Determine the minimum value of a positive integer N such that the 2nd, 3rd, 4th and 5th digits in order immediately following the decimal point (reading left to right) in the base ten expansion of √N is 2013.

*** For an extra challenge, solve this puzzle without using a computer program.

The program listing below yields these first 20 results.

  4532          67.32013071882734
 13976         118.2201336490532
 23019         151.7201370945861
 31620         177.820133843162
 42568         206.3201395889408
 56664         238.0420130985285
 57322         239.4201328209472
 69127         262.9201399664925
 96547         310.7201313079023
 100375        316.8201382488178
 116636        341.5201311782367
 121536        348.6201371120148
 137137        370.3201317778983
 147011        383.4201350998667
 149862        387.1201363917925
 165584        406.9201395851525
 169102        411.220135693767
 186555        431.9201315058144
 197776        444.7201367152155
 199827        447.0201337747552


OPEN "c:\qb64\work\2013.txt" FOR OUTPUT AS #1

DEFDBL A-Z
n = 2: done = 0:
DO

 a = SQR(n)
 b = INT(a)
 c = a - b
 sh$ = STR$(c)
 sh$ = MID$(sh$, 4, 4)
 LOCATE 1 + done, 40: PRINT
 IF sh$ = "2013" THEN
  PRINT n, a
  PRINT #1, n, a
  done = done + 1
 END IF
 n = n + 1

LOOP WHILE done <> 20
CLOSE 1


I have tried to use Excel as a tool to come up with something by Trial and Error.
The closest to which I can arrive is the following long multiplication algorithm.
The idea is to firstly arrive at ".9" (or better still ".9999") in the product [N].

              z  y . x  2  0  1  3
              z  y . x  2  0  1  3
              -------------------
             3z 3y  3x  6  0  3  9
           z  y  x   2  0  1  3  0
         etc
the product line would be:
 100*z*z+10*2yz+(y*y+2xz)+(4z+2xy)/10+x*x+4y)/100+(4x+2z)/1000+ .....

With z and y being 9 the ceiling for N would be 10000, and allowing all digits from 0 to 9 for x, y and z we have 1000 trials!

Posted by brianjn on 2013-01-09 18:58:40