Question

A, b, c and d are positive integers, such that a+b+ab = 76, c+d+cd = 54. Find (a+b+c+d)·a·b·c·d.

Answer 1

Answer:

The answer is "72000".

Step-by-step explanation:

Given:

$\to a+b+ ab = 76 \\\\\to c+d+cd = 54$

A & b are interchangeable here now and must both be equal.  

b is unusual, so ab is strange if an is strange

$\ odd + \ odd + \ odd = \ odd$ Whereas 76 are also  

Even if an is odd, ab is even

$\to odd + even + even = odd$

So a & b must be uniform

$\to a+b+ ab = 76\\\\\to a (1 + b) = 76 - b \\\\ \to a = \frac{(76 - b)}{(b + 1)}\\\\b = 2 \\\\a = \frac{74}{3}$  not possible

$b = 4 \\\\ a =\frac{72}{5}$    not possible

$b = 6\\ \\a = \frac{70}{7} = 10\\\\( 6 , 10 ) \ is \ one \ solution \\\\ b = 8 \\\\ a = \frac{68}{9} \ not\ possible$

$\to c+d+ cd = 54$

The same must be valid and interchangeable as above

$\to c = \frac{(54 - d)}{(d + 1)}\\\\d = 2 \\\\ c = \frac{52}{3} \ not \ possible \\\\d = 4 \\\\ c = \frac{50}{5} \ =\ 10 \\\\( 4, 10)$solution

$d = 6 \\\\ c = \frac{48}{7} \\ 4, 10 \\\\\to a + b + c + d = 6 + 10 + 4 + 10 = 30 \\\\\to abcd = 6 \times 10 \times 4 \times 10 = 2400\\\\(a+b+c+d)\cdot a \cdot b \cdot c \cdot d = 30 \times 2400 = 72000\\\\\to (a+b+c+d) \cdot a \cdot b \cdot c \cdot d = 72000$