In the equation 4/5 + 5/275 = 9/11
note that 4+5=9 but one of the fractions actually reduces.
Find a similar equation of the form
a/d + b/e = c/f
Where:
a+b=c,
all six numbers are different positive integers,
none of the fractions reduces,
and the largest of the six is minimum.
5 Highallow=25
10 dim Used(Highallow)
20 for A=1 to Highallow
30 Used(A)=1
40 for D=1 to Highallow
50 if Used(D)=0 then
60 :Used(D)=1
70 :if gcd(A,D)=1 then
80 :F1=A//D
90 :for B=1 to Highallow
100 :if Used(B)=0 then
110 :Used(B)=1
120 :for E=1 to Highallow
130 :if Used(E)=0 then
140 :Used(E)=1
150 :if gcd(B,E)=1 then
160 :F3=F1+B//E
170 :C=num(F3):F=den(F3)
175 :if C<=Highallow and F<=Highallow then
180 :if Used(C)=0 and Used(F)=0 and A+B=C and B>A then
181 :print F1;B//E;F3;tab(23);max(A,B,C,D,E,F)
184 :endif
185 :endif
310 :endif
320 :Used(E)=0
330 :endif
340 :next E
350 :Used(B)=0
360 :endif
370 :next B
380 :endif
390 :Used(D)=0
400 next D
405 Used(A)=0
410 next A
finds all such cases where no integer larger than 25 is used:
1//2 3//10 4//5 10
1//2 5//14 6//7 14
1//2 7//18 8//9 18
1//2 9//22 10//11 22
1//6 2//15 3//10 15
1//6 4//21 5//14 21
2//3 4//21 6//7 21
2//3 5//24 7//8 24
3//2 9//10 12//5 12
3//2 15//14 18//7 18
3//20 9//4 12//5 20
4//3 8//21 12//7 21
5//3 10//21 15//7 21
5//6 20//21 25//14 25
5//12 10//3 15//4 15
7//6 14//15 21//10 21
7//12 14//3 21//4 21
8//3 16//21 24//7 24
reformatted and sorted:
largest
integer
1/2 + 3/10 = 4/5 10
3/2 + 9/10 = 12/5 12
1/2 + 5/14 = 6/7 14
1/6 + 2/15 = 3/10 15
5/12 + 10/3 = 15/4 15
1/2 + 7/18 = 8/9 18
3/2 + 15/14 = 18/7 18
3/20 + 9/4 = 12/5 20
1/6 + 4/21 = 5/14 21
2/3 + 4/21 = 6/7 21
4/3 + 8/21 = 12/7 21
5/3 + 10/21 = 15/7 21
7/12 + 14/3 = 21/4 21
7/6 + 14/15 = 21/10 21
1/2 + 9/22 = 10/11 22
2/3 + 5/24 = 7/8 24
8/3 + 16/21 = 24/7 24
5/6 + 20/21 = 25/14 25
The first, of course, is the answer, involving no integer larger than 10: 1/2 + 3/10 = 4/5.
|
|
Posted by Charlie
on 2014-01-13 17:40:03 |
computer solution
