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 ?
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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
$$A*B*C*D=2^3*3^2*7*5$$
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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 $$D\ne1$$
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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
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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
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