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 Point to 2013 (Posted on 2013-01-09)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 No Subject | Comment 4 of 7 |
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

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