@Dave great one.
On Sun, Feb 27, 2011 at 3:33 AM, Dave <dave_and_da...@juno.com> wrote:
> Another optimization: Since the sum of the digits of n^2 is 45, n^2 is
> divisible by 9. Thus, we need consider only n values that are
> divisible by 3. Thus, the outer for-loop can be written
>
> for( n = 31992 ; n < 99381 ; n+=3 )
>
> Dave
>
> On Feb 23, 7:02 pm, Dave <dave_and_da...@juno.com> wrote:
>> Try this:
>>
>> int i,k,n;
>> long long j,nsq;
>> for( n = 31623 ; n < 100000 ; ++n )
>> {
>> nsq = (long long)n * (long long)n;
>> j = nsq;
>> k = 0;
>> for( i = 0 ; i < 10; ++i )
>> {
>> k |= (1 << (j % 10));
>> j /= 10;
>> }
>> if( k == 01777 )
>> printf("%i %lli\n",n,nsq);
>> }
>>
>> It finds 76 answers in the blink of an eye, the first being 32043^2
>> and the last being 99066^2.
>>
>> Dave
>>
>> On Feb 22, 3:17 pm, bittu <shashank7andr...@gmail.com> wrote:
>>
>>
>>
>> > How to find a number of 10 digits (non repeated digits) which is a
>> > perfect square? perfect square examples: 9 (3x3) 16 (4x4) 25(5x) etc.
>> > Ten digit number example 1,234,567,890
>>
>> > Thanks & Regards
>> > Shashank- Hide quoted text -
>>
>> - Show quoted text -
>
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Thanks & Regards,
Gaurav Gupta
Associate Software Engineer
IBM Software Lab India
Email: gauravgupta[at]in[dot]ibm[dot]com
Ph No. : +91-7676-999-350
"Quality is never an accident. It is always result of intelligent
effort" - John Ruskin
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