NB: italic font denotes a concatenation, as opposed to a product.

Math solution:

Note: abcabc=1001 x abc

The prime factors of 1001 are 7, 11, 13. 

The factors of 1001 are: 7,11,13,91,143 

Since 1001 = f1 x c1 where f1 is a factor and c1 is its co-factor, 

f1 x 1001 / f1^2 = c1

Each factor yields a solution, but only c1=7, f1= 143 yields a>0

e.g.: 143143/143^2=7, while 091091/91^2 = 11

Note also that this doesn't work again soon. That is, amongst 101, 1001, 10001, 100001, etc.,  1001 is unusual to have a factor, 143, with only one less digit. Many of the others are prime or composite of only two primes.  

Computer solution:

rabbit-3:~ lord$ sqq

143143 / 143^2 =   7.000

rabbit-3:~ lord$ more sqq.f

        program elvis

        implicit none

        integer a,b,c,abc,abc2,abcabc,ir

        real*8 r

           do a=1,9

                do b=0,9

                if(a.eq.b)go to 3

                   do c=0,9

                   if(c.eq.a.or.c.eq.b)go to 2

                   abc=100*a+10*b+c

                   abc2=abc**2

                   abcabc=1001*abc

                   r=(1.*abcabc)/(1.*abc2)

                   ir=r

                   if(abs(1*ir-r).lt.0.0001)print1, abcabc,abc,r

1                  format(i6,' / ',i3,'^2 =',f8.3)

2                  enddo

3               enddo

           enddo

        end

(There are no additional solutions when a,b,c are allowed to be non-unique, so long as a>0)

Edited on September 22, 2018, 10:06 am

Posted by Steven Lord on 2018-09-22 08:05:10