How do I factor each of these?

Number Theory Level 2

$\begin{aligned} ab+cd&=&38 \\ ac+bd&=&34\\ ad+bc&=&43\\ \end{aligned}$

Let $a,b,c$ and $d$ positive integers such that they satisfy the system of equations above. Find $a+b+c+d$ .

4 solutions

Paola Ramírez
Feb 8, 2015

$ac+bd+ad+bc=34+43$

$ac+bd+ad+bc=77$

$(a+b)(c+d)=77$

$\Rightarrow a+b=11$ and $c+d=7$ or $a+b=7$ and $c+d=11$ .

Note that: $a+b$ can not be $1$ becase $a$ and $b$ are positivve integers consequently $c+d$ can not be $77$ .

$\therefore \boxed{a+b+c+d=7+11=18}$

Of course, you will also have to prove that a positive integer solution does exist.

Shourya Pandey - 6 years, 4 months ago

(c+d)(a+b)=77----> (7)(11), 7+11=18;
(a+d)(c+b)=72----> (12)(6),12+6=18;
(a+c)(b+d)=81----> (9)(9) ,9+9=18 .

Ahmed Kamal - 6 years, 4 months ago

Prabhasa K
Feb 10, 2015

Resolving Shourya Pandey's doubt with another tedious method of solving:

Subtract equations in a pair. You get 3 sets of products.
(a-c)(d-b)=5 has only 2 possible combinations. Find (a-d)+(b-c) from this condition.
Use (a-d)(b-c)=4 & solve for (a-d) and (b-c), all the terms in step 1 are evaluated.
Substitute b,c,d in terms of a in any equation.
a=7 is a solution. Also, b=4, c=2, d=5. Hence positive integer solution exists.

Gamal Sultan
Feb 11, 2015

ab + cd = 38 ........................... (1)

ac + bd = 34 ............................(2)

ad + bc = 43 ............................(3)

Adding (2), (3) we get

(a + b)(c + d) = 77

Then

a + b = 7 or 11

And

c + d = 11 or 7

In both cases we obtain

a + b + c + d = 18

Neeraj Agarwal
Feb 11, 2015

Add 1 and 2 , 2 and 3 , 3 and 1.
we get,
(a+b)(c+d)=77,(a+c)(b+d)=81, (a+d)(b+c)=72.

11 7 =77, 9 9 =81, 12 6 =72

So, 11+7 = 9+9= 12+6=18