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 Four Egyptian Fractions (Posted on 2013-11-13)
Find positive integer solutions to the equation 1/a + 1/b + 1/c + 1/d = 1
(a ≤ b ≤ c ≤ d). How many solutions exist?

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 analytical solution | Comment 1 of 6

*edit*
*fixed some typos and miscalculated in case 1 thus missing solutions with 4<=b<=6*

since a,b,c,d are all postive then we have
a<=b<=c<=d implies 1/a>=1/b>=1/c>=1/d
thus the sum of the inverses is bounded below and above respectively
by 4/d and 4/a
now in order for 1 to be in this interval we need 4/a>=1 thus a<=4
so that gives 4 possible values for a, namely 2,3, and 4.  We can skip a=1 because
having any of the variables equal to 1 would require the remaining inverses to sum to zero
which is not possible since they are all strictly positive.

case 1: a=2
1/b+1/c+1/d=1/2
by similar reasoning as above, this sum is bounded above by 3/b and thus we need
b<=6
case 1.1: a=2,b=2
this gives 1/c+1/d=0 which has no solutions
case 1.2: a=2,b=3
this gives 1/c+1/d=1/6
again 2/c>=1/6, c<=12
testing each of c=2 to c=12 gives the following solutions
(a,b,c,d): (2,3,7,42), (2,3,8,24), (2,3,9,18), (2,3,10,15), (2,3,12,12)
case 1.3: a=2,b=4
this gives 1/c+1/d=1/4
2/c>=1/4, c<=8
(a,b,c,d): (2,4,5,20), (2,4,6,12), (2,4,8,8)
case 1.4: a=2,b=5
1/c+1/d=3/10
2/c>=3/10, c<=20/3=6.66
(a,b,c,d): (2,5,5,10)
case 1.5: a=2,b=6
1/c+1/d=1/3
2/c>=1/3, c<=6
(a,b,c,d): (2,6,6,6)

case 2: a=3
1/b+1/c+1/d=2/3
3/b>=2/3, b<=9/2 -> 3=a<=b<=4.5
case 2.1: a=3, b=3
1/c+1/d=1/3
2/c>=1/3, c<=6
again testing all possible values of c gives
(a,b,c,d): (3,3,4,12), (3,3,6,6)
case 2.2: a=3, b=4
1/c+1/d=5/12
2/c>=5/12, 4=b<=c<=24/5=4.8
thus the only possible value for c is 4 giving
(a,b,c,d): (3,4,4,6)

case 3: a=4
1/b+1/c+1/d=3/4
3/b>=3/4,4=a<=b<=4
thus the only possible value for b is 4
which gives
1/c+1/d=1/2
2/c>=1/2, 4=b<=c<=4, again c=4, d=4
thus giving
(a,b,c,d): (4,4,4,4)

so in total there are 14 solutions namely
(a,b,c,d):
(2,3,7,42)
(2,3,8,24)
(2,3,9,18)
(2,3,10,15)
(2,3,12,12)
(2,4,5,20)
(2,4,6,12)
(2,4,8,8)
(2,5,5,10)
(2,6,6,6)
(3,3,4,12)
(3,3,6,6)
(3,4,4,6)
(4,4,4,4)

Edited on November 13, 2013, 1:14 pm

Edited on November 13, 2013, 1:15 pm
 Posted by Daniel on 2013-11-13 12:07:45

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