+0  
 
0
967
5
avatar+1760 

Four positive integers $a$$b$$c$, and $d$ satisfy
\begin{align*} ab + a + b &= 524, \\ bc + b + c &= 146, \\ cd + c + d &= 104, \\ \end{align*}
and $abcd=8!$.

What is $a - d$?

 

(b) In Part a of this problem, you found the values of integers that satisfied the system of equations. In this part of the problem you will write up a full solution that describes how to solve this problem.

Four positive integers $a$$b$$c$, and $d$ satisfy
\begin{align*} ab + a + b &= 524, \\ bc + b + c &= 146, \\ cd + c + d &= 104, \\ \end{align*}
and $abcd=8!$.

What are $a$$b$$c$, and $d?$

Mellie  Jul 7, 2015

Best Answer 

 #1
avatar+91045 
+20

Four positive integers a, b, c, d satisfy

 

 

ab+a+b = 524 

bc+b+c = 146 

cd+c+d = 104 

abcd = 8! = $$2^7*3^2*7*5$$

 

What is   a-d ?


----------------------------------------

 

Can    a,b,c or d   be odd?

Assume a is odd

Let a=2N+1

(2N+1)b+(2N+1)+b=524

2Nb+b+2N+1+b=524

2Nb+2b+2N=523

2(2Nb+b+N)=523

Since N and b are integers there is a contradiction here.

Therefore a is even

By the same logic b,c and d are all even and therefore all are divisible by 2.

 

Let 2A=a,   2B=b,     2C=c,    2D=d

 

The equations become

 

2AB+A+B=262      It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73        It can be seen that B+C is odd      therefore with B & D one is odd and one is even

2CD+C+D=52        It can be seen that C+D is even    therefore C and D are both odd or both even

 

This means that the powers of 2 must be shared into two bundles.

They belong either to A and B          or              to    C and D

 $$A*B*C*D=2^3*3^2*7*5$$

 

-----------------------

 

Can A,B,C, OR D  be 1 ?

Assume A=1

2B+1+B=262

3B=261

B=87 = 3*29  no that is impossible so   $$B\ne1$$    and using the same logic   $$A\ne1$$

 

Assume C=1

2D+1+D=52

3D=51

D=17     again that is impossible so  $$c\ne1$$   and using the same logic   $$D\ne1$$

 

--------------------------

 

 

 

 

2AB+A+B=262       It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73         It can be seen that B+C is odd      therefore with B & D one is odd and one is even

2CD+C+D=52         It can be seen that C+D is even    therefore C and D are both odd or both even

$$A*B*C*D=2^3*3^2*7*5$$ 

 

 

Can A,B,C, OR D  be 2 ?

Assume A=2

4B+2+B=262

5B=260

B=52=2^2*13      no that is impossible so   $$A\ne 2$$   and using the same logic   $$B\ne2$$

 

Assume C=2

4D+2+D=52

5D=50

D=10      That is possible – so maybe C=2 and D=10   OR    D=2 and C=10

 

If D=2 then C=10

20B+B+10=73

21B=63

B=3

If B=3  then

6A+A+3=262

7A=259

A=37     That is impossible      so            $$D\ne2\qquad B\ne3 \qquad C\ne10$$    

 

If C=2  then D=10

4B+B+2=73

5B=71     That is impossible       so           $$D\ne10 \qquad C\ne 2$$       

 

So none of A,B,C,D are 1 or 2

 

 

  $$A*B*C*D=2^3*3^2*7*5$$              (A and B are paired) (C and D are paired)

So one of the even numbers is 2 * some other factor

And the one belonging to the matching pair must be 4 or 4 times some other factor

 

-----------------------------------------------------------------

 

2AB+A+B=262       It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73         It can be seen that B+C is odd      therefore B, D  one is odd and one is even

2CD+C+D=52         It can be seen that C+D is even    therefore C and D are both odd or both even

$$A*B*C*D=2^3*3^2*7*5$$ 

 

One pair has to have factors just from 3,3,5 and 7 and not all of these can be used because at least of of them belongs with a 2

( This is a total of 12-2=10 factors )

 

Possible odd numbers         3, 5, 7, 9, 15, 21, 35, 45, 63, 105                 1 and 3*3*5 are not included

 

Visual scan of these  (Note that A,B,C, and D are all 3 or bigger)

[Example of scanning method 45*6=270    and 270>262 so 45 is definitely too big]

105, 63 and 45 are all too big to be any of them.

35, 21, 15, can be A but not B or C or D

9 could be B or A

3 or 5 or 7 could be any of them.

Remember:  (A and B are paired) (C and D are paired)

 

 

2AB+A+B=262       It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73         It can be seen that B+C is odd      therefore B, D  one is odd and one is even

2CD+C+D=52         It can be seen that C+D is even    therefore C and D are both odd or both even

 

 

So

If C and D are the odd ones then they must be two of  3, 5 or 7

 

Let’s see if 3 and 5 of those work.

2CD+C+D=52        

2*3*5+3+5=38   Nope that doesn’t work so

 

Lets see if 3 and 7 work

2*3*7+3+7=52        BINGO

SO  

C and D could be 3 and 7

 

 

Could it work if C and D are the even ones?

Let C=2X     and D=2Y

2CD+C+D=52        

8XY+2X+2Y=52

4XY+X+Y=26

Possibilities for X and Y are  products of  2,3,3,7,5

If they were 2 and 3 we’d have   4*2*3+2+3=29

No it doesn’t work - there aren’t any smaller ones so    C and D must be 3 and 7

 

 

 

2AB+A+B=262       It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73         It can be seen that B+C is odd      therefore B, D  one is odd and one is even

2CD+C+D=52         It can be seen that C+D is even    therefore C and D are both odd or both even

 

If C=3 then

2*B*3+B+3=73

7B=70

B=10      that looks promising

 

If C=7 then

2*B*7+B+7=73

15B=66         Well, that is not right.

 

So    C=3, D=7, and B=10

 

2*A*10+A+10=262

21A = 252

A=12

 

 

So             A=12,     B=10,    C=3,   and   D=7

 

So            a=24,     b=20,    c=6      and     d=14

 

So             a-d =  24-14 = 10

Melody  Jul 8, 2015
Sort: 

5+0 Answers

 #1
avatar+91045 
+20
Best Answer

Four positive integers a, b, c, d satisfy

 

 

ab+a+b = 524 

bc+b+c = 146 

cd+c+d = 104 

abcd = 8! = $$2^7*3^2*7*5$$

 

What is   a-d ?


----------------------------------------

 

Can    a,b,c or d   be odd?

Assume a is odd

Let a=2N+1

(2N+1)b+(2N+1)+b=524

2Nb+b+2N+1+b=524

2Nb+2b+2N=523

2(2Nb+b+N)=523

Since N and b are integers there is a contradiction here.

Therefore a is even

By the same logic b,c and d are all even and therefore all are divisible by 2.

 

Let 2A=a,   2B=b,     2C=c,    2D=d

 

The equations become

 

2AB+A+B=262      It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73        It can be seen that B+C is odd      therefore with B & D one is odd and one is even

2CD+C+D=52        It can be seen that C+D is even    therefore C and D are both odd or both even

 

This means that the powers of 2 must be shared into two bundles.

They belong either to A and B          or              to    C and D

 $$A*B*C*D=2^3*3^2*7*5$$

 

-----------------------

 

Can A,B,C, OR D  be 1 ?

Assume A=1

2B+1+B=262

3B=261

B=87 = 3*29  no that is impossible so   $$B\ne1$$    and using the same logic   $$A\ne1$$

 

Assume C=1

2D+1+D=52

3D=51

D=17     again that is impossible so  $$c\ne1$$   and using the same logic   $$D\ne1$$

 

--------------------------

 

 

 

 

2AB+A+B=262       It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73         It can be seen that B+C is odd      therefore with B & D one is odd and one is even

2CD+C+D=52         It can be seen that C+D is even    therefore C and D are both odd or both even

$$A*B*C*D=2^3*3^2*7*5$$ 

 

 

Can A,B,C, OR D  be 2 ?

Assume A=2

4B+2+B=262

5B=260

B=52=2^2*13      no that is impossible so   $$A\ne 2$$   and using the same logic   $$B\ne2$$

 

Assume C=2

4D+2+D=52

5D=50

D=10      That is possible – so maybe C=2 and D=10   OR    D=2 and C=10

 

If D=2 then C=10

20B+B+10=73

21B=63

B=3

If B=3  then

6A+A+3=262

7A=259

A=37     That is impossible      so            $$D\ne2\qquad B\ne3 \qquad C\ne10$$    

 

If C=2  then D=10

4B+B+2=73

5B=71     That is impossible       so           $$D\ne10 \qquad C\ne 2$$       

 

So none of A,B,C,D are 1 or 2

 

 

  $$A*B*C*D=2^3*3^2*7*5$$              (A and B are paired) (C and D are paired)

So one of the even numbers is 2 * some other factor

And the one belonging to the matching pair must be 4 or 4 times some other factor

 

-----------------------------------------------------------------

 

2AB+A+B=262       It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73         It can be seen that B+C is odd      therefore B, D  one is odd and one is even

2CD+C+D=52         It can be seen that C+D is even    therefore C and D are both odd or both even

$$A*B*C*D=2^3*3^2*7*5$$ 

 

One pair has to have factors just from 3,3,5 and 7 and not all of these can be used because at least of of them belongs with a 2

( This is a total of 12-2=10 factors )

 

Possible odd numbers         3, 5, 7, 9, 15, 21, 35, 45, 63, 105                 1 and 3*3*5 are not included

 

Visual scan of these  (Note that A,B,C, and D are all 3 or bigger)

[Example of scanning method 45*6=270    and 270>262 so 45 is definitely too big]

105, 63 and 45 are all too big to be any of them.

35, 21, 15, can be A but not B or C or D

9 could be B or A

3 or 5 or 7 could be any of them.

Remember:  (A and B are paired) (C and D are paired)

 

 

2AB+A+B=262       It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73         It can be seen that B+C is odd      therefore B, D  one is odd and one is even

2CD+C+D=52         It can be seen that C+D is even    therefore C and D are both odd or both even

 

 

So

If C and D are the odd ones then they must be two of  3, 5 or 7

 

Let’s see if 3 and 5 of those work.

2CD+C+D=52        

2*3*5+3+5=38   Nope that doesn’t work so

 

Lets see if 3 and 7 work

2*3*7+3+7=52        BINGO

SO  

C and D could be 3 and 7

 

 

Could it work if C and D are the even ones?

Let C=2X     and D=2Y

2CD+C+D=52        

8XY+2X+2Y=52

4XY+X+Y=26

Possibilities for X and Y are  products of  2,3,3,7,5

If they were 2 and 3 we’d have   4*2*3+2+3=29

No it doesn’t work - there aren’t any smaller ones so    C and D must be 3 and 7

 

 

 

2AB+A+B=262       It can be seen that A+B is even     therefore A and B are both odd or both even

2BC+B+C=73         It can be seen that B+C is odd      therefore B, D  one is odd and one is even

2CD+C+D=52         It can be seen that C+D is even    therefore C and D are both odd or both even

 

If C=3 then

2*B*3+B+3=73

7B=70

B=10      that looks promising

 

If C=7 then

2*B*7+B+7=73

15B=66         Well, that is not right.

 

So    C=3, D=7, and B=10

 

2*A*10+A+10=262

21A = 252

A=12

 

 

So             A=12,     B=10,    C=3,   and   D=7

 

So            a=24,     b=20,    c=6      and     d=14

 

So             a-d =  24-14 = 10

Melody  Jul 8, 2015
 #2
avatar+18715 
+15

Four positive integers $a$$b$$c$, and $d$ satisfy
\begin{align*} ab + a + b &= 524, \\ bc + b + c &= 146, \\ cd + c + d &= 104, \\ \end{align*}
and $abcd=8!$.

 

I. We calculate a:

$$\small{\text{$
\begin{array}{lrcl}
(1) & ab+a+b &=& 524 \\\\
& b(1+a)+a &=& 524\\\\
& b &=& \dfrac{524-a} {1+a} \\\\
(2) & bc + b + c &=& 146 \\\\
& b(1+c)+c &=& 146 \\\\
& b &=& \dfrac{146-c}{1+c}\\\\
\\
\hline
\\
(1) = (2) & b = \dfrac{524-a} {1+a} &=& \dfrac{146-c}{1+c}\\\\
& \dfrac{524-a} {1+a} &=& \dfrac{146-c}{1+c}\\\\
& (1+c)(524-a) &=& (1+a)(146-c)\\\\
& (524-a)+c(524-a) &=& 146(1+a)-c(1+a)\\\\
& c(524-a +1+a) &=& 146(1+a)-(524-a)\\\\
& c\cdot 525 &=& 146(1+a)-524+a\\\\
& c\cdot 525 &=& 146 +146a-524+a\\\\
& c\cdot 525 &=& -378+147a\\\\
& c\cdot 525 &=& -378+147a\\\\
& \mathbf{c }& \mathbf{=}& \mathbf{ -0.72 + 0.28a } \qquad | \qquad \cdot 25\\\\
& \mathbf{25c }& \mathbf{=}& \mathbf{ -18+ 7a } \\\\
& \mathbf{-25c+7a }& \mathbf{=}& \mathbf{ 18 }
\end{array}
$}}$$

 

First we have find the solution for -25c + 7a = -1. And then we can multiply by (-18)

There is only a solution when gcd( -25,7) = -1 or in other words -25 and 7 are relatively prime.

(gcd = greatest common divisor)

 

Using the Euclidian algorithm to calcutate gcd(-25,7)

$$\small{\begin{array}{|r|r|r|r|}
\hline&&&\\
& & q& r \\
\hline &&&\\
-25 & 7 & -3 & -4 \\
7 & -4 & -1 & 3 \\
-4 & 3 & -1 & \textcolor[rgb]{1,0,0}{-1} \\
3 & -1 & -3 & 0 \\
&&&\\
\hline\end{array}}$$

 

The greatest common divisor gcd(-25,7) $$\textcolor[rgb]{1,0,0}{=-1}$$

 

Using Extended Euclidean algorithm to calculate the first solution

 $$\small{\begin{array}{|r|r|r|r||r|r|}
\hline&&&&&\\
& & q& r &c&a\\
\hline &&&&&\\
-25 & 7 & -3 & -4 & \textcolor[rgb]{1,0,0}{2} & 1-(-3)(2) = \textcolor[rgb]{1,0,0}{7}\\
7 & -4 & -1 & 3 & 1 & 1 -(-1)\cdot 1 = 2\\
-4 & 3 & -1 & -1 & 1 & 0 - (-1)\cdot 1 = 1 \\
3 & -1 & -3 & 0 & 0 & 0\cdot 3 -1 = 1\\
&&&&&\\
\hline\end{array}}$$
 
The first solution $$a_0$$ and $$c_0$$: -25c + 7a = -1
 
$$\small{\text{$
\begin{array}{rcl}
-25(2) + 7(7) &=& -1 \qquad | \quad \cdot(-18) \\
-25[2\cdot (-18) ] + 7 [7\cdot (-18) ] &=& 18 \\
-25\cdot (-36) + 7 \cdot (-126) &=& 18 \\
\end{array}
$}}$$
 
$$\small{\text{ The first solution is $ (-36,-126) \qquad -25 \cdot (\textcolor[rgb]{1,0,0}{ -36} ) + 7 \cdot (\textcolor[rgb]{1,0,0}{-126}) = 1 $}}\\\\$$
 

Next positive solution:

$$\small{\text{$
\begin{array}{lcl}
\boxed{
C\cdot c_0 + A \cdot a_0 = -1 }\\\\
\mathrm{First~ solution~~} (c_0,~ a_0)\\
\mathrm{All~ solutions~~} \left(c_0+\dfrac{z\cdot A}{ gcd(C,A) } ,~ a_0 - \dfrac{z\cdot C}{ gcd(C,A) }\right)\\
\end{array}
$}}\\\\$$

 

 $$\small{\text{$
\begin{array}{lcl} \boxed{-25\cdot c + 7 \cdot a = 18 }\\\\
\mathrm{First~ solution~~} (c_0=-36,~ a_0=-126)\\
\mathrm{All~ solutions~~} \left(-36-\dfrac{z\cdot 7}{ -1 } ,~ -126 + \dfrac{z\cdot (-25) }{ -1 }\right)\\
\end{array}$}}\\\\$$

we have:

$$\small{\text{$
\begin{array}{lcl} \left\{~\left(~ -36 + 7\cdot z,~ -126 +25\cdot z\right)~|~z \in Z ~ \right\} \\
\end{array} $}}$$

 

c = -36+7*6 = 6

a = -126 + 25*6 = 24

 

$$\small{\text{$
\begin{array}{lcl}
b &=& \dfrac{524-a} {1+a} \\\\
b &=& \dfrac{524-24} {1+24} \\\\
b &=& \dfrac{500} {25} \\\\
\mathbf{b} &\mathbf{=}& \mathbf{20}
\end{array}
$}}$$

 

$$\small{\text{$
\begin{array}{lcl}
cd+c+d &=& 104\\\\
d &=& \dfrac{104-c} {1+c} \\\\
d &=& \dfrac{104-6} {1+6} \\\\
d &=& \dfrac{98} {7} \\\\
\mathbf{d} &\mathbf{=}& \mathbf{14}
\end{array}
$}}$$

 

check:

$$a\cdot b \cdot c \cdot d = 24 \cdot 20 \cdot 6 \cdot 14 = 8! = 40320 \qquad \text{ okay }$$

 

heureka  Jul 8, 2015
 #3
avatar+78741 
+18

ab + a + b = 524   →   b(1 + a)  =  [524 - a]    →   b =  [524 - a] / [ 1 + a]

 

bc + b + c =  146   →   [524 - a] / [ 1 + a]c +  [524 - a] / [ 1 + a] + c = 146   →

c (  [524 - a] / [ 1 + a] + 1 )  =  146 -  [524 - a] / [ 1 + a]    →

c (  [ 1 + a + 524 -  a] / [ 1 + a]  =  ( [ 146 + 146a - 524 + a] / [1 + a]     →

c (525)   =  (147a - 378)     →   c   =  (147a - 378)/ (525) 

 

cd + c + d   =  104    →    (147a - 378)/ (525)d  +  (147a - 378)/ (525) + d = 104     →

d( [ 147a - 378]/ (525)  + 1 )  =  104 -  (147a - 378)/ (525)    →

d ( 525 - 378 + 147a) / (525)  =  (54600 + 378  - 147a) / (525)    →

d (147 + 147a) =  (54978 - 147a)   →   147d (1 + a) =  (54978 - 147a)   →

d (1 + a)  = (374 - a)  →   d =   (374 - a) / (1 +a)

 

Therefore

abcd =  8! = 40320

a*  [524 - a] / [ 1 + a]  *  (147a - 378)/ (525) * (374 - a) / (1 +a)  = 40320 

a * (524 -a) (147a - 378)(374 - a)  = 40320 (525) (1 + a)^2   ...  simplify

147a^4 - 132384a^3 + 29147916a^2 -74078928a    =  40320 (525)(1 + a)^2

21a( 7a^3 - 6304a^2 + 1387996a - 3527568) = 40320 (525)(1 + a)^2

a ( 7a^3 - 6304a^2 + 1387996a - 3527568) = 40320(25)(1 + a)^2

7a^4 - 6304a^3 + 1387996a^2 - 3527568a = 1008000a^2 + 2016000a + 1008000

7a^4 - 6304a^3 + 379996a^2 - 5543568a - 1008000 = 0   factor

(a - 24) (7a^3 -6136a^2 +232732a + 42000)   = 0

a = 24  is the only integer solution

b =  [524 - 24] / [ 1 + 24]  = 500/25 = 20

c =  (147(24) - 378)/ (525) = 6

d =  (374 - 24) / (1 +24)  = 350 / 25  = 14

 

And  24 * 20 * 6 * 14 =  40320

 

And a - d = 24 - 14  = 10

 

 

  

CPhill  Jul 8, 2015
 #4
avatar+26329 
+15

Here's yet another approach!

 

 integers 1

 integers 2

.

Alan  Jul 8, 2015
 #5
avatar+18715 
+18

 

$$\small{\text{$
\begin{array}{l}
\small{\text{here is an easier way, because: }}
0.28+0.72 = 1
\end{array} $}} \\\\
\small{\text{$
\begin{array}{rcl}
\mathbf{c }& \mathbf{=}& \mathbf{ -0.72 + 0.28a } \\
c &=& 0.28a-(1-0.28) \\
c &=& 0.28a - 1 + 0.28 \\
(1+c) &=& 0.28\cdot (1+a) \qquad | \qquad \text{ we set } C= 1+c \text{ and } A = 1+a\\
C &=& 0.28\cdot A \qquad | \qquad \cdot 25 \quad \text{ to get the lowest integer }\\
25C &=& 7A\\
25C-7A &=& 0 \qquad | \qquad \text{ we see } gcd{(25,7)}= 1 \\
&& \text{We have immediately the solution}\\
25 \cdot \textcolor[rgb]{1,0,0}{7} - 7\cdot \textcolor[rgb]{1,0,0}{25} &=& 0 \qquad | \qquad C = 7 \text{ and } A = 25\\
&& \text{and back}\\
C = 1+c &=& 7 \text{ so } \mathbf{c} = 7-1 \mathbf{=6} \\
A = 1+a &=& 25 \text{ so } \mathbf{a}= 25-1 = \mathbf{24} \\
\end{array} $}}$$

 

excursion:

We have also a function for a:

$$\small{
\begin{array}{rcl}
7a^4-6304a^3+379996a^2-5543568a-1008000&=&0\\
a_1 &=& -0.18 \\
a_2 &=& 24 \\
a_3 &=& 39.92 \\
a_4 &=& 836.83\\\\
\end{array}
}$$

heureka  Jul 9, 2015

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