+885 Four positive integers \(a\), \(b\), \(c\), and \(d\) have a product of 8! and satisfy \(\begin{align*} &ab+a+b=524,\\ &bc+b+c=146, \text{ and}\\ &cd+c+d=104. \end{align*}\)What is\(a-d\) ?
Since you wanted the answer FAST, here is a FAST answer!!!.
Hint: Solve using 4 simultaneous equations:
Answer: a = 24 and b = 20 and c = 6 and d = 14
+9673 ab + a + b = 524
ab + a + b + 1 = 525
(a+1)(b+1) = 525.
bc + b + c = 146
(b+1)(c+1) = 147.
cd + c +d = 104
(c+1)(d+1) = 105.
525 = 3 * 5^2 * 7
147 = 3 * 7^2
105 = 3*5*7
Most likely b+1 = 21, c+1 = 7 by finding GCD.
so b = 20, c = 6.
Now we find a and d using the values of b and c and the equations bolded.
21(a+1) = 525
a+1 = 25
a = 24
7(d+1) = 105
d+1 = 15
d = 14
So, a-d = 24 - 14 = 10
+129852 abcd = 8! = 40320
ab + a + b = 524 (1)
bc + b + c = 146 (2)
cd + c + d = 104 (3)
Manipulating (1) and (2) we have that
(a + 1) (b + 1) = 525 ⇒ (b + 1) = 525 / (a + 1)
(b + 1) (c + 1) = 147 ⇒ (b + 1) = 147 / (c + 1)
Which implies that
525 / (a + 1) = 147/ (c + 1)
525/147 = 25/7 = (a + 1) / ( c + 1)
Which implies that a = 24 and c = 6
Which implies that
b + 1 = 525/25 ⇒ b = 525/25 - 1 = 21 - 1 = 20
And manipulating (3), we have
(c + 1) (d + 1) = 105
(7) (d + 1) = 105
d + 1 = 15
d = 14
So
a - d = 24 - 14 = 10
