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Hello i got this question on my aptitude exam.I tried dividing both side by 5.But was unable to get the answer

if(60-a)(60-b)(60-c)(60-d)(60-e)=1025 what is the value of a+b+c+d=? hint: 1025 is divisible by 5

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    $\begingroup$ Have you stated the problem correctly? Were you actually asked for the sum of a, b, c, d and e? $\endgroup$ – mikado Sep 16 '16 at 15:41
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eqn = (60 - a) (60 - b) (60 - c) (60 - d) (60 - e) == 1025;

Assuming that all variables are restricted to positive integers

solns = FindInstance[{eqn,
   a > 0, b > 0, c > 0, d > 0, e > 0},
  {a, b, c, d, e}, Integers, 3]

(*  {{a -> 59, b -> 61, c -> 59, d -> 59, e -> 1085}, 
     {a -> 61, b -> 59, c -> 59, d -> 19, e -> 85}, 
     {a -> 61, b -> 59, c -> 59, d -> 59, e -> 1085}}  *)

Verifying that solns satisfy eqn

eqn /. solns

(*  {True, True, True}  *)

From the symmetry, there are other obvious additional solutions.

The requested sums are

(a + b + c + d) /. solns

(*  {238, 198, 238}  *)
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If you factor $1025=5^2\cdot 41$ and assume that the variables need to be positive integers less than $60$, you can either have the factors be $1,1,5,5,41$ or $1,1,1,25,41$. That leads to variables $59,59,55,55,19$ or $59,59,59,35,19$. If you were asked for the sum of four of them, you have several answers. If you were asked for the sum of all five, it is either $247$ or $231$. If you allow the variables to exceed $60$, you need an even number of the factors to be negative, but the possibilities expand a bunch. Please make sure you have the correct problem, or punch the poser in the nose if this is the way it was asked.

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The equation you give is

eqn = (60 - a) (60 - b) (60 - c) (60 - d) (60 - e) == 1025;

I have assumed the additional constraints that the variables are non-negative and less than 60. In addition, I specify an ordering to avoid solutions that are trivial rearrangements.

solns = FindInstance[{eqn, 60 > a >= b >= c >= d >= e >= 0}, {a, b, c,
    d, e}, Integers, 3]
(* {{a -> 59, b -> 59, c -> 55, d -> 55, e -> 19}, {a -> 59, b -> 59, 
  c -> 59, d -> 35, e -> 19}} *)

eqn /. solns
(* {True, True} *)

It appears that there are two solutions for the sum. Note that I've assumed that we should consider the sum of all five variables, to avoid different answers arising from trivial permutations.

a + b + c + d + e /. solns
(* {247, 231} *)

EDIT

This answer was written before the question was migrated from the Mathematica site.

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