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challenge for programmable calculators
12-24-2013, 07:45 PM (This post was last modified: 12-24-2013 10:07 PM by Thomas Klemm.)
Post: #36
RE: challenge for programmable calculators
(12-24-2013 05:43 PM)cruff Wrote:  I wish I could have proved that \(a\) must be \(1\), which would have further limited the search for n by a factor of 7, but that escapes my aging brain at the moment.
That's brilliant! We can now use Gerson's idea and solve the quadratic equation for b:
\[ ab^2+(1+a^2-9a)b+11a+1=0\]
The determinant \((1+a^2-9a)^2-4a(11a+1)\) is only positive for a = 1 which leads to the well known solutions b = 3 and b = 4.

Here's a program for the HP-42S:
Code:

00 { 56 Byte Prgm }
01 1.007
02 STO 00
03 LBL 00
04 11
05 RCL 00
06 IP
07 STO 01
08 STO 02
09 1/X
10 +
11 STO* 02
12 9
13 LASTX
14 RCL+ 01
15 -
16 2
17 /
18 ENTER
19 ENTER
20 X^2
21 R^
22 -
23 X<0?
24 GTO 01
25 SQRT
26 STO- ST Z
27 +
28 RCL 02
29 STO+ ST Z
30 +
31 9
32 STO* ST Z
33 *
34 STOP
35 LBL 01
36 ISG 00
37 GTO 00
38 END

Cheers
Thomas
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Messages In This Thread
Proof using number theory - cruff - 12-24-2013, 05:43 PM
RE: challenge for programmable calculators - Thomas Klemm - 12-24-2013 07:45 PM
RE: challenge for programmable calculators - radwilliams - 12-24-2013, 05:57 PM



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