Here is a rather inelegant, but one-line Mathematica solution:

Select[Flatten[
  Array[If[IntegerQ[
       Sqrt[Extract[{1, 4, 9}, #1]*10^4 + 
         Extract[{1, 4, 9}, #2]*10^3 + Extract[{1, 4, 9}, #3]*10^2 + 
         Extract[{1, 4, 9}, #4]*10^1 + Extract[{1, 4, 9}, #5]*10^0]] &&
       OrderedQ[{Extract[{1, 4, 9}, #1], Extract[{1, 4, 9}, #2], 
        Extract[{1, 4, 9}, #3], Extract[{1, 4, 9}, #4], 
        Extract[{1, 4, 9}, #5]}], 
     Extract[{1, 4, 9}, #1]*10^4 + Extract[{1, 4, 9}, #2]*10^3 + 
      Extract[{1, 4, 9}, #3]*10^2 + Extract[{1, 4, 9}, #4]*10^1 + 
      Extract[{1, 4, 9}, #5]*10^0, 0] &, {3, 3, 3, 3, 3}]], # != 0 &]

The program returns {11449} .

1. Every digit is square number so the options are only 1,4,9.
2. It is an ordered number,so there are only few cases 1 * 9
3. we can write the combinations and get the square root of each to find the correct answer

The smallest possible 5 digit number, where each digit is a square number is 11111. The square root of this is approximately 105.4. Set up a spreadsheet to calculate the squares of numbers greater than 105. The answer is 11449, which is the square of 107.

This is Java:

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    public class Run {
      public static void main(String[] args) {
        for (int iterator = 10000; iterator < 99999; iterator++) {
            if (checkroot(iterator)) {
                if (checkdigits(iterator)) {
                    if (checkorder(iterator)) {
                        System.out.println("\t\t\t\t\t\tthis is it: " + iterator + " !!! ");}}}}
    }

    private static boolean checkorder(int number) {
        boolean result = true;
        int digit = 0;
        for (int i = 0; i < 5; i++) {
            digit = (number % 10);
            result = result && (digit >= (number / 10) % 10);
            number = number / 10;}
        return result;
    }
    private static boolean checkdigits(int number) {
        int remember = number;
        boolean result = true;
        int digit = (number % 10);
        for (int i = 0; i < 5; i++) {
            result = result && checkroot(digit);
            number = number / 10;
            digit = number % 10;}
        if (result) System.out.println("digits are perfect square for: " + remember);
        return result;
    }
    private static boolean checkroot(int number) {
        double root = Math.sqrt(number);
        return (root % 1 == 0);
    }
} 

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from math import *
for i in range(10**4,10**5):
    nums=[]
    for j in str(i):
        nums.append(int(j))
    if sqrt(i).is_integer() and all(nums[i] <= nums[i+1] for i in xrange(len(nums)-1)) and sqrt(int(str(i)[0])).is_integer()and sqrt(int(str(i)[1])).is_integer() and sqrt(int(str(i)[2])).is_integer() and sqrt(int(str(i)[3])).is_integer() and sqrt(int(str(i)[4])).is_integer():
        print i

Freaky Square :P

Java solution -

for(int i = 10_000; i < 100_000; i++)
        if(Math.round(Math.sqrt(i))==Math.sqrt(i))
            if (String.valueOf(i).contains("2") || 
                String.valueOf(i).contains("3") || 
                String.valueOf(i).contains("5") || 
                String.valueOf(i).contains("6") ||
                String.valueOf(i).contains("7") || 
                String.valueOf(i).contains("8") ){}
            else{
                System.out.println(i);
            }

}

This returns -

10000
10404
11449
14400
19044
40000
40401
44100
44944
90000

You can now manually pick the ordered number. I intentionally did this to keep the algorithm short but you can also pick up the number programmatically by converting it into a array.

The most basic solution I could think of:

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    int i, j, k, l, m, n;
    int temp;

    for(i = 10000 ; i < 100000 ; i++){

        temp = sqrt(i);

        if( i == temp*temp){

               //extracting the digits of the number

        j = i%10;                        //5th digit
        k = (i/10)%10;                 //4th digit
        l = (i/100)%10;               //3rd digit
        m = (i/1000)%10;             //2nd digit
        n = (i/10000);                //1st digit

       //checking all the conditions (I don't like using && command, so I did separated if cases)

        if( n <= m ){
            if ( m <= l ){
                if( l <= k ){
                    if( k <= j ){
                        temp = sqrt(n);
                        if( temp*temp == n ){
                            temp = sqrt(m);
                            if( temp*temp == m ){
                                temp = sqrt(l);
                                if( temp*temp == l ){
                                    temp = sqrt(k);
                                    if( temp*temp == k ){
                                        temp = sqrt(j);
                                        if( temp*temp == j ){
                                            temp = sqrt(i);
                                            if( temp*temp == i ){
                                                printf("%d\n", i);  }}}}}}}}}}}}

    return 0;
} 

that's my solution ,it's simple DFS :

def dfs(n,num): if n==5: if is_sqr(num) : s=str(num) if not not( s[1]>=s[0] and s[2]>=s[1] and s[3]>=s[2] \ and s[4]>=s[3] ): print num elif n==0: for i in range(1,4): dfs(n+1,a[i]) else: n+=1 for i in range(4): dell=num*10+a[i] dfs(n,dell)

Here's my python Solution:

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import math
num = 0
for a in range(10):
    if(a==4 or a==9 or a==1):
        for b in range(a, 10):
            if(b==0 or b==4 or b==9or b==1):
                for c in range(b, 10):
                    if(c==0 or c==4 or c==9or c==1):
                        for d in range(c, 10):
                            if(d==0 or d==4 or d==9or d==1):
                                for e in range(d, 10):
                                    if(e==0 or e==4 or e==9or e==1):
                                        num = a*10000+b*1000+c*100+d*10+e
                                        if((math.sqrt(num))== (int)(math.sqrt(num))):
                                            print num

Output = 11449

java code:

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    public class brilliant201409291605{
        public static void main(String[] args){
            int[] a = new int[10];
            a[0] = 1;
            a[1] = 4;
            a[4] = 9;
            int[] b = new int[5];
            for(int i=0;i<5;i++) b[i] = 0;
            for(int i=0;i<625;i++){
                for(int j=4;j>=0;j--){
                    if(b[j]==9){
                        b[j]=0;
                        continue;
                    }
                    b[j] = a[b[j]];
                    break;
                }
                if(b[0]==0) continue;
                boolean c = false;
                for(int j=0;j<4;j++){
                    if(b[j]>b[j+1]){
                        c = true;
                        break;
                    }
                }
                if(c) continue;
                int d = 0;
                for(int j=0;j<5;j++){
                    d *= 10;
                    d += b[j];
                }
                if(Math.sqrt(d)-((int)(Math.sqrt(d)))==0) System.out.println(d);
            }
        }
    }

output:

11449
Press any key to continue . . .

My C code for the problem. :)

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#include <stdio.h>
#include <math.h>
int main()
{
        float a,b,c,d,e;
        float num;

                for(a=1;a<=9;a++)
                        if(a==4 || a==9 || a==1)
                        for(b=a;b<=9;b++)
                                if(b==0 || b==4 || b==9|| b==1)
                                for(c=b;c<=9;c++)
                                        if(c==0 || c==4 || c==9|| c==1)
                                        for(d=c;d<=9;d++)
                                                if(d==0 || d==4 || d==9|| d==1)
                                                for(e=d;e<=9;e++)
                                                        if(e==0 || e==4 || e==9|| e==1)
                                                        {
                                                               num = a*10000+b*1000+c*100+d*10+e;
                                                               if(sqrt(num)== (int)sqrt(num) )
                                                                printf("%.0f", num);
                                                        }

        return 0;
}

Each no are square then squres no. In one digit is 1,4,&9 then it says its ordered no. Then we orderd 1, 4 & 9 in five digits its become a no 11449 the is a square of 107.. Thats the ans...

The most constricting condition is that the digits must each be square, so that means 1, 4, or 9 (or 0... but that would not work as this an ordered integer). I solved it by guessing that the answer would probably have all three and there are only a few such examples and I just used calculator to quickly spot the one with an integer square root. Hmmm what if you lack a calculator!? Then you first assume that the 5-digit number starts with 1. Then you know that the square root has to be between 100 (10.000) and 142 (root of 20,000 is 100 times root 2). And if you assume the number ends with 9, it the root has to end with 7. So try squaring 107, 117, 127, or 137. Happily the first one works. (If none of those had worked, and we would next have assumed the number starts with 4 and tried roots between 200 and 224 and so 207 and 217. The number cannot start with 9 since 99,999 is three squared times 11,111... but if 11,111 were a square then both 11,111 and 99,999 would be answers and there is only one answer. So if none of those had worked we would have known the number ends with 4 and been forced to go back and test 102, 108, 112, 118, 122, 128, 132, 138, 202, 208, 212, and 218)

I don't propose this as a "Solution" but here is how i did it:

The only viable digits are 1, 4 and 9.

5 digit square numbers start from the square of 100. They seem to end somewhere near the square of 300. Set up an column of numbers from 100 to 350 in excel. Calculate their square in the next column.

Now, soon as I did this, I saw the seventh number from the the top fits the bill. Answer is 11449, square of 107.

If it was not this apparent, what would I have done? Take out all the numbers that do not start with 1, 4 or 9.

The only viable number that starts with 9 is 99999. the square root is not a whole number. Take out all numbers starting with 9.

take out all numbers starting with 4 but have a lesser digit next to it. Then take out all numbers that have any digit except 4 or 9 as the second digit from left....and so on....

I am glad i didn't have to do this. Doesn't sound particularly interesting or elegant...

Is there any solution that does not involve such exhaustive checking?

You know, you forgot 44100, which is 210 squared. 0 is a square integer too.

include<conio.h>

include<stdio.h>

void main()

{ long int temp[220];

int i,j,a,m,n,o,p,f1,f2,f3,f4,f5;

a=1;

for(i=100;i<317;i++)

{
    temp[a]=i*i;

    a++;
}

for(j=1;j<218;j++)

   {
    m=temp[j]/10;    

    f1=temp[j]%10;      

    n=m/10;

    f2=m%10;

    o=n/10;

    f3=n%10;

    p=o/10;

    f4=o%10;

    f5=p;

    if(f1==0 || f1==1 || f1==4 || f1== 9)

    {
        if(f2==0 || f2==1 || f2==4 || f2== 9)

        {
            if(f3==0 || f3==1 || f3==4 || f3== 9)

            {
                if(f4==0 || f4==1 || f4==4 || f4== 9)

                {
                    if(f5==1 || f5==4 || f5== 9)

                    {
                    printf("Number is %lu \n",temp[j]);
                    }
                }
            }
        }
    }
}
getch();

}

Here is a Python 3.4.1 solution:

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    import math

    count = 0

    #Method to determine a square number
    def isSquare(n):
        if math.sqrt(n) == int(n**(1/2)):
            return True
        else:
            return False

    #This method determines if number is ordered
    def ordered(n):
        if sorted(list(str(n))) == list(str(n)):
            return True
        else:
            return False

    #Inspect all 5-digit numbers
    for x in range (10000, 100000): 
        if isSquare(x):    #check if it is a square
            for a in str(x):   
                #each digit must be a square
                if isSquare(int(a)):
                    #count = 5; if all digits are squares
                    count += 1 

            if count == 5:
                #check if the number is ordered
                if ordered(x):   
                    print (x)    #print out the result

        count = 0   #reset count

The program returns 11449 as the answer.