1
$\begingroup$

For the formula:

(a/210)*0 + (b/210)*1 + (c/210)*2 + (d/210)*3 + (e/210)*4 = 1/2 + 1/3 + 1/5 + 1/7,

where 0<=a,b,c,d,e<=210,

and a+b+c+d+e=210,

how can I find all solutions for a,b,c,d,e?

$\endgroup$

1 Answer 1

1
$\begingroup$

The result depends on the domain of variables.

Reduce[(a/210)*0 + (b/210)*1 + (c/210)*2 + (d/210)*3 + (e/210)*4 == 
1/2 + 1/3 + 1/5 + 1/7 && 0 <= {a, b, c, d, e} <= 210 && 
a + b + c + d + e == 210 && d <= e, {a, b, c, d, e}, Reals]

((a == 0 && ((b == 173 && c == 37 && d == 0) || (173 < b <= 976/5 && 1/3 (976 - 5 b) <= c <= 1/2 (593 - 3 b) && d == 593 - 3 b - 2 c) || (976/5 < b < 593/3 && 0 <= c <= 1/2 (593 - 3 b) && d == 593 - 3 b - 2 c) || (b == 593/3 && c == 0 && d == 0))) || (0 < a <= 173/ 2 && ((b == 173 - 2 a && c == 1/3 (976 - 7 a - 5 b) && d == 593 - 4 a - 3 b - 2 c) || (173 - 2 a < b < 1/5 (976 - 7 a) && 1/3 (976 - 7 a - 5 b) <= c <= 1/2 (593 - 4 a - 3 b) && d == 593 - 4 a - 3 b - 2 c) || (1/5 (976 - 7 a) <= b < 1/3 (593 - 4 a) && 0 <= c <= 1/2 (593 - 4 a - 3 b) && d == 593 - 4 a - 3 b - 2 c) || (b == 1/3 (593 - 4 a) && c == 0 && d == 593 - 4 a - 3 b))) || (173/2 < a <= 976/ 7 && ((0 <= b < 1/5 (976 - 7 a) && 1/3 (976 - 7 a - 5 b) <= c <= 1/2 (593 - 4 a - 3 b) && d == 593 - 4 a - 3 b - 2 c) || (1/5 (976 - 7 a) <= b < 1/3 (593 - 4 a) && 0 <= c <= 1/2 (593 - 4 a - 3 b) && d == 593 - 4 a - 3 b - 2 c) || (b == 1/3 (593 - 4 a) && c == 0 && d == 593 - 4 a - 3 b))) || (976/7 < a <= 593/ 4 && ((0 <= b < 1/3 (593 - 4 a) && 0 <= c <= 1/2 (593 - 4 a - 3 b) && d == 593 - 4 a - 3 b - 2 c) || (b == 1/3 (593 - 4 a) && c == 0 && d == 593 - 4 a - 3 b)))) && e == 1/4 (247 - b - 2 c - 3 d)

describes an infinite set of the solutions over the reals. The same description for the rationals. The command

FindInstance[(a/210)*0 + (b/210)*1 + (c/210)*2 + (d/210)*3 + (e/210)*4 == 
1/2 + 1/3 + 1/5 + 1/7 &&  0 <= {a, b, c, d, e} <= 210 && 
a + b + c + d + e == 210 &&  d <= e, {a, b, c, d, e}, Integers, 50000];
Dimensions[%]

{47988, 5}

produces a finite set of 47988 solutions over the integers.

$\endgroup$
1
  • $\begingroup$ I tried another one with one more variable and ran out of RAM: x = 2310; FindInstance[(a/x)*0 + (b/x)*1 + (c/x)*2 + (d/x)*3 + (e/x)*4 + (f/ x)*5 == 1/2 + 1/3 + 1/5 + 1/7 + 1/11 && 0 <= {a, b, c, d, e, f} <= x && a + b + c + d + e + f == x && e <= f, {a, b, c, d, e, f}, Integers, 5000000]; Dimensions[%] $\endgroup$
    – Jamie M
    Commented Apr 7 at 23:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.