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 Neither 3 nor 6 (Posted on 2011-03-11)
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 No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 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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