Replying to myself, I should have printed i*i instead of i near the
end of the code:
printf("%i\n",i*i);
Dave
On Jun 27, 11:47 pm, Dave <dave_and_da...@juno.com> wrote:
> @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 ??- Hide quoted text -
>
> - Show quoted text -
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