# Solutions for Problem 9

Find the only Pythagorean triplet, {a, b, c}, for which a + b + c = 1000, then find the product of a, b and c

## Ruby by tomchristmas

Runtime: 1432ms (eek!)

```1.upto(500){|c|
c.upto(499){|b|
a = 1000 - b - c
puts "#{a * b * c} (#{a}^2 + #{b}^2 = #{c}^2)" if c **2 == a ** 2 + b ** 2
}
}
```

## Python by Althalus

Runtime: 43 ms

```from math import sqrt
import time
start_time = time.time()

#These ranges don't make much sense, but it does satisfy the requirement a < b
brange = range(1,499)
arange = range(1,488)
for a in arange:
for b in brange:
if sqrt(a**2 + b**2) + a + b == 1000:
print(a*b*sqrt(a**2 + b**2))

run_time = time.time() - start_time
print (run_time)
exit()
```

## Caml by SanguineV

Runtime: 41.991 ms

```(* Given a limit, an a and a list of b's, check if any of them satisfy:
* a^2 + b^2 = c^2 && a + b + c = lim
* if not, return 0. *)
let rec check lim a = function
| b::bs ->
let c = sqrt (a *. a +. b *. b) in
if a +. b +. c = lim && c = floor c
then a *. b *. c
else check lim a bs
| [] -> 0.
;;

(* Given a total and a list of numbers, see if any two numbers in the list
* will be the a and b for the equation we want to solve. *)
let rec findABC tot = function
| a::aas ->
let res = check tot a aas in
if res = 0.
then findABC tot aas
else res
| [] -> 0.
;;

(* Generate floats from n down to m *)
let rec floatsN2M n m =
if n = m
then []
else n::(floatsN2M (n -. 1.) m)
;;

(* Find the solution. *)
Printf.printf "a * b * c = %f\n" (findABC 1000. (floatsN2M 499. 1.));;
```

## Excel VBA by mmaster

Runtime: Less than a second

```   Sub ProblemNine()
Dim Limit As Integer
Limit = 1000

Dim a, b, c As Integer

For a = 1 To Limit - 2
For b = a + 1 To Limit - 1
c = Limit - (a + b)
If a ^ 2 + b ^ 2 = c ^ 2 Then
MsgBox "a = " & a & ", b = " & b & ", c = " & c & ", T = " & a * b * c
Exit Sub
End If
Next b
Next a

End Sub
```