a , b , c are positive integers satisfying the equations
5 a + 5 b + 2 a b 5 b + 5 c + 2 b c 5 c + 5 a + 2 c a = 9 2 = 1 3 6 = 2 4 4
Find 7 a + 8 b + 9 c .
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
A long method : Get value of b and c from equation first and third and form quadratic.
Quadratic Diophantine Equations - Solve by Simon's Favorite Factoring Trick is sufficient to solve this problem.
Consider the expansion, ( 5 + 2 a ) ( 5 + 2 b ) = 2 5 + 1 0 b + 1 0 a + 4 a b = 2 5 + 2 ( 5 a + 5 b + 2 a b ) = 2 5 + 2 ( 9 2 ) = 2 0 9 . Analogously, we can have the following system of equations:
( 5 + 2 a ) ( 5 + 2 b ) ( 5 + 2 b ) ( 5 + 2 c ) ( 5 + 2 a ) ( 5 + 2 c ) = = = 2 5 + 2 ( 9 2 ) = 2 0 9 2 5 + 2 ( 1 3 6 ) = 2 9 7 2 5 + 2 ( 2 4 4 ) = 5 1 3 .
Multiplying all these equations and take its positive square root:
( 5 + 2 a ) ( 5 + 2 b ) ( 5 + 2 c ) = 5 6 4 3
Solving for a , b and c :
5 + 2 c 5 + 2 a 5 + 2 b = = = 5 6 4 3 ÷ 2 0 9 = 2 7 ⇒ c = 1 1 5 6 4 3 ÷ 2 9 7 = 1 9 ⇒ a = 7 5 6 4 3 ÷ 5 1 3 = 1 1 ⇒ b = 3 .
The desired answer is 7 a + 8 b + 9 c = 7 ( 7 ) + 8 ( 3 ) + 9 ( 1 1 ) = 1 7 2 .
5a+5b+2ab=92 ........》
1
.
5c+5b+2bc=136 ........》
2
.5c+5a+2ca=244 ........》
3
.
On subtracting
1
from
2
We get
. 5c+5b+2bc-(5a+5b+2ab)=136-92
Or
5(c-a)+2b(c-a)=44
Or
(5+2b)(c-a)=11×4 (by factorizing 44 into suitable factors)
therefore b=3 & c-a=4 or c=a+4》
4
_
_
Similarly on subtracting
2
from
3
and factorizing gives
(5+2c)(a-b)=108 .
Now on substituting the value of
4
in the above equation we get
2
a
2
+ 7a -147=0
Or
(a-7)(2a+21)=0
_for positive value a=7 and therefore c=11(from 4) & required answer of the given expression is
1
7
2
_
N
o
t
e
In the first equation 11×4 is the only expansion that would satisfy the requirements of the question which demands for positive integral value.
( 2 a + 5 2 / 2 ) ( 2 b + 5 2 / 2 ) = 9 2 + 2 5 / 2
( 2 b + 5 2 / 2 ) ( 2 c + 5 2 / 2 ) = 1 3 6 + 2 5 / 2
( 2 c + 5 2 / 2 ) ( 2 a + 5 2 / 2 ) = 2 4 4 + 2 5 / 2
a ′ b ′ = 2 0 9 / 2
b ′ c ′ = 2 9 7 / 2
c ′ a ′ = 5 1 3 / 2
b ′ = 2 9 7 / ( 2 c ′ )
c ′ = 5 1 3 / ( 2 a ′ )
a ′ 2 9 7 / ( 2 c ′ ) = 2 0 9 / 2
a ′ 2 9 7 / ( 2 ∗ 5 1 3 / ( 2 a ′ ) ) = 2 0 9 / 2
a ′ 2 9 7 / ( 5 1 3 / a ′ ) = 2 0 9 / 2
a ′ 2 = 3 6 1 / 2
a ′ = 3 6 1 / 2 = 1 9 2 / 2
b ′ = 2 0 9 / ( 2 ∗ ( 1 9 2 / 2 ) ) = 2 0 9 / ( 1 9 2 ) = 1 1 2 / 2
c ′ = 5 1 3 / ( 2 ∗ ( 1 9 2 / 2 ) ) = 5 1 3 / ( 1 9 2 ) = 2 7 2 / 2
2 a + 5 2 / 2 = 1 9 2 / 2
2 b + 5 2 / 2 = 1 1 2 / 2
2 c + 5 2 / 2 = 2 7 2 / 2
a = 7
b = 3
c = 1 1
7 a + 8 b + 9 c = 1 7 2
Problem Loading...
Note Loading...
Set Loading...
⎩ ⎪ ⎨ ⎪ ⎧ 5 a + 5 b + 2 a b = 9 2 5 b + 5 c + 2 b c = 1 3 6 5 c + 5 a + 2 c a = 2 4 4 . . . ( 1 ) . . . ( 2 ) . . . ( 3 )
( 1 ) : 5 a + 5 b + 2 a b ⇒ 2 5 ( a + b ) + a b a b = 9 2 = 4 6 = 4 6 − 2 5 ( a + b )
Since a and b are positive integers, then a + b is even.
\(\begin{array} {} \text{If } a + b = 2 & \Rightarrow ab = 1 & \Rightarrow 46 - \frac{5}{2}(a+b) = 41 & \color{red}{\text{No solutions for } a \text{ and } b} \\ \text{If } a + b = 4 & \Rightarrow ab = \{3,4 \} & \Rightarrow 46 - \frac{5}{2}(a+b) = 36 & \color{red}{\text{No solutions for } a \text{ and } b} \\ \text{If } a + b = 6 & \Rightarrow ab = \{6,8,9 \} & \Rightarrow 46 - \frac{5}{2}(a+b) = 31 & \color{red}{\text{No solutions for } a \text{ and } b} \\ \text{If } a + b = 8 & \Rightarrow ab = \{7,12,15,16 \} & \Rightarrow 46 - \frac{5}{2}(a+b) = 26 & \color{red}{\text{No solutions for } a \text{ and } b} \\ \text{If } a + b = 10 & \Rightarrow ab = \{9,16, \color{blue}{21}, 24, 25 \} & \Rightarrow 46 - \frac{5}{2}(a+b) = \color{blue}{21} & \color{blue}{\Rightarrow a, b = 3, 7} \end{array} \)
Comparing ( 2 ) and ( 3 ) , we note that b < a , therefore, b = 3 and a = 7 .
⇒ { ( 2 ) : ( 3 ) : 1 5 + 5 c + 6 c = 1 3 6 5 c + 3 5 + 1 4 c = 2 4 4 ⇒ c = 1 1 ⇒ c = 1 1
Therefore, 7 a + 8 b + 9 c = 1 7 2