Find all positive integers x, y, z so that:
- x2 + y = z and:
- Base ten expansion of each of x and y has precisely n digits and that of z has precisely 2n digits, where n > 1.
- All the digits of x are the same, all the digits of y are the same and all the digits of z are the same.
Prove that there are no others.
After some single-digit z's, the program produced
x y and z
2 7 11
3 2 11
4 6 22
5 8 33
6 8 44
7 6 55
8 2 66
9 7 88
33 22 1111
66 88 4444
88 33 7777
333 222 111111
666 888 444444
3333 2222 11111111
6666 8888 44444444
33333 22222 1111111111
66666 88888 4444444444
333333 222222 111111111111
666666 888888 444444444444
3333333 2222222 11111111111111
6666666 8888888 44444444444444
It looks like all 3's for x and all 2's for y produce all 1's for z, and all 6's for x and all 8's for y produce all 4's for z, for as large values of n as you like.
DefDbl A-Z
Dim crlf$
Private Sub Form_Load()
Text1.Text = ""
crlf$ = Chr(13) + Chr(10)
Form1.Visible = True
DoEvents
xt = 0
For xl = 1 To 7
xt = 10 * xt + 1
For xd = 1 To 9
x = xt * xd
yt = 0
For yl = 1 To 15
yt = 10 * yt + 1
For yd = 1 To 9
y = yt * yd
z = x * x + y
zs$ = LTrim$(Str(z))
good = 1
For i = 2 To Len(zs$)
If Mid(zs$, i, 1) <> Left(zs$, 1) Then good = 0: Exit For
Next
If good Then Text1.Text = Text1.Text & Str(x) & Str(y) & Str(z) & crlf$
Next
Next
Next
Next
End Sub
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Posted by Charlie
on 2014-05-11 16:07:53 |
computer exploration
