question:


a+b+c=10

a²+b²+c²=38

a³+b³+c³=160

Find the values of a,b and c.

Please tell me the answer with explanation.


Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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#2 2016-11-17 23:32:37

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

Hi;

I got this far:

Here are the divisors of 280

{1, 2, 4, 5, 7, 8, 10, 14, 20, 28, 35, 40, 56, 70, 140, 280}

It is easy to finish now.

Welcome to the forum.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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#3 2016-11-18 18:00:51

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#4 2016-11-18 19:08:00

iamaditya
Member
From: Planet Mars
Registered: 2016-11-15
Posts: 744

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

Hi, thickhead yours is the right solution. Thanks for helping me

But thickhead I couldn't get how did you simplified from equation (2) to equation(4) i.e.
ab=30/c and a+b=10-c
so  c³-10c²+31c-30=0

In (2) first term ab is clear ;second and third terms have c common i.e.c(a+b)=c(10-c). i hope you can simplify further.


Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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#5 2016-11-18 19:45:03

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

In (2) first term ab is clear ;second and third terms have c common i.e.c(a+b)=c(10-c). i hope you can simplify further.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#6 2016-11-20 04:03:08

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

Hi iamaditya,(really it is false;it is you who is aditya)
Suppose only first and second equations are given with a,b,c as positive integers. Can you evolve a method to find the numbers? Not purely hit and try method (Ram bharosa method) but with some logic in it.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#7 2016-11-20 23:35:42

iamaditya
Member
From: Planet Mars
Registered: 2016-11-15
Posts: 744

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

If only the first 2 equations i.e. a+b+c=10 and a²+b²+c²=38 are given then it is impossible to work out as there are three unknown variables and only 2 equations.


Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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#8 2016-11-20 23:53:51

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

But then there is the condition that they are only natural numbers.That limits the scope of the solution and gives only one solution.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#9 2016-11-21 15:29:47

Monox D. I-Fly
Member
Registered: 2015-12-02
Posts: 952

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

iamaditya wrote:

If only the first 2 equations i.e. a+b+c=10 and a²+b²+c²=38 are given then it is impossible to work out as there are three unknown variables and only 2 equations.

By trial and error, I find the solutions are 2, 3, and 5, though I don't know how to do it algebraically.

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#10 2016-11-21 15:41:20

thickhead
Member
Registered: 2016-04-16
Posts: 1,086

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

Trial and error no dobt,but systematically. e.g. if you assume a<b<c  since there is nothing to choose among them, c has very limited value.To start with c<=6. once c=6 fails you come to c=5.


{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}
(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha
{Gods rejoice at those places where ladies are respected.}

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#11 2016-11-23 23:18:26

iamaditya
Member
From: Planet Mars
Registered: 2016-11-15
Posts: 744

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

Yeah thickhead you are right.


Practice makes a man perfect.
There is no substitute to hard work
All of us do not have equal talents but everybody has equal oppurtunities to build their talents.-APJ Abdul Kalam

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#12 2016-11-24 03:02:05

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Re: a+b+c=10 a²+b²+c²=38 a³+b³+c³=160 find a,b,c

By trial and error, I find the solutions are 2, 3, and 5, though I don't know how to do it algebraically.

That guy in post #2 whittles down the possibilities.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Offline

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