Cocassi Function

$ab+bc+ca=\frac{32+65+77}{2}=87$ $abc=\sqrt{(87-32)*(87-65)*(87-77)}=\boxed{110}$

$\begin{cases} a(b+c)=32 \\ b(c+a)=65 \\ c(a+b)=77 \end{cases}$

$\Rightarrow \begin{cases} ab+ca=32 \\ bc+ab=65 \\ ca+bc=77 \end{cases}$

Equation 2 - equation 1:

$\Rightarrow \begin{cases} bc-ca= 33 \\ ca+bc=77 \end{cases}$

Add the two equations together:

$\Rightarrow 2bc = 110 \quad \Rightarrow bc = 55$

Substituting $bc = 55$ in the appropriate equations, we get $ab = 10$ and $ca = 22$ .

Therefore,

$ab \times bc \times ca= 10 \times 55 \times 22$

$(abc)^2 = 2^25^211^2 = (110)^2\quad \Rightarrow abc = \boxed {110}$

process of elimination and a real simple trial and error works here too. one of the equations kinda gives this away.

only one combination of numbers gets you 77 and that is 7 and 11.

that means c=either 7 or 11 and (a+b) is the other.

using the value of c as 11 in the first equation makes b only possible as being 5 and a only possible as being 2. these values prove out on the middle equation as well.

2x5x11 = 110

Let ab=X

bc=Y

ac=Z

$Then, X+Z=32$

$X+Y=65$

$Y+Z=77$

$X+Y+Z=(32+65+77)/2=87$

$Hence, X=87-77=10$

$Y=87-32=55$

$Z=87-65=22$

$XYZ=(ab)(bc)(ac)=(abc)^{2}=55*22*10=11^{2}*10^{2}$

$abc=11*10=\boxed{110}$

I did trial and error: 65 had to be 5 x 13 and 77 = 7 x 11, noting that two of those four numbers, 5 and 11, add to 16, a factor of 32...QED

my method was a more novel one even though it was risky...I didn't solve it by algebra but by reasoning...for that I had to take the risk of assuming that a, b and c are integers!!! I started from the last equation and deduced that c could either be 11 or 7 as 77 has only these as its integral factors. Therefore b+a would be either 7 or 11. then I turned to the second equation and saw that b could either be 13 or 5 but if b would be 13 then b+a not= 7 and b+a not =11 as both a and b are positive hence we get b=5 and c+a=13... now as pointed out earlier c could either be 11 or 7 which gives a=2 or a=6... but looking at the first equation a can never be 6 b'cuz 6 is not a factor of 32, hence a=2.. so abc=2x5x11=110... It may look lengthy, but it is far more easier than solving the equations thing and certainly more professional..The only catch is the assumption that we took that all of them are integers which I was 95% sure would be the case!!!

simplify and perform addition and subtraction on equation

you will get ab: 10 bc: 55 ac: 22

multiply all, you ll get

ab bc ac = 55 22 10 ( 5 11 2 11 2*5)

take root of this

abc = 110

SOLVE : Given, a(b+c)=32 ---(1) b(c+a)=65 ---(2) c(a+b)=77 ---(3)

Now adding (1),(2),(3) => a(b+c) + b(c+a) +c(a+b) =32+65+77 =>2(ab+bc+ca) =174 [ ̇. ̇ a(b+c) + b(c+a) + c(a+b) = 2(ab+bc+ca) ] => ab+bc+ca = 87 We have to find the value of abc . abc = √ ̅ ̅ ̅(a²b²c²) = √ ̅(bc.ca.ab) = √ ̅ ̅ [{ab+bc+ca-a(b+c)} {ab+bc+ca-b(c+a)} {ab+bc+ca-c(a+b)}] = √ ̅ ̅ { (87-32) (87-65) (87-77) } = √ ̅ (55x22x10) = √ ̅ ̅ (12100) = 110

Yet another way. My idea was to factor the numbers and check for integer solutions, which I found right away, as there were not many combinations to try:

$\begin{cases} a(b+c)=2^5 \\ b(c+a)=5\cdot 13 \\ c(a+b)=7\cdot 11 \end{cases}$

In the 3rd equation, if $c=11$ , then $a+b=7$ . In the second, if $b=5$ , then $a$ must be $2$ and this value also satisfies $a+c=13$ (So far, so good!). We finally check if these values satisfy the first equation: $2\cdot(5+11)=32$ (They do!).

Therefore, $abc=2\cdot5\cdot11=\boxed{110}$

My solution won't work for every simultaneous equation, but it's quite simple for this one, and doesn't require the use of a calculator. First split each of the totals into their prime factors.

32 => 2^5

65 => 13 * 5

77 => 11 * 7

We notice that 65 and 77 have only two factors. This means that b must be 13 or 5 and c must be 11 or 7.

Using equation 1, b+c must be power of 2, so we can deduce that b must be 5 and c must be 11 to give b+c = 16.

a * 16 = 32: a = 2

a * b * c = 2 * 5 * 11 = 110

a(b+c)=32

b(c+a)=65

c(a+b)=77

that means that a is 2 or 4 or 8 or 16 or 32

and b is or 5 or 13

and c is 7 or 11

so b must be 5 and c must be 11 why ?

bec the sum of b + c must be a factor of 32 and 5 + 11 = 16 that an factor of 32

so a must be 2 why ?

bec 32 / (b + c) = a so 32 / 16 = 2

now abc = 2 * 5 * 11 = 110

a(b+c)= 32 = ab + ac

b(c+a)= 65 = bc+ac

c(a+b)= 77 = ac+bc

2ab+2ac+2bc=32+65+77=174

2ab+2ac+2bc-ab-ac-ab-ac= 2bc=110

bc= 55

2ab+2ac+2bc-ab-bc-ab-bc= 2ac=44

ac= 22

2ab+2ac+2bc-bc-ac-bc-ac= 2ab=20

ab= 10

ac*bc = 1210

abc*c when divided by ab = 1210 divided by 10

= c*c= 121

c= 11

therefore

ab c = 10 11

abc= 110_

by solving the three equations : ab = 10 bc = 55 ac=22 c=22/a b=10/a a=2 b=5 c=11 then abc = 110

expand equations into (1),(2),(3) find (2)-(1),(3)-(2),(3)-(1) and assume them as (4),(5),(6) find(3)+(6) to get bc=55 find (1)+(5) to get ac=22 divide bc and ac to get b/a=5/2--->b=5/2a substitute b=5/2a in (5) to get a=2 multiply a*bc to get abc=110 therefore, abc=110

a(b+c)=32 (1) b(c+a)=65 (2) c(a+b)=77 (3)

Cộng cả 3 vế 3 đẳng thức đã cho theo vế ta được : 2(ab+bc+ca)=174 Từ đó suy ra : ab+bc+ca=87 (4)

Từ (1)+(4) ta được : bc+a(b+c)=87 => bc=87-a(b+c)=87-32=55 Tương tự : (2)+(4) : ca=22 (3)+(4) : ab=10 Nhân các vế lại với nhau ta được : (abc)^2=55x22x10=12100 => abc=110 (ok)

ab+ac=32 ab=32-ac bc+ba=65
ca+cb=77 ac=77-cb

bc=32-ac=65 bc=32+cb-77=65 bc=55 55+ba=65 ba=10 ca=22 (55/b)(10/b)=22 22 b b=550 b=5 c=11 a=2 abc=110

a(b+c)=32, i.e. ab+ac=32 ...(1)

b(c+a)=65, i.e. bc+ab=65 ...(2)

c(a+b)=77, i.e. ac+bc=77 ...(3)

subtracting (2) from (1)

ac-bc = -33 ...(4)

adding (4) and (3)

2ac=44

ac=22

from (1) ab=10

from (3) bc=55

now, ac * ab * bc = 22 * 10 * 55 = 12100

therefore, (abc)^2 = 12100

therefore, abc = 110

$When\quad I\quad removed\quad the\quad brackets\quad I\quad got\\ ab+ca=32\quad ...(B)\\ bc+ab=65\quad ...(C)\\ ca+bc=77\quad ...(A)\\ Nothing\quad special?May\quad be.\\ I\quad got\quad value\quad of\quad 2ab+2ca+2bc\quad staring\quad at\quad me\\ And\quad values\quad of\quad ab,\quad bc\quad \& \quad ac\quad smiling\quad \quad at\quad me.\\ what\quad to\quad do\quad with\quad ab,\quad bc\quad \& \quad ac?multiply?\\ Yeah!That's\quad the\quad solution\quad dissolving\quad the\quad problem.\\ using:\\ ab+ca+bc=87.\quad \& \\ ab+A=87;bc+B=87;ca+C=87,\\ got\quad ab,bc\& ca(10,55\& 22\quad respectively)\\ \\ Now,lets\quad dissolve\quad it:\quad multiplying\quad ab,bc\& ca\\ we\quad get\quad this:({ abc) }^{ 2 }=22\times 10\times 55\\ giving\quad abc=110.Tada!\\ \\$

Note that from subtracting the three equations:

ac + cb - bc - ab - ab - ac = 77 - 65 - 32 = -20

ab = 10

So from first equation then:

10 + ac = 32 => ac = 22

Second equation:

bc + 10 = 65 => bc = 55

Now:

(ab)(ac)(bc) = 10 22 55

abc = sqrt(10 22 55) = 110

$\begin{cases}a(b+c)=32\\b(c+a)=65\\c(a+b)=77\end{cases}$

adding $1st\,$ with $2nd\,$ and subtracting $3rd\,$ :

$ab=10;\,$ (i)

adding $1st\,$ and $3rd\,$ and subctracting $2nd\,$ :

$ac=22;\,$ (ii)

adding $2nd\,$ and $3rd\,$ and subctracting $1st\,$ :

$bc=55;\,$ (iii)

from (i): $a=\frac {10} b ;\,$ substituting in (ii) and (iii):

$\frac {10c} b = 22; c = \frac {22b} {10};$ multiplying by $b\,$ and remembering (iii):

$\frac {22b^2} {10} = 55;\,{b^2} = 25;\, b = 5;\,$ (iv)

substituting in (iii) and (i):

$c = 11;\, a = 2;$

$a, b, c = (2, 5, 11);\, abc = \boxed {110}$

Add all 3 solutions, we'll get 2(ab+bc+ca)= 174 hence ab+bc+ca=87............equation-4 now, subtract, eq-1 from eq-4 eq-2 from eq-4 eq-3 from eq-4, which gives bc=55 ac=22 ab=10 From bc and ac "c is common" and in 55 and 22 "11 is common factor" so, c=11 a=2, b=5 hence, abc=110

Adding all eqtn and dividing by 2 we get ab+ac+bc=87 Using this in all the eqn we get ab=10 bc=55 ac=22 ab*ac=a²bc=220 Hence,a²=4 a=2 And so on, b=5 c=11 So, abc=110

Adding the equations,

2(ab+bc+ca) = 174,

ab+bc+ca = 87.

Subtracting the three given equations from this equation to obtain bc = 55, ac = 22, and ab = 10. Multiplying these three new equations to yield

(abc)^2 = 5 * 11 * 2 * 11* 2 * 5,

abc = 5 * 11 * 2 = 110.

Note: There are two correct answers, 110 and -110, since a=-2, b=-5, c=-11 is solution too.

a(b+c)=32 b(c+a)=65 c(a+b)=77 now add those 3 eq ab+bc+ac= 174/3=87 abc=((87-32)*(87-65)8(87-77))1/2 abc=110

subtracting eq 3 of 2, eq 3 of 1 and eq 2 of 1 we get 4: a(c-b)=12 5: b(c-a)=45 6: c(b-a)=33 so we get that 1<=a<b<c from eq 3 we know that c must be 7 or 11, and eq 6 tells us that c is 3 or 11, so c=11 similarly, eq 2 give us b=5 or 13 ; eq 5 says that b can't be worth 13, so b=5 now we may activate eq 1 and get that a=2 a b c=110

a(b+c)=3 ...equ(1) b(c+a)=6 ...equ(2) c(a+b)=77 ...equ(3) from equ(1)+equ(2)+equ(3)... ab+bc+ca=87 ...equ(4) solving equ(4) & equ(1)..we get bc=55 ...equ(5) simlary by solving equ(2) & equ(3) with equ(4) we get ab=10 ...equ(6) ca=22 ...equ(7) Now multiply equ(5),equ(6) & equ(7) { a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }=12100 abc=sqrt { 12100 } abc=110

There are so many ways to solve the system equations. (i) trial and error method: solve eq (i) first a=2,b=5,c=11 hence first condition is fulfilled and equations too. so (abc) =2 5 11=110. (ii) simultaneous equations: Note that subtracting equations (2) and (3) from the first equation, we get ac + cb - bc - ab - ab - ac = 77 - 65 - 32 = -20 ab = 10 So from first equation then: 10 + ac = 32 => ac = 22 Second equation: bc + 10 = 65 => bc = 55 Now (ab)(ac)(bc) = 10 22 55 (abc)^2= (10 22 55) abc = sqrt(10 22 55) = 110

c(a+b)-a(b+c)=77-32=45 b(c-a)=45 and b(c+a)=65 (c-a)/(c+a)=45\65=9\13 (c-a)+(c+a)/(c-a)-(c+a)=9+13/9-13=>2c/2a=22/4=11/2 c=11,a=2 then b=5 so abc=110

Multiplying given equations,and sutracting then adding we get 2bc =110 so bc ==55,substituting this value we getvalues of a=2,b=5 ans c = 11 therefore,abc =110. also by clue from value 77 equation we can guess value of c =11 therefore a+b will be 7 ,then we get a=2 and b =5 Ans K.K.GARG,india

to all its an easy one but interesting one try my previous question and follow me if you like my problems

(ab+bc+ca)=(32+65+77)/2 =87,

b(a+c)=87-c*a,

5(11+2)=87-(2*11),

65=65,

a=11, b=5, c=2,

abc=a b c =11x5x2=110.

a=2,b=5,c=11