Four positive integers , , , and satisfy
and .
What is ?
(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 , , , and satisfy
and .
What are , , , and
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
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
Four positive integers , , , and satisfy
and .
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
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:
$$a\cdot b \cdot c \cdot d = 24 \cdot 20 \cdot 6 \cdot 14 = 8! = 40320 \qquad \text{ okay }$$
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
$$\small{\text{$
\begin{array}{l}
\small{\text{here is an easier way, because: }}
0.28+0.72 = 1
\end{array} $}} \\\\
\small{\text{$
\begin{array}{rcl}
\mathbf{c }& \mathbf{=}& \mathbf{ -0.72 + 0.28a } \\
c &=& 0.28a-(1-0.28) \\
c &=& 0.28a - 1 + 0.28 \\
(1+c) &=& 0.28\cdot (1+a) \qquad | \qquad \text{ we set } C= 1+c \text{ and } A = 1+a\\
C &=& 0.28\cdot A \qquad | \qquad \cdot 25 \quad \text{ to get the lowest integer }\\
25C &=& 7A\\
25C-7A &=& 0 \qquad | \qquad \text{ we see } gcd{(25,7)}= 1 \\
&& \text{We have immediately the solution}\\
25 \cdot \textcolor[rgb]{1,0,0}{7} - 7\cdot \textcolor[rgb]{1,0,0}{25} &=& 0 \qquad | \qquad C = 7 \text{ and } A = 25\\
&& \text{and back}\\
C = 1+c &=& 7 \text{ so } \mathbf{c} = 7-1 \mathbf{=6} \\
A = 1+a &=& 25 \text{ so } \mathbf{a}= 25-1 = \mathbf{24} \\
\end{array} $}}$$
excursion:
We have also a function for a:
$$\small{
\begin{array}{rcl}
7a^4-6304a^3+379996a^2-5543568a-1008000&=&0\\
a_1 &=& -0.18 \\
a_2 &=& 24 \\
a_3 &=& 39.92 \\
a_4 &=& 836.83\\\\
\end{array}
}$$