@Bhavesh: Check the squares of the integers from
ceiling(sqrt(123456789)) to floor(sqrt(987654321)) to see which ones
contain all nine nonzero digits. Since the sum of the nine nonzero
digits is 45, a satisfactory square will be a multiple of 9, and
therefore, we only need consider the squares of integers that are
multiples of 3. Something like this should do the trick:
int i, j, k;
for( i = 11112 ; i < 31426 ; i += 3 )
{
j = i * i;
k = 0;
do
{
k |= 1 << (j % 10);
j /= 10;
} while( j > 0 );
if( k = 0x3FE ) // 11 1111 1110 in binary.
printf("%i\n",i);
}
Dave
On Jun 27, 11:01 pm, Bhavesh agrawal <agr.bhav...@gmail.com> wrote:
> All the nine digits are arranged here so as to form four square numbers:
>
> 9 81 324 576
>
> How would you put them together so as to form single smallest possible
> square number and a single largest possible square number..
>
> 139854276 and 923187456 are the answers given everywhere but how to proceed
> this ??
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