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Neither 3 nor 6 (Posted on 2011-03-11) Difficulty: 4 of 5
0, 25, 2025, 13225…are squares that remain squares if every digit in the number defining them is augmented by 1.
Let's call them squarish numbers.

a. List two more samples of squarish numbers.
b. Prove that all such numbers are evenly divisible by 25.
c. Why are there neither 3-digit nor 6-digit squarish numbers?
d. Prove that between 10^k and 10^(k+1) there is at most one squarish number.

See The Solution Submitted by Ady TZIDON    
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re: Part a only (spoiler) | Comment 2 of 6 |
(In reply to Part a only (spoiler) by Charlie)

list
   10   Repu=1:Rep9=9*Repu
   20   for N=1 to 99999999
   30    Sq=N*N
   40    while Sq>Rep9
   50      Repu=Repu*10+1
   60      Rep9=9*Repu
   70    wend
   80    S$=cutspc(str(Sq))
   90    if instr(S$,"9")=0 then
  100      :Tst=Sq+Repu
  110      :Sr=int(sqrt(Tst)+0.5)
  120      :if Sr*Sr=Tst then
  130        :print Sq;N,Tst;Sr
  140   next N
OK

run
 25  5   36  6
 2025  45        3136  56
 13225  115      24336  156
 4862025  2205   5973136  2444
 60415182025  245795     71526293136  267444
 207612366025  455645    318723477136  564556
OK

verifies that there are no 13, 14, 15 or 16-digit squarish numbers either (as 99999999^2 is a 16-digit number that starts with a 9), in addition to the 8, 9 and 10 I already noted.


  Posted by Charlie on 2011-03-11 13:54:36
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