Four positive integers , , , and satisfy

Question

Four positive integers , , , and satisfy

0

7320

5

+1836

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

5⁺⁴ Answers

#1

+118703

+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 Online Users

Melody · Answer 1 · 2015-07-08T06:15:21

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

heureka · Answer 2 · 2015-07-08T02:44:29

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:

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

CPhill · Answer 3 · 2015-07-08T03:42:50

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

Alan · Answer 4 · 2015-07-08T04:23:56

Here's yet another approach!

integers 1

integers 2

.

heureka · Answer 5 · 2015-07-09T04:11:22

$\small{\text{$ \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:

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

.

Four positive integers , , , and satisfy

0 users composing answers..

Best Answer

5⁺⁴ Answers

2 Online Users

Top Users

Sticky Topics

Four positive integers , , , and satisfy

0 users composing answers..

Best Answer

5+4 Answers

2 Online Users

Top Users

Sticky Topics

5⁺⁴ Answers