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# help! fast!

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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$$a-d$$ ?

Dec 28, 2017

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Hint: Solve using 4 simultaneous equations:

Answer: a = 24 and b = 20 and c = 6 and d = 14

Dec 29, 2017
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thanks so much, guest!

Dec 29, 2017
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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

Dec 29, 2017
#4
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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   Dec 29, 2017
edited by CPhill  Dec 29, 2017