Count in factors: Difference between revisions
m (→{{header|Ruby}}: optparse provides -h option, so use -h instead of -?.) |
(→{{header|Euphoria}}: Euphoria example added) |
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writefln("%2d = %s", i, productStr(factorize(i))) ; |
writefln("%2d = %s", i, productStr(factorize(i))) ; |
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}</lang> |
}</lang> |
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=={{header|Euphoria}}== |
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<lang euphoria>function factorize(integer n) |
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sequence result |
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integer k |
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if n = 1 then |
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return {1} |
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else |
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k = 2 |
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result = {} |
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while n > 1 do |
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while remainder(n, k) = 0 do |
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result &= k |
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n /= k |
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end while |
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k += 1 |
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end while |
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return result |
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end if |
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end function |
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sequence factors |
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for i = 1 to 22 do |
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printf(1, "%d: ", i) |
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factors = factorize(i) |
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for j = 1 to length(factors)-1 do |
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printf(1, "%d * ", factors[j]) |
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end for |
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printf(1, "%d\n", factors[$]) |
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end for</lang> |
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Output: |
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<pre>1: 1 |
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2: 2 |
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3: 3 |
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4: 2 * 2 |
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5: 5 |
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6: 2 * 3 |
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7: 7 |
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8: 2 * 2 * 2 |
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9: 3 * 3 |
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10: 2 * 5 |
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11: 11 |
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12: 2 * 2 * 3 |
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13: 13 |
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14: 2 * 7 |
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15: 3 * 5 |
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16: 2 * 2 * 2 * 2 |
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17: 17 |
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18: 2 * 3 * 3 |
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19: 19 |
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20: 2 * 2 * 5 |
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21: 3 * 7 |
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22: 2 * 11 |
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</pre> |
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=={{header|Go}}== |
=={{header|Go}}== |
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<lang go>package main |
<lang go>package main |
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Revision as of 19:37, 24 August 2011
You are encouraged to solve this task according to the task description, using any language you may know.
Write a program which counts up from 1, displaying each number as the multiplication of its prime factors. For the purpose of this task, may be shown as itself.
For examle, is prime, so it would be shown as itself. is not prime; it would be shown as . Likewise, 2144 is not prime; it would be shown as .
c.f. Prime decomposition
Ada
count.adb: <lang Ada>with Ada.Command_Line; with Ada.Text_IO;
procedure Count is
type Number_List is array (Positive range <>) of Positive;
function Decompose (N : Natural) return Number_List is
Size : Natural := 0;
M : Natural := N;
K : Natural := 2;
begin
if N = 1 then
return (1 => 1);
end if;
-- Estimation of the result length from above
while M >= 2 loop
M := (M + 1) / 2;
Size := Size + 1;
end loop;
M := N;
-- Filling the result with prime numbers
declare
Result : Number_List (1 .. Size);
Index : Positive := 1;
begin
while N >= K loop -- Divisors loop
while 0 = (M mod K) loop -- While divides
Result (Index) := K;
Index := Index + 1;
M := M / K;
end loop;
K := K + 1;
end loop;
return Result (1 .. Index - 1);
end;
end Decompose;
procedure Put (List : Number_List) is
begin
for Index in List'Range loop
Ada.Text_IO.Put (Integer'Image (List (Index)));
if Index /= List'Last then
Ada.Text_IO.Put (" x");
end if;
end loop;
end Put;
N : Natural := 1; Max_N : Natural := 15;
begin
if Ada.Command_Line.Argument_Count = 1 then
Max_N := Integer'Value (Ada.Command_Line.Argument (1));
end if;
loop
Ada.Text_IO.Put (Integer'Image (N) & ": ");
Put (Decompose (N));
Ada.Text_IO.New_Line;
N := N + 1;
exit when N > Max_N;
end loop;
end Count;</lang>
output:
1: 1 2: 2 3: 3 4: 2 x 2 5: 5 6: 2 x 3 7: 7 8: 2 x 2 x 2 9: 3 x 3 10: 2 x 5 11: 11 12: 2 x 2 x 3 13: 13 14: 2 x 7 15: 3 x 5
C
Code includes a dynamically extending prime number list. The program doesn't stop until you kill it, or it runs out of memory, or it overflows. <lang C>#include <stdio.h>
- include <stdlib.h>
typedef unsigned long long ULONG;
ULONG get_prime(int idx) {
static long n_primes = 0, alloc = 0;
static ULONG *primes = 0;
ULONG last, p;
int i;
if (idx >= n_primes) {
if (n_primes >= alloc) {
alloc += 16; /* be conservative */
primes = realloc(primes, sizeof(ULONG) * alloc);
}
if (!n_primes) {
primes[0] = 2;
primes[1] = 3;
n_primes = 2;
}
last = primes[n_primes-1];
while (idx >= n_primes) {
last += 2;
for (i = 0; i < n_primes; i++) {
p = primes[i];
if (p * p > last) {
primes[n_primes++] = last;
break;
}
if (last % p == 0) break;
}
}
}
return primes[idx];
}
int main() {
ULONG n, x, p;
int i, first;
for (x = 1; ; x++) {
printf("%lld = ", n = x);
for (i = 0, first = 1; ; i++) {
p = get_prime(i);
while (n % p == 0) {
n /= p;
if (!first) printf(" x ");
first = 0;
printf("%lld", p);
}
if (n <= p * p) break;
}
if (first) printf("%lld\n", n);
else if (n > 1) printf(" x %lld\n", n);
else printf("\n");
}
return 0;
}</lang> Output:<lang>1 = 1 2 = 2 3 = 3 4 = 2 x 2 5 = 5 6 = 2 x 3 7 = 7 8 = 2 x 2 x 2 9 = 3 x 3 10 = 2 x 5 11 = 11 12 = 2 x 2 x 3 13 = 13 14 = 2 x 7 . . .</lang>
D
<lang d>import std.stdio, std.string, std.algorithm, std.array, std.conv;
int[] factorize(int n) in {
assert(n > 0);
} body {
if (n == 1) return [1];
int[] result;
int m = n, k = 2;
while (n >= k) {
while (m % k == 0) {
result ~= k;
m /= k;
}
k++;
}
return result;
}
void main() {
foreach (i; 1 .. 22)
writeln(i, ": ", join(array(map!text(factorize(i))), " * "));
}</lang> Output:
1: 1 2: 2 3: 3 4: 2 * 2 5: 5 6: 2 * 3 7: 7 8: 2 * 2 * 2 9: 3 * 3 10: 2 * 5 11: 11 12: 2 * 2 * 3 13: 13 14: 2 * 7 15: 3 * 5 16: 2 * 2 * 2 * 2 17: 17 18: 2 * 3 * 3 19: 19 20: 2 * 2 * 5 21: 3 * 7
Alternative version
Library uiprimes is a homebrew library to generate prime numbers upto the maximum 32bit unsigned integer range 2^32-1, by using a pre-generated bit array of Sieve of Eratosthenes (a dll in size of ~256M bytes :p ).
<lang d>import std.stdio, std.math, std.conv, std.algorithm,
std.array, std.string, import xt.uiprimes ;
pragma(lib, "uiprimes.lib") ;
// function _factorize_ included in uiprimes.lib ulong[] factorize(ulong n) {
if(n == 0) return [] ;
if(n == 1) return [1] ;
ulong[] res ;
uint limit = cast(uint) (1 + sqrt(n)) ;
foreach(p; Primes(limit)) {
if(n == 1) break ;
if(0UL == (n % p ))
while((n > 1) && (0UL == (n % p ))) {
res ~= p ;
n = n / p ;
}
}
if(n > 1)
res ~= [n] ;
return res ;
}
string productStr(T)(T[] nums) {
return array(map!q{to!string(a)}(nums)).join(" x ") ;
}
void main() {
foreach(i;1..21)
writefln("%2d = %s", i, productStr(factorize(i))) ;
}</lang>
Euphoria
<lang euphoria>function factorize(integer n)
sequence result
integer k
if n = 1 then
return {1}
else
k = 2
result = {}
while n > 1 do
while remainder(n, k) = 0 do
result &= k
n /= k
end while
k += 1
end while
return result
end if
end function
sequence factors for i = 1 to 22 do
printf(1, "%d: ", i)
factors = factorize(i)
for j = 1 to length(factors)-1 do
printf(1, "%d * ", factors[j])
end for
printf(1, "%d\n", factors[$])
end for</lang>
Output:
1: 1 2: 2 3: 3 4: 2 * 2 5: 5 6: 2 * 3 7: 7 8: 2 * 2 * 2 9: 3 * 3 10: 2 * 5 11: 11 12: 2 * 2 * 3 13: 13 14: 2 * 7 15: 3 * 5 16: 2 * 2 * 2 * 2 17: 17 18: 2 * 3 * 3 19: 19 20: 2 * 2 * 5 21: 3 * 7 22: 2 * 11
Go
<lang go>package main
import "fmt"
func main() {
fmt.Println("1: 1")
for i := 2; ; i++ {
fmt.Printf("%d: ", i)
var x string
for n, f := i, 2; n != 1; f++ {
for m := n % f; m == 0; m = n % f {
fmt.Print(x, f)
x = "×"
n /= f
}
}
fmt.Println()
}
}</lang> Output:
1: 1 2: 2 3: 3 4: 2×2 5: 5 6: 2×3 7: 7 8: 2×2×2 9: 3×3 10: 2×5 ...
Haskell
<lang Haskell> primes = [2] ++ filter is_prime [3..]
is_prime num = foldl (\acc x -> acc && (num `mod` x /= 0)) True $ takeWhile (<= floor ((fromIntegral num)/2)) primes
-- spf = smallest prime factor spf num = head.take 1 $ dropWhile (\x -> (num `mod` x) /= 0) primes
factors num
|is_prime num = [num] |otherwise = (spf num) : factors (floor ((fromIntegral num) / (fromIntegral (spf num))))
map factors [1..] </lang>
Icon and Unicon
<lang Icon>procedure main() write("Press ^C to terminate") every f := [i:= 1] | factors(i := seq(2)) do {
writes(i," : [")
every writes(" ",!f|"]\n")
}
end
link factors</lang>
Output:
1 : [ 1 ] 2 : [ 2 ] 3 : [ 3 ] 4 : [ 2 2 ] 5 : [ 5 ] 6 : [ 2 3 ] 7 : [ 7 ] 8 : [ 2 2 2 ] 9 : [ 3 3 ] 10 : [ 2 5 ] 11 : [ 11 ] 12 : [ 2 2 3 ] 13 : [ 13 ] 14 : [ 2 7 ] 15 : [ 3 5 ] 16 : [ 2 2 2 2 ] ...
J
Solution:Use J's factoring primitive, <lang j>q:</lang> Example (including formatting):<lang j> ('1 : 1',":&> ,"1 ': ',"1 ":@q:) 2+i.10 1 : 1 2 : 2 3 : 3 4 : 2 2 5 : 5 6 : 2 3 7 : 7 8 : 2 2 2 9 : 3 3 10: 2 5 11: 11</lang>
Java
<lang java>public class CountingInFactors{
public static void main(String[] args){
for(int i = 1; i<= 10; i++){
System.out.println(i + " = "+ countInFactors(i));
}
for(int i = 9991; i <= 10000; i++){
System.out.println(i + " = "+ countInFactors(i));
}
}
private static String countInFactors(int n){
if(n == 1) return "1";
StringBuilder sb = new StringBuilder();
n = checkFactor(2, n, sb);
if(n == 1) return sb.toString();
n = checkFactor(3, n, sb);
if(n == 1) return sb.toString();
for(int i = 5; i <= n; i+= 2){
if(i % 3 == 0)continue;
n = checkFactor(i, n, sb);
if(n == 1)break;
}
return sb.toString();
}
private static int checkFactor(int mult, int n, StringBuilder sb){
while(n % mult == 0 ){
if(sb.length() > 0) sb.append(" x ");
sb.append(mult);
n /= mult;
}
return n;
}
}</lang> Output:
1 = 1 2 = 2 3 = 3 4 = 2 x 2 5 = 5 6 = 2 x 3 7 = 7 8 = 2 x 2 x 2 9 = 3 x 3 10 = 2 x 5 9991 = 97 x 103 9992 = 2 x 2 x 2 x 1249 9993 = 3 x 3331 9994 = 2 x 19 x 263 9995 = 5 x 1999 9996 = 2 x 2 x 3 x 7 x 7 x 17 9997 = 13 x 769 9998 = 2 x 4999 9999 = 3 x 3 x 11 x 101 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
PARI/GP
<lang parigp>fnice(n)={ my(f,s="",s1); if (n < 2, return(n)); f = factor(n); s = Str(s, f[1,1]); if (f[1, 2] != 1, s=Str(s, "^", f[1,2])); for(i=2,#f[,1], s1 = Str(" * ", f[i, 1]); if (f[i, 2] != 1, s1 = Str(s1, "^", f[i, 2])); s = Str(s, s1) ); s }; n=0;while(n++, print(fnice(n)))</lang>
Perl
<lang perl>use utf8; sub factors { my $n = shift; my $p = 2; my @out;
while ($n >= $p * $p) { while ($n % $p == 0) { push @out, $p; $n /= $p; } $p = next_prime($p); } push @out, $n if $n > 1 || !@out; @out; }
sub next_prime { my $p = shift; do { $p = $p == 2 ? 3 : $p + 2 } until is_prime($p); $p; }
sub is_prime { factors(shift) == 1 }
print "$_ = ", join(" × ", factors($_)), "\n" for (1 .. 1000);</lang>
Perl 6
<lang perl6># Define a lazy list of primes.
- Uses the ... sequence operator with a lambda that calculates
- the next available prime by using some of the existing list
- as test divisors, so we never divide by anything that isn't
- known to be a prime already. This is quite fast.
my @primes := 2, 3, -> $n is copy {
repeat { $n += 2 } until $n %% none do for @primes -> $p {
last if $p > sqrt($n);
$p;
}
$n;
} ... *;
- Finds the factors of the given argument.
multi factors(1) { 1 } multi factors(Int $remainder is copy) {
gather for @primes -> $factor {
# if remainder < factor², we're done
if $factor * $factor > $remainder {
take $remainder if $remainder > 1;
last;
}
# How many times can we divide by this prime?
while $remainder %% $factor {
take $factor;
last if ($remainder div= $factor) === 1;
}
}
}
- An infinite loop, from 1 incrementing upward.
- calls factor() with each of 1, 2, 3, etc., receives an
- array containing that number's factors, and then
- formats and displays them.
say "$_: ", factors($_).join(" × ") for 1..*;</lang>
The first twenty numbers:
1: 1 2: 2 3: 3 4: 2 × 2 5: 5 6: 2 × 3 7: 7 8: 2 × 2 × 2 9: 3 × 3 10: 2 × 5 11: 11 12: 2 × 2 × 3 13: 13 14: 2 × 7 15: 3 × 5 16: 2 × 2 × 2 × 2 17: 17 18: 2 × 3 × 3 19: 19 20: 2 × 2 × 5
Here we use a multi declaration with a constant parameter to match the degenerate case. We use copy parameters when we wish to reuse the formal parameter as a mutable variable within the function. (Parameters default to readonly in Perl 6.) Note the use of gather/take as the final statement in the function, which is a common Perl 6 idiom to set up a coroutine within a function to return a lazy list on demand.
Note also the '×' above is not ASCII 'x', but U+00D7 MULTIPLICATION SIGN. Perl 6 does Unicode natively.
PicoLisp
This is the 'factor' function from Prime decomposition#PicoLisp. <lang PicoLisp>(de factor (N)
(make
(let (D 2 L (1 2 2 . (4 2 4 2 4 6 2 6 .)) M (sqrt N))
(while (>= M D)
(if (=0 (% N D))
(setq M (sqrt (setq N (/ N (link D)))))
(inc 'D (pop 'L)) ) )
(link N) ) ) )
(for N 20
(prinl N ": " (glue " * " (factor N))) )</lang>
Output:
1: 1 2: 2 3: 3 4: 2 * 2 5: 5 6: 2 * 3 7: 7 8: 2 * 2 * 2 9: 3 * 3 10: 2 * 5 11: 11 12: 2 * 2 * 3 13: 13 14: 2 * 7 15: 3 * 5 16: 2 * 2 * 2 * 2 17: 17 18: 2 * 3 * 3 19: 19 20: 2 * 2 * 5
PureBasic
<lang PureBasic>Procedure Factorize(Number, List Factors())
Protected I = 3, Max
ClearList(Factors())
While Number % 2 = 0
AddElement(Factors())
Factors() = 2
Number / 2
Wend
Max = Number
While I <= Max And Number > 1
While Number % I = 0
AddElement(Factors())
Factors() = I
Number / I
Wend
I + 2
Wend
EndProcedure
If OpenConsole()
NewList n()
For a=1 To 20
text$=RSet(Str(a),2)+"= "
Factorize(a,n())
If ListSize(n())
ResetList(n())
While NextElement(n())
text$ + Str(n())
If ListSize(n())-ListIndex(n())>1
text$ + "*"
EndIf
Wend
Else
text$+Str(a) ; To handle the '1', which is not really a prime...
EndIf
PrintN(text$)
Next a
EndIf</lang>
1= 1 2= 2 3= 3 4= 2*2 5= 5 6= 2*3 7= 7 8= 2*2*2 9= 3*3 10= 2*5 11= 11 12= 2*2*3 13= 13 14= 2*7 15= 3*5 16= 2*2*2*2 17= 17 18= 2*3*3 19= 19 20= 2*2*5
Python
This uses the functools.lru_cache standard library module to cache intermediate results. <lang python>from functools import lru_cache
primes = [2, 3, 5, 7, 11, 13, 17] # Will be extended
@lru_cache(maxsize=2000) def pfactor(n):
if n == 1:
return [1]
n2 = n // 2 + 1
for p in primes:
if p <= n2:
d, m = divmod(n, p)
if m == 0:
if d > 1:
return [p] + pfactor(d)
else:
return [p]
else:
if n > primes[-1]:
primes.append(n)
return [n]
if __name__ == '__main__':
mx = 5000
for n in range(1, mx + 1):
factors = pfactor(n)
if n <= 10 or n >= mx - 20:
print( '%4i %5s %s' % (n,
if factors != [n] else 'prime',
'x'.join(str(i) for i in factors)) )
if n == 11:
print('...')
print('\nNumber of primes gathered up to', n, 'is', len(primes))
print(pfactor.cache_info())
</lang>
Sample output
1 prime 1 2 prime 2 3 prime 3 4 2x2 5 prime 5 6 2x3 7 prime 7 8 2x2x2 9 3x3 10 2x5 ... 4980 2x2x3x5x83 4981 17x293 4982 2x47x53 4983 3x11x151 4984 2x2x2x7x89 4985 5x997 4986 2x3x3x277 4987 prime 4987 4988 2x2x29x43 4989 3x1663 4990 2x5x499 4991 7x23x31 4992 2x2x2x2x2x2x2x3x13 4993 prime 4993 4994 2x11x227 4995 3x3x3x5x37 4996 2x2x1249 4997 19x263 4998 2x3x7x7x17 4999 prime 4999 5000 2x2x2x5x5x5x5 Number of primes gathered up to 5000 is 669 CacheInfo(hits=3935, misses=7930, maxsize=2000, currsize=2000)
Ruby
Starting with Ruby 1.9, 'prime' is part of the standard library and provides Integer#prime_division.
<lang ruby>require 'optparse' require 'prime'
maximum = 10 OptionParser.new do |o|
o.banner = "Usage: #{File.basename $0} [-m MAXIMUM]"
o.on("-m MAXIMUM", Integer,
"Count up to MAXIMUM [#{maximum}]") { |m| maximum = m }
o.parse! rescue ($stderr.puts $!, o; exit 1)
($stderr.puts o; exit 1) unless ARGV.size == 0
end
- 1 has no prime factors
puts "1 is 1" unless maximum < 1
2.upto(maximum) do |i|
# i is 504 => i.prime_division is [[2, 3], [3, 2], [7, 1]]
f = i.prime_division.map! do |factor, exponent|
# convert [2, 3] to "2 x 2 x 2"
([factor] * exponent).join " x "
end.join " x "
puts "#{i} is #{f}"
end</lang>
Example:
$ ruby prime-count.rb -h
Usage: prime-count.rb [-m MAXIMUM]
-m MAXIMUM Count up to MAXIMUM [10]
$ ruby prime-count.rb -m 10000 | (head -n 10; tail -n 10)
1 is 1
2 is 2
3 is 3
4 is 2 x 2
5 is 5
6 is 2 x 3
7 is 7
8 is 2 x 2 x 2
9 is 3 x 3
10 is 2 x 5
9991 is 97 x 103
9992 is 2 x 2 x 2 x 1249
9993 is 3 x 3331
9994 is 2 x 19 x 263
9995 is 5 x 1999
9996 is 2 x 2 x 3 x 7 x 7 x 17
9997 is 13 x 769
9998 is 2 x 4999
9999 is 3 x 3 x 11 x 101
10000 is 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5
Seed7
<lang seed7>$ include "seed7_05.s7i";
const proc: writePrimeFactors (in var integer: number) is func
local
var boolean: laterElement is FALSE;
var integer: checker is 2;
begin
while checker * checker <= number do
if number rem checker = 0 then
if laterElement then
write(" * ");
end if;
laterElement := TRUE;
write(checker);
number := number div checker;
else
incr(checker);
end if;
end while;
if number <> 1 then
if laterElement then
write(" * ");
end if;
laterElement := TRUE;
write(number);
end if;
end func;
const proc: main is func
local
var integer: number is 0;
begin
writeln("1: 1");
for number range 2 to 2147483647 do
write(number <& ": ");
writePrimeFactors(number);
writeln;
end for;
end func;</lang>
Output:
1: 1 2: 2 3: 3 4: 2 * 2 5: 5 6: 2 * 3 7: 7 8: 2 * 2 * 2 9: 3 * 3 10: 2 * 5 11: 11 12: 2 * 2 * 3 13: 13 14: 2 * 7 15: 3 * 5 . . .
Tcl
This factorization code is based on the same engine that is used in the parallel computation task. <lang tcl>package require Tcl 8.5
namespace eval prime {
variable primes [list 2 3 5 7 11]
proc restart {} {
variable index -1 variable primes variable current [lindex $primes end]
}
proc get_next_prime {} {
variable primes variable index if {$index < [llength $primes]-1} { return [lindex $primes [incr index]] } variable current while 1 { incr current 2 set p 1 foreach prime $primes { if {$current % $prime} {} else { set p 0 break } } if {$p} { return [lindex [lappend primes $current] [incr index]] } }
}
proc factors {num} {
restart set factors [dict create] for {set i [get_next_prime]} {$i <= $num} {} { if {$num % $i == 0} { dict incr factors $i set num [expr {$num / $i}] continue } elseif {$i*$i > $num} { dict incr factors $num break } else { set i [get_next_prime] } } return $factors
}
# Produce the factors in rendered form
proc factors.rendered {num} {
set factorDict [factors $num] if {[dict size $factorDict] == 0} { return 1 } dict for {factor times} $factorDict { lappend v {*}[lrepeat $times $factor] } return [join $v "*"]
}
}</lang> Demonstration code: <lang tcl>set max 20 for {set i 1} {$i <= $max} {incr i} {
puts [format "%*d = %s" [string length $max] $i [prime::factors.rendered $i]]
}</lang>
Visual Basic .NET
<lang vbnet>Module CountingInFactors
Sub Main()
For i As Integer = 1 To 10
Console.WriteLine("{0} = {1}", i, CountingInFactors(i))
Next
For i As Integer = 9991 To 10000
Console.WriteLine("{0} = {1}", i, CountingInFactors(i))
Next
End Sub
Private Function CountingInFactors(ByVal n As Integer) As String
If n = 1 Then Return "1"
Dim sb As New Text.StringBuilder()
CheckFactor(2, n, sb)
If n = 1 Then Return sb.ToString()
CheckFactor(3, n, sb)
If n = 1 Then Return sb.ToString()
For i As Integer = 5 To n Step 2
If i Mod 3 = 0 Then Continue For
CheckFactor(i, n, sb)
If n = 1 Then Exit For
Next
Return sb.ToString() End Function
Private Sub CheckFactor(ByVal mult As Integer, ByRef n As Integer, ByRef sb As Text.StringBuilder)
Do While n Mod mult = 0
If sb.Length > 0 Then sb.Append(" x ")
sb.Append(mult)
n = n / mult
Loop
End Sub
End Module</lang> Output:
1 = 1 2 = 2 3 = 3 4 = 2 x 2 5 = 5 6 = 2 x 3 7 = 7 8 = 2 x 2 x 2 9 = 3 x 3 10 = 2 x 5 9991 = 97 x 103 9992 = 2 x 2 x 2 x 1249 9993 = 3 x 3331 9994 = 2 x 19 x 263 9995 = 5 x 1999 9996 = 2 x 2 x 3 x 7 x 7 x 17 9997 = 13 x 769 9998 = 2 x 4999 9999 = 3 x 3 x 11 x 101 10000 = 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5