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

The answer is "72000".

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 = ((76 - b))/((b + 1))\\\\b = 2 \\\\a = (74)/(3)` not possible

`b = 4 \\\\ a =(72)/(5)` not possible

`b = 6\\ \\a = (70)/(7) = 10\\\\( 6 , 10 ) \ is \ one \ solution \\\\ b = 8 \\\\ a = (68)/(9) \ not\ possible \\\\`

`\to c+d+ cd = 54`

The same must be valid and interchangeable as above

`\to c = ((54 - d))/((d + 1))\\\\d = 2 \\\\ c = (52)/(3) \ not \ possible \\\\d = 4 \\\\ c = (50)/(5) \ =\ 10 \\\\( 4, 10) \\\\`solution

`d = 6 \\\\ c = (48)/(7) \\ 4, 10 \\\\\to a + b + c + d = 6 + 10 + 4 + 10 = 30 \\\\\to abcd = 6 * 10 * 4 * 10 = 2400\\\\(a+b+c+d)\cdot a \cdot b \cdot c \cdot d = 30 * 2400 = 72000\\\\\to (a+b+c+d) \cdot a \cdot b \cdot c \cdot d = 72000\\\\`