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[?] Calculate this simple equation 2^x * 5^y = 5000 ?

knudl budl

Posted 7 years ago

Hello, I', trying to find the correct way to calculate a "rather" simple equation 2^x * 5^y = 5000 , where x, y would be integers and the program would start with x=1, y=1, check for correctness and if incorrect, add an integer to x, check again... for so long until it finds the right x and y (x=3,y=4). I have no idea how to solve that, my programming skills are very bad. Any help much appreciated!

POSTED BY: knudl budl

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Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 7 years ago

I'd take advantage of unique factorization of integers for this. FactorInteger[5000] /. {{2, x_} :> x, {5, y_} :> y} (* Out[270]= {3, 4} *)

POSTED BY: Daniel Lichtblau

Robert Hanlon

Robert Hanlon, Retired

Posted 7 years ago

Use `FindInstance` FindInstance[2^x5^y == 5000 && x > 0 && y > 0, {x, y}, Integers] ( {{x -> 3, y -> 4}} ) Or use `Solve` Solve[2^x5^y == 5000 && x > 0 && y > 0, {x, y}, Integers] (* {{x -> 3, y -> 4}} ) Or `Reduce` xmax = x /. Solve[(2^x5^y == 5000 /. y -> 1) && x > 0, x][[1]] // Floor (* 9 ) Reduce[2^x5^y == 5000 && 0 < x <= xmax && y > 0, {x, y}, Integers] // ToRules (* {x -> 3, y -> 4} ) Alternatively, ymax = y /. Solve[(2^x5^y == 5000 /. x -> 1) && y > 0, y][[1]] // Floor (* 4 ) Reduce[2^x5^y == 5000 && 0 < x && 0 < y <= ymax, {x, y}, Integers] // ToRules (* {x -> 3, y -> 4} *)

Use FindInstance

FindInstance[2^x*5^y == 5000 && x > 0 && y > 0, {x, y}, Integers]

(*  {{x -> 3, y -> 4}}  *)

Or use Solve

Solve[2^x*5^y == 5000 && x > 0 && y > 0, {x, y}, Integers]

(*  {{x -> 3, y -> 4}}  *)

Or Reduce

xmax = x /. Solve[(2^x*5^y == 5000 /. y -> 1) && x > 0, x][[1]] // Floor

(*  9  *)

Reduce[2^x*5^y == 5000 && 0 < x <= xmax && y > 0, {x, y}, Integers] // ToRules

(*  {x -> 3, y -> 4}  *)

Alternatively,

ymax = y /. Solve[(2^x*5^y == 5000 /. x -> 1) && y > 0, y][[1]] // Floor

(*  4  *)

Reduce[2^x*5^y == 5000 && 0 < x && 0 < y <= ymax, {x, y}, Integers] // ToRules

(*  {x -> 3, y -> 4}  *)

POSTED BY: Robert Hanlon

Frank Kampas

Frank Kampas, Physicist at Large Consulting

Posted 7 years ago

In Mathematica, looping is inefficient, It's better to do it all at once. In[17]:= Table[{x, y, 2^x5^y == 5000}, {x, 0, 5}, {y, 0, 5}] Out[17]= {{{0, 0, False}, {0, 1, False}, {0, 2, False}, {0, 3, False}, {0, 4, False}, {0, 5, False}}, {{1, 0, False}, {1, 1, False}, {1, 2, False}, {1, 3, False}, {1, 4, False}, {1, 5, False}}, {{2, 0, False}, {2, 1, False}, {2, 2, False}, {2, 3, False}, {2, 4, False}, {2, 5, False}}, {{3, 0, False}, {3, 1, False}, {3, 2, False}, {3, 3, False}, {3, 4, True}, {3, 5, False}}, {{4, 0, False}, {4, 1, False}, {4, 2, False}, {4, 3, False}, {4, 4, False}, {4, 5, False}}, {{5, 0, False}, {5, 1, False}, {5, 2, False}, {5, 3, False}, {5, 4, False}, {5, 5, False}}} It also possible to select out the answer. In[15]:= Select[Flatten [Table[{x, y}, {x, 0, 5}, {y, 0, 12}], 1], 2^#[[1]]5^#[[2]] == 5000 &] Out[15]= {{3, 4}}

In Mathematica, looping is inefficient, It's better to do it all at once.

In[17]:= Table[{x, y, 2^x*5^y == 5000}, {x, 0, 5}, {y, 0, 5}]

Out[17]= {{{0, 0, False}, {0, 1, False}, {0, 2, False}, {0, 3, 
   False}, {0, 4, False}, {0, 5, False}}, {{1, 0, False}, {1, 1, 
   False}, {1, 2, False}, {1, 3, False}, {1, 4, False}, {1, 5, 
   False}}, {{2, 0, False}, {2, 1, False}, {2, 2, False}, {2, 3, 
   False}, {2, 4, False}, {2, 5, False}}, {{3, 0, False}, {3, 1, 
   False}, {3, 2, False}, {3, 3, False}, {3, 4, True}, {3, 5, 
   False}}, {{4, 0, False}, {4, 1, False}, {4, 2, False}, {4, 3, 
   False}, {4, 4, False}, {4, 5, False}}, {{5, 0, False}, {5, 1, 
   False}, {5, 2, False}, {5, 3, False}, {5, 4, False}, {5, 5,
   False}}}

It also possible to select out the answer.

In[15]:=  Select[Flatten [Table[{x, y}, {x, 0, 5}, {y, 0, 12}], 1], 
 2^#[[1]]*5^#[[2]] == 5000 &]

Out[15]= {{3, 4}}

POSTED BY: Frank Kampas

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