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Hi all,
A month has elapsed, and March, 7th is a pretty good day for another
S&SMC, this time S&SMC#14, Cooking Conjectures !
Preamble
Number Theory is a fascinating branch of Mathematics, where we
mostly deal with such basic, fundamental entities as integer numbers.
Despite the apparent simplicity, experience shows that it's
actually very easy to make integerrelated conjectures that
seem extremely simple on the outside, yet are nearly intractable.
A good example would be the infamous Fermat Last Theorem (FLT),
which remained a conjecture for several centuries despite the
strongest, most strenuous attempts at proving it.
FLT was at long last proved, thus leaving its conjecture status (despite its name) to
actually become a theorem proper, but many other important
conjectures are still awaiting for either some clever demonstration, which
will break new barriers and advance knowledge,
or else a counterexample which proves them false, which doesn't usually advance
knowledge an inch but at least gives
peace of mind and conclusively stops further costly, timewasting attempts to try
and prove them. When such
a counterexample is found, we say that the conjecture is
cooked, i.e., unsound, false.
For this very S&SMC#14, put on your chef's hat
and let's try and cook some nice conjectures for dinner !
The Challenge
For each of these three plausible conjectures, you must find a
cook, i.e., the lowest counterexample that falsifies them. In
all cases you are done if you manage to find but one counterexample, within the
stated ranges. Optimizing for maximum speed will be first priority, then
for program size and simplicity. These are the conjectures
to cook:
Conjecture 1: Welldone
Some famous mathematician of old noted the fact
that, apparently, you could find any number of integer solutions
to the equations:
a^{2} + b^{2} = c^{2} (e.g.: 3^{2} + 4^{2} = 5^{2})
a^{3} + b^{3} + c^{3} = d^{3} (e.g.: 3^{3} + 4^{3} + 5^{3} = 6^{3})
...
where the number of terms added up is the same as the power, but
he was able to prove that there were no nontrivial integer
solutions to:
a^{3} + b^{3} = c^{3}
where the number of terms added up was less than the power.
He thus conjectured that such equations as:
a^{4} + b^{4} + c^{4} = d^{4}
a^{5} + b^{5} + c^{5} + d^{5} = e^{5}
...
etc., would also have no nontrivial solutions at all. Now,
you must prove him wrong by finding a counterexample for
the 5thpower case, i.e., write a program that finds one
solution of:
a^{5} + b^{5} + c^{5} + d^{5} = e^{5}
for nonzero, positive integer values a, b, c, d, e less
than 150. Such a solution eluded mathematicians for
two centuries till it was found in 1966 using
tremendous computing power for the time. Just duplicate
the feat using your small HP handheld calculator and
your programming ingenuity.
Conjecture 2: Medium
Another less wellknown mathematician stated the
following conjecture:
"Every positive integer greater than 5 is the sum of a prime and a power"
For instance:
...
1234 = 991 + 3^{5}
1235 = 79 + 34^{2}
1236 = 11 + 35^{2}
1237 = 337 + 30^{2}
1238 = 13 + 35^{2}
...
etc. You must find the smallest counterexample N
which cannot be expressed as the sum of a
prime and a power, limiting your
search to the range from N = 6 to 2000. Of
course the prime numbers are 2,3,5,... (so 1 is not
considered to be a prime), and the powers are
1 ^{1} = 1, 2 ^{2} = 4, 2 ^{3} = 8, 3 ^{2} = 9, 2 ^{4}= 16, ...,
(so 0 is not considered to be a power for the
purposes of decomposition into the sum of a prime
and a power, but 1 is).
Conjecture 3: Rare
Finally, a more recent and amusing conjecture is
that all positive integer numbers can be made into a palindromic number
(i.e., one which reads the same from right or from
left, such as 123474321) by reversing their digits and adding the
result to the original number, then repeating these steps
until you get a palindromic number. For instance, 78 gets
palindromic in 4 cycles:
cycle 1: 78 + 87 = 165
cycle 2: 165 + 561 = 726
cycle 3: 726 + 627 = 1353
cycle 4: 1353 + 3531 = 4884, palindromic
You must find the smallest alleged counterexample N, for
N up to 200, which fails to produce a palindromic result
after M cycles, where M should go up to 200 cycles minimum,
preferably up to 1000 cycles. Any number N which fails
to produce a palindromic result after M cycles, for
suitably large M (say 200, 500, or 1000) will be considered a
counterexample for the purposes of this challenge.
Your program must ask for the maximum number of cycles
to perform, M, and must output any and all values of
N up to 200 which do not produce a palindromic result.
Be aware that the numbers involved will get very big
very soon. Your program must cater for this, without
ever producing overflow or losing significant digits.
Caveat Emptor
The usual caveats apply, namely:
 Do not post just solutions, actual code is mandatory.
If you won't post code to accompany your alleged solutions at the time of posting them, then do not post the solutions, period. Wait till you can post both.
 Code must be for an HP calculator preferably, but other vintage handhelds are acceptable as well. Posting code written in Visual Basic, Java, FORTRAN, or any other PC language will be considered unpolite and disrespectful.
 Try to achieve a proper balance between you manually doing most
of the work and having a "PRINT (solution)" program which
does nothing of interest, and you contributing no enlightening ideas
of your own
and having instead
a dumb, bruteforce program which delivers the goods but
takes ages to run. The ideal is to use some clever ingenuity to significantly speed up the program while still letting it do the hard work
as it should. Remember the old computerese proverb:
"The Program was made for man, not man for the Program"
Anyway, you'll need some reasonable algorithmic and
programming ingenuity if you don't want to kill your
batteries in one go.
I'll post my own original solutions next Monday, which will be
three simple programs 7, 8, and 9lines long for the HP71B.
Though far from being the last word in stateoftheart programming, they're didactically simple and they're
fast and they get the work done reasonably quickly. I'll post
background and relevant comments as well.
Let's see yours.
Best regards from V.
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</lurk mode> Regarding challenge 3, I "know" that the answer is 196.
<lurk mode>
very much appreciating your challenges (without ever submitting code...)
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Hi Valentin,
Nice challenges and nicely formulated. I immediately started thinking about them, but, unpleasantly enough, I can't think of any other solution than to scan possibilities, which is disapproved in the first place. So for the moment no contributions from my side.
Still, as I was anxious to know the answer, I did write a bruteforce program for my 32SII for the first challenge to see how fast it would run. Well, fast enough, but the amount of computations will take a few days to complete, despite the fact that I don't examine combinations that I already have computed but in different order.
Obviously this #14 is not my cup of tea, but I like reading it and I'm looking forward to your approaches. As always.
groeten,
Bram
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Hi, Bram:
Bram posted:
"Obviously this #14 is not my cup of tea, but I like reading it and I'm looking forward to your approaches. As always."
Thanks for your interest and kind words but don't despair so soon. Even if Conjecture 1 would seem hard, it actually isn't that much, and you still have Conjectures 2 & 3 which are easier to cook.
However, the HP32SII is probably not the ideal calc for this #14, because of its extremely limited amount of RAM. After you've entered some clever programming you'll be left with too few bytes for necessary variables, and even the program itself can't be much more refined than a pure bruteforce search in that little RAM.
If you can, I suggest you try instead one of these models: HP41CV/CX, HP42S, HP71B, HP48/49 series. Even if you don't have any of these models available, you can always get Emu71 or other free emulators from the net, and try your might with that.
Don't give up, I know you can succeed ! :) You might also consider that if all interested people simply get the lazy bug and decide to just wait and see whatever solutions I will eventually post, I might consider the response a total failure and stop altogether posting any further challenges ... in other words, if I do work hard to produce them in the first place, then interested people are expected to work as well trying to solve them for good, else no deal.
This isn't intended as a showcase of my abilities but an interaction between all of us who care for this, where we'll get to see some fun math and interesting programming techniques for a variety of models, and hopefully enjoy it all and learn something valuable in the process.
Best regards from V.
Posts: 620
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Hello!
Thank you for this nice challenge, really something worth thinking about whike driving to work and back...
Quote: Bram: Nice challenges and nicely formulated. I immediately started thinking about them, but, unpleasantly enough, I can't think of any other solution than to scan possibilities, which is disapproved in the first place. So for the moment no contributions from my side.
The same for me, I'm afraid. Nevertheless (I'm an engineer, not a mathematician ;) ), I tried a brute force "attack" on challenge no.1 with a little Cprogramme on my PowerBook (I wish I had this Casio pocket calculator with "C" programming language, that would make my attempt a valid entry...) and it took 18 seconds (*) to find the result(s) (**). Which means: definitely no answer within my remaining lifetime on any vintage programmable calculator!
The only alternatives that come to mind would be an iterative approach or an optimisation (simplex method?) approach, but both would not fit into any of my programmable calculators.
Looking forward to see the real smart solutions!
Greetings, Max
(*) and 27 Minutes, if searching up to 300 instead of 150 ... which probably tranlates to a millenium on the hp41
(**) my very first real brutebrutebrute approach would have taken several hours, so at least some thinking had to go into it.
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Hi, Maximilian:
Maximilian posted:
"Thank you for this nice challenge, really something worth thinking about whike driving to work and back... "
You're welcome, glad you find it interesting.
"I tried a brute force "attack" on challenge no.1 [...] and it took 18 seconds [...] which means: definitely no answer within my remaining lifetime on any vintage programmable calculator!"
Believe me, it's not as hard as it seems. Obviously launching a pure bruteforce search with the equivalent of four or five fornext loops is certain to take ages, as I mentioned in my original posting. But some clever refinements plus careful reading of the given conditions can make all the difference in the world, to the point where a vintage handheld, and certainly such models as the modern 48/49 series, can solve it in a few hours at most, if not mere minutes.
"Looking forward to see the real smart solutions! "
I think you're overestimating the real difficulty. That's usually the way with my challenges: most people find them extremely difficult at first, till they realize that they're actually quite manageable, even easy given the right approach. Else, I wouldn't post them, it's no fun to ask solutions to challenges that would forcibly require the use of a fast, fullfledged computer.
Go ahead and if you don't manage to cope with Conjecture 1, try
Conjectures 2 & 3 instead, they're probably easier (that's why I labelled them "Welldone" (hardest), "Medimum" (soso hard) and "Rare" (easiest) :)
Thanks for your interest and best regards from V.
Posts: 123
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Hi,
Any chance that you could put your C solution online? I'd be happy to port it to the HP49 series and make it a valid entry (yes, you can use C with these, and the result is very fast).
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Hi!
Quote:
Any chance that you could put your C solution online? I'd be happy to port it to the HP49 series and make it a valid entry (yes, you can use C with these, and the result is very fast).
I have uploaded it here:
http://www.bombie.de/tmp/hochfuenf.c
Greetings, Max
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Hi Max,
a PB2000C version is in your mailbox.
Marcus
Posts: 3,283
Threads: 104
Joined: Jul 2005
Hi Max & Valentin,
Quote:
(I wish I had this Casio pocket calculator with "C" programming language, that would make my attempt a valid entry...) and it took 18 seconds (*) to find the result(s) (**)
I *do* have a PB2000C and I was able to drain a set of expensive batteries with the program without ever coming near a solution...
So I tried to do something about it and I shortened the loops as good as I could. Here is the program:
/*
Short&Sweet Math Challenges No.14 http://www.hpmuseum.org/
Version for Casio PB2000C
*/
/* IMIN: starting point */
#define IMIN 150
/* IMAX: highest number to search */
#define IMAX 150
/* There is no command line in Casio C */
main()
{
double a5 = 0.0;
double b5 = 0.0;
double c5 = 0.0;
double d5 = 0.0;
double e5 = 0.0;
double sum = 0.0;
double sumc = 0.0;
double sumd = 0.0;
double diff = 0.0;
double z = 0.0;
double vsmall = 1.0E10;
int imin = 0;
int imax = 0;
int i = 0;
int ia = 0;
int ib = 0;
int ic = 0;
int id = 0;
int ie = 0;
double *powtab = NULL;
int lpowtab = 0;
int output = 0;
/* get the parameters */
printf( "imin=" );
scanf( "%d", &imin );
if ( imin == 0 ) imin = IMIN;
printf( "imax=" );
scanf( "%d", &imax );
if ( imax == 0 ) imax = IMAX;
printf( "output=" );
scanf( "%d", &output );
printf("SSMC #14.1: Searching up to %d\n", imax);
/* power table to save multiplictions */
lpowtab = imax + 1;
powtab = (double *) malloc(lpowtab * sizeof(powtab[0]));
if (powtab == NULL) {
printf("*** ERROR ***\nMemory allocation failed.\n");
return;
}
for (i = 1 ; i < imin ; i++) {
z = (double) i;
powtab[i] = z * z * z * z * z;
}
/* outermost loop for "e" */
for (ie = imin ; ie <= imax ; ie++) {
z = (double) ie;
e5 = powtab[ie] = z * z * z * z * z;
if (output) {
clrscr();
}
printf("e=%4d e^5=%15.0f\n", ie, e5);
/* compute lower bound of "d" loop */
z = e5  3 * powtab[ie1];
id = z <= 2 ? 1 : (int) pow( z, 0.2 );
if (output) {
gotoxy(12,1);
printf("%3d",id);
}
for (; id < ie; id++) {
d5 = powtab[id];
if (output) {
gotoxy(12,1);
printf("%3d",id);
gotoxy(16,1);
printf("%15.0f",d5);
}
/* compute lower bound of "c" loop */
z = e5  3 * d5;
ic = z <= 2 ? 1 : (int) pow( z, 0.2 );
if (output) {
gotoxy(0,1);
printf(" %3d",ic);
}
for (; ic <= id; ic++) {
c5 = powtab[ic];
sumc = d5 + c5;
if (output) {
gotoxy(8,1);
printf("%3d",ic);
gotoxy(16,1);
printf("%15.0f",sumc);
}
/* check if c^5 + d^5 exeeds e^5 */
if ( sumc > e5 ) break;
/* compute lower bound of "b" loop */
z = e5  d5  2 * c5;
ib = z <= 2 ? 1 : (int) pow( z, 0.2 );
if (output) {
gotoxy(0,1);
printf(" %3d",ib);
}
for (; ib <= ic; ib++) {
b5 = powtab[ib];
sumd = sumc + b5;
if (output) {
gotoxy(4,1);
printf("%3d",ib);
gotoxy(16,1);
printf("%15.0f",sumd);
}
/* check if b^5 + c^5 + d^5 exeeds e^5 */
if ( sumd > e5 ) break;
/* compute lower bound of "a" loop */
z = e5  sumd;
ia = z <= 2 ? 1 : (int) pow( z, 0.2 );
if (output) {
gotoxy(0,1);
printf("%3d",ia);
}
for (; ia <= ib; ia++) {
a5 = powtab[ia];
sum = sumd + a5;
diff = sum  e5;
if (vsmall < diff && diff < vsmall) {
if (output) clrscr();
printf("SOLUTION: a=%4d b=%4d c=%4d\n d=%4d e=%4d\n",
ia, ib, ic, id, ie);
beep(1);
if (output) getchar();
}
else if ( sum > e5 ) {
/* a^5 + b^5 + c^5 + d^5 exeeds e^5 */
break;
}
else {
if (output) {
gotoxy(0,1);
printf("%3d",ia);
gotoxy(16,1);
printf("%15.0f",sum);
}
}
}
}
}
}
}
free(powtab);
powtab = NULL;
printf("\nFinished\n");
}
I'm relucant to post the result, because I've only tested the program above with time consuming screen output enabled and even after giving it a whole night it didn't reach more than e=70. At the time of this writing, the software is running again, this time with a restricted set of values for e to check ;). I hope that it will come to a solution before Sunday.
To provide a solution for an HP calculator of old, I ported it back to DOS Borland C and let it run on my venerable HP 200lx. The machine seems to be much faster than the Casio, having reached e=65 after about 15 or 20 minutes. The Casio runs interpreted code (just like BASIC) while the HP runs a compiled DOS program.
I strongly believe that the search algorithm must be improved greatly to have a chance on a slow calculator.
Marcus
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After a few hours of hard computational work, the HP 200lx was finally able to deliver:
27^5+84^5+110^5+133^5=144^5
My poor Casio has at least arrived at showing the same result after being told to start with e=144...
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Hi Valentin,
this time I had some spare time to try to solve at least one of your challenges. Here is a BASIC program for the third conjecture.
The program
(Indentation and empty lines added later in the listing)
100 REM S&SMC 14, Problem 3
110 INPUT "Number of cycles=";NC
120 INPUT "From=";N1,"To=";N2
130 DIM D(1,199):REM digits
140 FOR N=N1 TO N2
160 REM Split N in digits
170 ID=0:L=0
180 T=N
190 D(ID,L)=T10*INT(T/10)
200 T=INT(T/10)
210 L=L+1
220 IF T<>0 THEN 190
225 L=L1
230 REM Cycle loop
240 FOR C=1 TO NC
250 GOSUB 1000:REM display
260 REM Add and Test
270 F=1:CY=0
280 FOR I=0 TO L
290 D1=D(ID,I):D2=D(ID,LI)
300 F=F AND (D1=D2)
310 IF F AND (I>=LI)THEN I=L:C=NC:GOTO 350:REM Palindromic
320 S=D1+D2+CY:CY=0
330 IF S>9 THEN S=S10:CY=1
340 D(1ID,I)=S
350 NEXT I
360 ID=1ID:REM swap source&dest
365 IF CY=1 THEN L=L+1:D(ID,L)=1
370 NEXT C
380 IF F=0 THEN 410
390 NEXT N
400 END
1000 REM display
1005 PRINT N;C;":";
1010 FOR II=0 TO L
1020 PRINT CHR$(48+D(ID,II));
1030 NEXT II
1040 PRINT
1050 RETURN
I wrote the program on a TI74 with bare 8K RAM. The BASIC used here is pretty standard and I assume that a HP71 could run the same software with minor changes (: > @).
How is it done?
I didn't make the attempt to handle the sums as integers or real numbers because the number of digits available (10 to 14) wouldn't be enough. Splitting an array of 10 digit numbers in individual digits seemed to be too time consuming. So I traded memory for speed in this first attempt. Each individual digit is stored as a number in an array.
The array is defined with DIM D(1,199). TIBasic does not support single precision or integer variables so this is a real waste of memory. The array caters for two numbers with up to 200 digits. A posible modification could be to use two character strings (with a maximum length of 255 in most BASICs) or to PEEK/POKE directly into memory.
The first dimension is used to switch between the source and the destination number. There is only one source number, read from left to right for the first summand and from right to left for the second. The other number is the sum. After each cycle, the roles are changed (see line 360).
While adding, two things happen. A comparison is made between the two digits added (line 300) and the result of the comparison is ANDed into a flag (F). If the flag is still set after half of the digits are processed, the number is a palindrome and the rest of the two inner loops are skipped (line 310.)
Secondly, a carry (CY) is set whenever the sum is greater than 9. This carry is than added in the next step of the digit loop (lines 320340.)
The results
The program took about half an hour to find the result 196 with 200 cycles. The size of the array should be enough for more than 400 cycles (the longest number I got has 88 digits). The following output was created with a modified version of the software that can PRINT to a datafile on the PC (via the PC interface).
1 1 :1
2 1 :2
3 1 :3
4 1 :4
5 1 :5
6 1 :6
7 1 :7
8 1 :8
9 1 :9
10 1 :10
10 2 :11
11 1 :11
12 1 :12
12 2 :33
13 1 :13
13 2 :44
14 1 :14
14 2 :55
15 1 :15
15 2 :66
16 1 :16
16 2 :77
17 1 :17
17 2 :88
18 1 :18
18 2 :99
19 1 :19
19 2 :110
19 3 :121
20 1 :20
20 2 :22
21 1 :21
21 2 :33
22 1 :22
23 1 :23
23 2 :55
24 1 :24
24 2 :66
25 1 :25
25 2 :77
26 1 :26
26 2 :88
27 1 :27
27 2 :99
28 1 :28
28 2 :110
28 3 :121
29 1 :29
29 2 :121
30 1 :30
30 2 :33
31 1 :31
31 2 :44
32 1 :32
32 2 :55
33 1 :33
34 1 :34
34 2 :77
35 1 :35
35 2 :88
36 1 :36
36 2 :99
37 1 :37
37 2 :110
37 3 :121
38 1 :38
38 2 :121
39 1 :39
39 2 :132
39 3 :363
40 1 :40
40 2 :44
41 1 :41
41 2 :55
42 1 :42
42 2 :66
43 1 :43
43 2 :77
44 1 :44
45 1 :45
45 2 :99
46 1 :46
46 2 :110
46 3 :121
47 1 :47
47 2 :121
48 1 :48
48 2 :132
48 3 :363
49 1 :49
49 2 :143
49 3 :484
50 1 :50
50 2 :55
51 1 :51
51 2 :66
52 1 :52
52 2 :77
53 1 :53
53 2 :88
54 1 :54
54 2 :99
55 1 :55
56 1 :56
56 2 :121
57 1 :57
57 2 :132
57 3 :363
58 1 :58
58 2 :143
58 3 :484
59 1 :59
59 2 :154
59 3 :605
59 4 :1111
60 1 :60
60 2 :66
61 1 :61
61 2 :77
62 1 :62
62 2 :88
63 1 :63
63 2 :99
64 1 :64
64 2 :110
64 3 :121
65 1 :65
65 2 :121
66 1 :66
67 1 :67
67 2 :143
67 3 :484
68 1 :68
68 2 :154
68 3 :605
68 4 :1111
69 1 :69
69 2 :165
69 3 :726
69 4 :1353
69 5 :4884
70 1 :70
70 2 :77
71 1 :71
71 2 :88
72 1 :72
72 2 :99
73 1 :73
73 2 :110
73 3 :121
74 1 :74
74 2 :121
75 1 :75
75 2 :132
75 3 :363
76 1 :76
76 2 :143
76 3 :484
77 1 :77
78 1 :78
78 2 :165
78 3 :726
78 4 :1353
78 5 :4884
79 1 :79
79 2 :176
79 3 :847
79 4 :1595
79 5 :7546
79 6 :14003
79 7 :44044
80 1 :80
80 2 :88
81 1 :81
81 2 :99
82 1 :82
82 2 :110
82 3 :121
83 1 :83
83 2 :121
84 1 :84
84 2 :132
84 3 :363
85 1 :85
85 2 :143
85 3 :484
86 1 :86
86 2 :154
86 3 :605
86 4 :1111
87 1 :87
87 2 :165
87 3 :726
87 4 :1353
87 5 :4884
88 1 :88
89 1 :89
89 2 :187
89 3 :968
89 4 :1837
89 5 :9218
89 6 :17347
89 7 :91718
89 8 :173437
89 9 :907808
89 10 :1716517
89 11 :8872688
89 12 :17735476
89 13 :85189247
89 14 :159487405
89 15 :664272356
89 16 :1317544822
89 17 :3602001953
89 18 :7193004016
89 19 :13297007933
89 20 :47267087164
89 21 :93445163438
89 22 :176881317877
89 23 :955594506548
89 24 :1801200002107
89 25 :8813200023188
90 1 :90
90 2 :99
91 1 :91
91 2 :110
91 3 :121
92 1 :92
92 2 :121
93 1 :93
93 2 :132
93 3 :363
94 1 :94
94 2 :143
94 3 :484
95 1 :95
95 2 :154
95 3 :605
95 4 :1111
96 1 :96
96 2 :165
96 3 :726
96 4 :1353
96 5 :4884
97 1 :97
97 2 :176
97 3 :847
97 4 :1595
97 5 :7546
97 6 :14003
97 7 :44044
98 1 :98
98 2 :187
98 3 :968
98 4 :1837
98 5 :9218
98 6 :17347
98 7 :91718
98 8 :173437
98 9 :907808
98 10 :1716517
98 11 :8872688
98 12 :17735476
98 13 :85189247
98 14 :159487405
98 15 :664272356
98 16 :1317544822
98 17 :3602001953
98 18 :7193004016
98 19 :13297007933
98 20 :47267087164
98 21 :93445163438
98 22 :176881317877
98 23 :955594506548
98 24 :1801200002107
98 25 :8813200023188
99 1 :99
100 1 :100
100 2 :101
101 1 :101
102 1 :102
102 2 :303
103 1 :103
103 2 :404
104 1 :104
104 2 :505
105 1 :105
105 2 :606
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Marcus
Edited: 9 Mar 2006, 8:24 a.m.
Posts: 68
Threads: 1
Joined: Jul 2005
Hi Valentin,
Thanks you for another S&SMC. I always find them fun to try, but lately it's been an even tougher challenge to find time to work on them.
I wrote a program in HP71B Basic to find a solution for the first part of the challenge. I was going to write it in Sharp Basic, but the Sharp handhelds I have only support 10 significant digits and the program needs to be able to handle integers with 11 digits.
I don't have an actual HP71B, so I used emu71. This was the first time I used this emulator and I agree that it is a great piece of software.
Here is my program. There are some small optimizations to reduce the number of cases that are tested (including starting the outer loop at 150 and working backward  thanks for the hint). However, it still took about 10 minutes to find the answer, even when running with emu71.
10 FOR E=150 TO 5 STEP 1 @ E5=E^5 @ FOR D=E1 TO 4 STEP 1 @ D5=D^5 @ C1=INT ((E5D5)^.2)
15 IF C1>D THEN D=0 @ GOTO 70
20 FOR C=C1 TO 3 STEP 1 @ C5=C^5 @ B1=INT((E5D5C5)^.2) @ IF B1>C THEN C=0 @ GOTO 60
30 FOR B=B1 TO 2 STEP 1 @ B5=B^5 @ A=INT((E5D5C5B5)^.2) @ IF A>B THEN B=0 @ GOTO 50
40 IF A^5+B5+C5+D5=E5 THEN 100
50 NEXT B
60 NEXT C
70 NEXT D
80 NEXT E
90 STOP
100 PRINT A;" ";B;" ";C;" ";D
110 PRINT E
It finds the solution 27^5 + 84^5 + 110^5 + 133^5 = 144^5.
I'll see if I can make time to try the other parts of the challenge this weekend.
Best Regards,
Eamonn
Posts: 2,761
Threads: 100
Joined: Jul 2005
Hello Valentin,
Here is my attempt to cook conjecture #1. Ok, it's in QBASIC, but the conversion to HP71B basic should be straightfoward. So, not to be completely off the rules, I tested it on the HP200LX. After 26 minutes and 15 seconds the answer was found:
144 133 110 84 27
The program could be ported to the HP42S but it would take even longer to run. Perhaps the 49G+ gets the same result in less than 10 minutes. Though the algorithm finds the right answer, it may still have flaws, which everybody is welcome to point me out. Of course, it is easier when we know the answer. If I hadn't seen Eamonn's and other people's solutions, I might not have even tried. Thanks all of you!
As always, I am looking forward to your solutions.
5 CLS
10 DEFDBL AH, OZ
15 DEFINT IN
25 DIM Q(150)
30 FOR I = 2 TO 150
35 T = I * I
40 Q(I) = T * T * I
45 NEXT
50 FOR I = 150 TO 6 STEP 1
55 E5 = Q(I)
60 LOCATE 1, 1: PRINT I
65 FOR J = I  1 TO 5 STEP 1
80 FOR K = J  1 TO 4 STEP 1
85 FOR L = K  1 TO 3 STEP 1
87 IF Q(J) + Q(K) > E5 THEN 110
90 FOR M = L  1 TO 2 STEP 1
91 IF Q(J) + Q(K) + Q(L) > E5 THEN 105
92 S = Q(J) + Q(K) + Q(L) + Q(M)
94 IF S < E5 THEN 110
98 IF S = E5 THEN 130
100 NEXT M
105 NEXT L
110 NEXT K
115 NEXT J
120 NEXT I
130 CLS: PRINT I, J; K; L; M

Edited to include Emu71 version:
25 STD @ DIM Q(150)
30 FOR I=2 TO 150 @ Q(I)=I^5 @ NEXT I
50 FOR I=150 TO 6 STEP 1
55 E5=Q(I) @ DISP I
65 FOR J=I1 TO 5 STEP 1
80 FOR K=J1 TO 4 STEP 1
85 FOR L=K1 TO 3 STEP 1
87 IF Q(J)+Q(K)>E5 THEN 110
90 FOR M=L1 TO 2 STEP 1
91 IF Q(J)+Q(K)+Q(L)>E5 THEN 105
92 S=Q(J)+Q(K)+Q(L)+Q(M)
94 IF S<E5 THEN 110 ELSE IF S=E5 THEN 130
100 NEXT M
105 NEXT L
110 NEXT K
115 NEXT J
120 NEXT I
130 PRINT I,J;K;L;M
Since this takes roughly one hour to find the answer (Emu71 @ 500MHz), I think my previous estimation of running time on the 49G+ is wrong: I'd say less than 10 hours instead of less than 10 minutes :(
Edited: 12 Mar 2006, 4:20 p.m.
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In order to test the algorithm, I made a slight modification in the QBasic program. So beginning with 580, the following results were obtained in about 5 minutes (Pentium III @ 500 MHz, QBX  Compiled QBASIC):
576 532 440 336 108
432 399 330 252 81
288 266 220 168 54
144 133 110 84 27
Exactly as expected, since (n*a)^5+(n*b)^5+(n*c)^5+(n*d)^5=(n*e)^5. Now, a question arises: besides {133, 110, 84, 24} and their multiples, are there any other fournumber sets (and their multiples) that 'cook' the conjecture? Or is this set unique?

I've just found the answer to this question. According to this paper, NEW RESULTS IN EQUAL SUMS OF LIKE POWERS, the set is unique. Perhaps I should attend to an introductory course on Number Theory :)
Edited: 12 Mar 2006, 6:22 p.m.
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I think I had better stick to my first idea:
85359^{5} = 85282^{5} + 28969^{5} + 3183^{5} + 55^{5} was found by J. Frye (J.C. Meyrignac, pers. comm., Sep. 9, 2004):
http://mathworld.wolfram.com/DiophantineEquation5thPowers.html
What will come next?
Gerson.
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Quote:
85359^{5} = 85282^{5} + 28969^{5} + 3183^{5} + 55^{5}
This is even beyond the accuracy of my double precision Sharp PCE500. The sum has 25 digits while the Sharp can handle 20 digit precision. You need at least 83 bits (unsigned binary) to store a number of that size.
Marcus
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But verified on the 49 series, in "exact" mode.
Regards, James
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James,
I simply forgot about the CAS machines which come with arbitrary precision integer arithmetic. The TI Voyage 200 can calculate the sum as well.
Marcus
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Joined: Jul 2005
Hi Valentin,
here is my attempt at the second problem in RPL on an HP49g+.
First Step: Computing Prime Numbers
The following program, named MKPR, computes a list of primes. The upper limit is given on the stack:
%%HP: T(3)A(R)F(.);
\<< { 2. } 'PRIME' STO 3. SWAP
FOR i 1. \> j
\<<
DO i PRIME j GET
IF DUP2 SQ \>=
THEN
IF / FP
THEN j 1. + 'j' STO 0.
ELSE 1.
END
ELSE DROP2 PRIME i + 'PRIME' STO 1.
END
UNTIL
END
\>> 2.
STEP
\>>
It computes the following list if 2000 is entered on the stack:
{ 2. 3. 5. 7. 11. 13. 17. 19. 23. 29. 31. 37. 41. 43. 47. 53. 59. 61. 67. 71. 73. 79. 83. 89. 97. 101. 103. 107. 109. 113. 127. 131. 137. 139. 149. 151. 157. 163. 167. 173. 179. 181. 191. 193. 197. 199. 211. 223. 227. 229. 233. 239. 241. 251. 257. 263. 269. 271. 277. 281. 283. 293. 307. 311. 313. 317. 331. 337. 347. 349. 353. 359. 367. 373. 379. 383. 389. 397. 401. 409. 419. 421. 431. 433. 439. 443. 449. 457. 461. 463. 467. 479. 487. 491. 499. 503. 509. 521. 523. 541. 547. 557. 563. 569. 571. 577. 587. 593. 599. 601. 607. 613. 617. 619. 631. 641. 643. 647. 653. 659. 661. 673. 677. 683. 691. 701. 709. 719. 727. 733. 739. 743. 751. 757. 761. 769. 773. 787. 797. 809. 811. 821. 823. 827. 829. 839. 853. 857. 859. 863. 877. 881. 883. 887. 907. 911. 919. 929. 937. 941. 947. 953. 967. 971. 977. 983. 991. 997. 1009. 1013. 1019. 1021. 1031. 1033. 1039. 1049. 1051. 1061. 1063. 1069. 1087. 1091. 1093. 1097. 1103. 1109. 1117. 1123. 1129. 1151. 1153. 1163. 1171. 1181. 1187. 1193. 1201. 1213. 1217. 1223. 1229. 1231. 1237. 1249. 1259. 1277. 1279. 1283. 1289. 1291. 1297. 1301. 1303. 1307. 1319. 1321. 1327. 1361. 1367. 1373. 1381. 1399. 1409. 1423. 1427. 1429. 1433. 1439. 1447. 1451. 1453. 1459. 1471. 1481. 1483. 1487. 1489. 1493. 1499. 1511. 1523. 1531. 1543. 1549. 1553. 1559. 1567. 1571. 1579. 1583. 1597. 1601. 1607. 1609. 1613. 1619. 1621. 1627. 1637. 1657. 1663. 1667. 1669. 1693. 1697. 1699. 1709. 1721. 1723. 1733. 1741. 1747. 1753. 1759. 1777. 1783. 1787. 1789. 1801. 1811. 1823. 1831. 1847. 1861. 1867. 1871. 1873. 1877. 1879. 1889. 1901. 1907. 1913. 1931. 1933. 1949. 1951. 1973. 1979. 1987. 1993. 1997. 1999. }
You should set numeric and approximate mode before creating the list. The list is hopefully correct; if not, please complain!
Second Step: Computing Powers
The next RPL program, named MKPOW, creates a list of powers as indicated in the challenge. The upper limit is entered on the stack:
%%HP: T(3)A(R)F(.);
\<< { 1. } 'POWR' STO \> n
\<< 2. n LN 2. LN /
FOR p 2. n p XROOT
FOR i POWR i p ^
IF DUP2 POS
THEN DROP2
ELSE + 'POWR' STO
END
NEXT
NEXT
\>> POWR SORT 'POWR' STO
\>>
Here is the result:
{ 1. 4. 8. 9. 16. 25. 27. 32. 36. 49. 64. 81. 100. 121. 125. 128. 144. 169. 196. 216. 225. 243. 256. 289. 324. 343. 361. 400. 441. 484. 512. 529. 576. 625. 676. 729. 784. 841. 900. 961. 1000. 1024. 1089. 1156. 1225. 1296. 1331. 1369. 1444. 1521. 1600. 1681. 1728. 1764. 1849. 1936. }
If anything is wrong with my list, post a comment, please!
Final Step: Search for Sums
Here is my attempt, named SSMC14, to find a number which cannot be represented as a sum of elements taken from each of the lists:
%%HP: T(3)A(R)F(.);
\<< 6. SWAP
FOR n
n 3 DISP
0. \> p
\<<
DO p 1. + 'p' STO n POWR p GET 
IF DUP 0. >
THEN PRIME SWAP POS
ELSE n HALT 1.
END
UNTIL
END
\>>
NEXT
\>>
If it finds a solution, it hits a breakpoint. The prgramm must be killed manually or can be continued to search for more solutions. The upper limit is entered on the stack. Both lists must be created beforehand.
Here is my result:
1549
Marcus
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Hi, Marcus:
Your result is correct. Timing ?
Best regards from V.
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Quote:
Your result is correct. Timing ?
Wow!
Creating the lists is a matter minutes (PRIME, 2:10) or Seconds (POWR, 0:04). The search itself took 25 minutes. All measurements were done on a 49g+
Marcus
Posts: 362
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Unfortunately I haven't had time to look at the SSMC but I would have used the NEXTPRIME command on the 49.
It works correctly up to a few millions.
Arnaud
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Arnaud,
NEXTPRIME might be worth a try to speed up building the PRIME list but we cannot gain more than about 2 minutes here. On the other hand, the programs posted should work on the lower end machines, too.
The 48S cannot sort the power list but the roles of the two lists in program SSMC14 can be reversed and the prime list is always sorted. This results in a performance penalty because the userRPL code must operate on the longer of the two lists and the builtin operation POS works on the shorter. The 48G has SORT and can use a sorted POWR list.
I'll probably do some tests on my 48 series machines, too.
Marcus
Posts: 3,283
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I revised the above solution slightly and added a controlling program to get some timing information.
MKPOW
%%HP: T(3)A(R)F(.);
\<< { 1. } 'POWR' STO
\> n
\<<
2
n LN 2 LN / @ log_{2} n as upper limit
FOR p
2
n p XROOT @ n^{1/p} as upper limit
FOR i
POWR i p ^
IF DUP2 POS @ check if already in list
THEN DROP2
ELSE + 'POWR' STO @ add to list
END
NEXT
NEXT
\>>
POWR SORT 'POWR' STO @ sort list (only on 48G and up)
\>>
MKPR
%%HP: T(3)A(R)F(.);
\<<
{ 2. 3. } 'PRIME' STO @ set up list with first elements
5 SWAP @ loop from 5 to limit on stack
FOR i
2 \> j @ j indexes into the list for divisors
\<<
DO
i @ number to test
PRIME j GET @ get divisor from list
IF DUP2 SQ \>= @ check if beyond reasonable limit
THEN
IF / FP @ check if divisible
THEN
'j' INCR DROP @ not divisible, try next divisor
0 @ continue with DOUNTIL
ELSE
1 @ leave DOUNTIL loop
END
ELSE
DROP2 @ drop i and divisor
PRIME i +
'PRIME' STO @ add new prime to list
1 @ leave DOUNTIL loop
END
UNTIL @ flag is already on the stack
END
\>>
2 STEP @ next odd number to try
\>>
SSMC14
%%HP: T(3)A(R)F(.);
\<<
FOR n @ both limits must be on stack!
n 3 DISP @ inform user
0 \> p @ p indexes into the POWR list
\<<
DO
n @ number to test
POWR 'p' INCR GET @ first summand from POWR list
 @ other summand
IF DUP 0 > @ test if both summands are positive
THEN
PRIME SWAP POS @ leave DOUNTIL if summand is a prime
ELSE
DROP n @ Heureka! Show just the result on the stack
1E99 'n' STO @ leave FORNEXT
1 @ leave DOUNTIL
END
UNTIL @ flag is already on the stack
END
\>>
NEXT
\>>
SSMC14V2
Slower but works with unsorted POWR list on HP48S
%%HP: T(3)A(R)F(.);
\<<
FOR n
n 3 DISP
0 \> p
\<<
DO
n
POWR 'p' INCR GET

IF DUP 0 >
THEN POWR SWAP POS
ELSE DROP n 1E99 'n' STO 1
END
UNTIL
END
\>>
NEXT
\>>
TIM
Timing loop
%%HP: T(3)A(R)F(.);
\<<
TIME
2000 MKPOW TIME 600 .1 BEEP
2000 MKPR TIME 600 .1 BEEP
6 2000 SSMC14 TIME 600 .1 BEEP
\>>
The software is currently executing on an HP48G, I'll post timing information later.
Marcus

Here are the execution times on a 48G:
MKPOW: 00:00:16 (h:m:s)
MKPR: 00:09:09
SSMC14: 02:05:10
This is roughly factor 4 compared to the 49g+.
Marcus
Edited: 13 Mar 2006, 8:16 a.m.
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Here is my Emu71 version for problem #1 slightly modified to avoid some unnecessary additions. Now it is about 50% faster:
>LIST
25 STD @ DIM Q(150) @ FOR I=2 TO 150 @ Q(I)=I^5 @ NEXT I
50 FOR I=150 TO 6 STEP 1 @ E5=Q(I) @ DISP I
65 FOR J=I1 TO 5 STEP 1 @ FOR K=J1 TO 4 STEP 1 @ FOR L=K1 TO 3 STEP 1
70 S1=Q(J)+Q(K) @ IF S1>E5 THEN 110
80 FOR M=L1 TO 2 STEP 1 @ S2=S1+Q(L) @ IF S2>E5 THEN 105
90 S3=S2+Q(M) @ IF S3<E5 THEN 110 ELSE IF S3=E5 THEN 130
100 NEXT M
105 NEXT L
110 NEXT K @ NEXT J
120 NEXT I
130 PRINT I,J;K;L;M
>
>RUN
150
149
148
147
146
145
144
144 133 110 84 27
Running time: 47 min 21 sec @ 500MHz (previously 1 hour 10 min 36 sec).
Regards,
Gerson.
Posts: 1,755
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Joined: Jan 2005
Hi all:
Thanks to all of you who were interested in this S&SMC#14, and most specially
to the outstanding ones (Marcus von Cube, Eamonn, Gerson W. Barbosa, Mr. Hohmann) who actually cared to take the time to develop and post their very clever solutions, I really hope you enjoyed it and consider the afforded time well spent. These are my original solutions, with comments.
Conjecture 1: Welldone
This is a particular case of a plausible conjecture issued by Euler in 1769,
which resisted every effort to prove or disprove it for almost two centuries until it was finally disproved in 1966 when
the first counterexample was found by Lander and Parkin, using a then quite
powerful CDC 6600 computer.
Now, we're asked to duplicate that feat, namely finding a nontrivial solution
to A^{5}+B^{5}+C^{5}+D^{5}
=E^{5} in nonzero integers, using our favorite handhelds.
A straightforward, dumb bruteforce search would require excessive running
times (flattening several sets of batteries in the process), but a few
refinements here and there will cut down the task by several orders of
magnitude. However, these refinements need a significant amount of RAM to
implement, which means only advanced models like the HP41CX, HP42S,
HP71B, and HP48/49 can be profitably used.
Here's my original 7line, 251byte solution for the HP71B
1 DESTROY ALL @ OPTION BASE 0 @ K=150 @ DIM P(K)
2 FOR I=0 TO K @ P(I)=I*I*I*I*I @ NEXT I @ R=.2 @ L=2^R @ M=3^R @ N=4^R
3 FOR E=K TO 1 STEP 1 @ T=P(E) @ DISP E
4 FOR D=E1 TO E/N STEP 1 @ F=TP(D) @ U=F^R
5 FOR C=INT(U) TO U/M STEP 1 @ G=FP(C) @ V=G^R @ FOR B=INT(V) TO V/L STEP 1
6 IF NOT FP((GP(B))^R) THEN A=(GP(B))^R @ DISP A;B;C;D;E @ END
7 NEXT B @ NEXT C @ NEXT D @ NEXT E
upon running, it eventually outputs:
>RUN
27 84 110 133 144
which is the soughtfor counterexample, as indeed
27^{5} + 84^{5} + 110^{5} + 133^{5} = 61917364224 = 144^{5}
The running time is a short 90 min. in an actual, physical HP71B, and only
13 seconds under Emu71 in my IBM laptop. To achieve such short times
the following techniques are used to speed the search, they key of which is to
iteratively reduce the problem to a similar yet easier one:
 first of all, a table of 5^{th}powers for integer arguments from 0 to 150 is
precomputed, so that during the search the extremely frequent
need of raising values to the 5^{th} power is reduced to indexing
an array element, which is significantly faster.
 some other needed constants are precomputed as well, which saves
additional time when they're used inside the search's nested
loops, namely 1/5, 2^{1/5}, 3^{1/5},
4^{1/5}
 the outer loop traverses all possible values for E in reverse
order, from the largest possible value to the smallest.
 for earch value of E, we must try to decompose E^{5} as the sum of
four 5^{th}powers. Assuming D is the
largest one,
we must search for
a decomposition of E^{5}D^{5} in three
5^{th}powers where, as D is the largest element by definition, we only need to
search for values of D in the limited range:
(E^{5}/4)^{1/5} <= D
< (E^{5})^{1/5} i.e. E/4^{1/5} <= D < E
which we traverse in reverse order as well. For each value of D,
we must now try to decompose E^{5}D^{5}
as the sum of three 5^{th}powers,
so we've reduced our problem to a similar but simpler one,
and the same argument is repeated till a single summand is
left, which we simply check to ascertain whether it is a
5^{th}power or not (i.e.: the fractional part of its fifth root equals 0).
When this condition is met, we've found a counterexample so
the program outputs it and stops.
Conjecture 2: Medium
We're asked for the smallest integer number from 6 to 2000 that can't be
decomposed as the sum of a prime and a nonzero power.
A bruteforce search would again require excessive running time, but
we can trade RAM for speed once more. This is my original 9line,
259byte solution for the HP71:
1 DESTROY ALL @ OPTION BASE 1 @ M=2000 @ D=LOG(M) @ N=INT(SQR(M))
2 INTEGER P(M),Q(310) @ FOR B=2 TO N
3 FOR E=2 TO INT(D/LOG(B)) @ P(B^E)=1 @ NEXT E @ NEXT B @ P(1)=1 @ Q(1)=2 @ I=2
4 FOR N=3 TO M1 STEP 2 @ FOR D=3 TO INT(SQR(N)) STEP 2 @ IF MOD(N,D)=0 THEN 6
5 NEXT D @ Q(I)=N @ I=I+1
6 NEXT N @ T=I1 @ FOR N=6 TO M
7 FOR D=1 TO T @ E=NQ(D) @ IF E<=0 THEN DISP N @ END ELSE IF P(E) THEN 9
8 NEXT D @ DISP N @ END
9 NEXT N
upon running, it outputs:
>RUN
1549
which is the smallest counterexample. The running
time is a fast 30 min.
in a physical HP71B, and only 6 seconds under Emu71 in my IBM laptop,
thanks to these simple yet efficient techniques:
 first of all, this challenge's actually far easier than it seems at first.
Actually, for the search to proceed fast and smoothly, we need but two things: (1) to have
all relevant prime numbers instantly at hand, and (2) to be able to test
extremely quickly whether a given integer is
a perfect power or not.
This is achieved as follows:
 all primes up to 2000 are precomputed and stored in an array so
that they can be retrieved very fast during the search
 in order to quickly check whether a given integer is a perfect power or
not, we stablish a 2000element 'flag' array where each element
has the value 1 or 0 depending on whether the index is a perfect
power or not, i.e., P(7)=0, P(8)=1 ( as 8=2^{3}), etc. This could be achieved using much less RAM by packing eight such 'flags' per byte in a string, for instance, but here speed is our top priority so we simply define an integer array and let each individual element act as a 'flag'.
 with primes and very fast powerdetection available, the main
search just traverses all possible values from 6 to 2000, in
ascending order. For each, every prime is subtracted in turn
and the result is checked to see if it is a perfect power,
skipping to the next value if it is. When no prime subtracted
will result in a perfect power remaining, the corresponding
value is output as a valid counterexample.
 by the way, the prime array is dimensioned to have 310 elements
because there are some 300+ prime numbers in the range from 2
to 2000 (actually, 303). If you've got INTEGRATE capabilities
in your HP model, you can very quickly compute an approximate
value for the required number of elements as follows:
approx. #primes up to M = INT(INTEGRAL(2,M,1,1/LN(IVAR))
so that the line:
2 INTEGER P(M),Q(310) @ ...
becomes:
2 INTEGER P(M),Q(INTEGRAL(2,M,1,1/LN(IVAR))) @ ...
and this allows you to expand the search to numbers greater than
2000 by simply changing the value assigned to M at line 1,
without having to consult tables for the proper size of
array Q. As the integration's uncertainty is specified as 1, (i.e: FIX 0
or SCI 0 for models that use the display setting instead), the integral
is computed very quickly.
Now, you'll agree with me that a line which dimensions an
array with a size defined by a nonelementary integral
isn't that frequent a sight.
Conjecture 3: Rare
This is a wellknown conjecture as yet unproved or disproved,
though everybody and his uncle feels that 196 is a true counterexample, as
it has failed to produce a palindrome after hundreds and hundreds of millions
of cycles. For instance, after 670,000,000 cycles you're dealing with numbers
280,000,000digit long(!!), yet no palindrome in sight ...
In order to implement this challenge, multiprecision addition
is mandatory, which can be done with arrays or else
with strings, which is the technique I've used in my original
8line, 364byte solution for the HP71B:
1 DIM M$[500],N$[500] @ INPUT "#Cmax=";M @ FOR I=0 TO 200 @ N$=STR$(I) @ K=0
2 CALL MADD(N$,M$,K) @ IF M$=REV$(M$) THEN 4 ELSE IF K<M THEN N$=M$ @ GOTO 2
3 DISP I;"fails (";M$;" after";M;"cycles)"
4 NEXT I @ DISP "OK" @ END
5 SUB MADD(A$,C$,K) @ DIM B$[500] @ L=LEN(A$) @ E=10^11 @ C$="" @ C=0
6 B$=REV$(A$) @ FOR I=L TO 1 STEP 11 @ C=VAL(A$[I10,I])+VAL(B$[I10,I])+C
7 D$=STR$(MOD(C,E)) @ C$=RPT$("0",11LEN(D$))&D$&C$ @ C=C DIV E @ NEXT I
8 C$=LTRIM$(STR$(C)&C$,"0") @ K=K+1 @ END SUB
The algorithm is completely straightforward and no special techniques are
needed, though both for speed and to avoid obscuring the inner works with
trivial utility routines, I've made use of several stringhandling
keywords (REV$, RPT$, LTRIM$) which are available in a number
of very common LEX files and ROMs (STRNGLEX, REVLEX, RPTLEX, JPC ROM, etc).
The program works as follows:
 the multiprecision results will be stored in strings (which in the
case of the HP71B would allow us to handle up to 65000digit numbers!),
initially dimensioned to hold up to 500digit numbers, more than enough
for up to 1000 cycles.
 to perform the multiprecision addition of a number and its mirror
image, a call to the MADD subprogram is made. This subprogram takes
the number as one of its string arguments pased by value, and returns
the result of the addition as another string argument passed by
reference. For instance:
>CALL MADD("78",M$,0) @ M$
165
>>CALL MADD("8263485213753244473212",M$,0) @ M$
10387229637326370316840
because 78+87 = 165 and 8263485213753244473212+2123744423573125843628
= 10387229637326370316840. Actually, the subprogram's code could
be inserted directly in the main program proper, so we could get
rid of the subprogram altogether and get a slightly faster, shorter
program (7 lines instead of 8) but 'outsourcing' particular tasks
to subprograms encourages modular programming and makes for clearer,
cleaner code and provides additional functionality as well.
After calling the subprogram, the main program just checks if the new
result is palindromic,
or if we've exhausted the number of cycles, and iterates as needed.
Upon running, it produces the following, for assorted maximum number of cycles
(10, 20, 40, 100, 200, 1000):
>RUN
#Cmax=10
89 fails (8872688 after 10 cycles)
98 fails (8872688 after 10 cycles)
167 fails (17050517 after 10 cycles)
177 fails (17794887 after 10 cycles)
187 fails (17735476 after 10 cycles)
196 fails (18211171 after 10 cycles)
OK
>RUN
#Cmax=20
89 fails (93445163438 after 20 cycles)
98 fails (93445163438 after 20 cycles)
187 fails (176881317877 after 20 cycles)
196 fails (70446464506 after 20 cycles)
OK
>RUN
#Cmax=40
196 fails (13305261530450734933 after 40 cycles)
OK
>RUN
#Cmax=100
196 fails (44757771534490515617290699271561508443627774644 after 100 cycles)
OK
>RUN
#Cmax=200
196 fails (910449546741765655298269802255629632301207255281
2103235826563197972803556567037646054008 after 200 cycles)
OK
>RUN
#Cmax=1000
196 fails (35346644392413689785837714402912114362859098083
41408344020861450405992918328457190349563871687
95800463971545914548326676428378028814710683108
50549641273388365259932008237493462055424091251
57901200166876923521977766210101074152201325440
26439582289914006246477437313605494900387117318
73088382467552483640965506947400858697069355944
18174493380829951504425811945379423290791058264
41012030342772858788740429334664452 after 1000 cycles)
OK
So it's clear that 196 is a very firm candidate
for a counterexample. Of course, it's not the only one, other candidates
include (up to N=2000): 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, 978, 986, 1495,
1497, 1585, 1587, 1675, 1677, 1765, 1767, 1855, 1857, 1945, 1947, and 1997.
That's all. Thanks again and
Best regards from V.
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Hello Valentin,
Unfortunatly, I had little time for your challenge, but the first part looked especially interesting for me, trying to reproduce a proof made on 1966 on a large computer with a small handheld of the '80s.
And your solution is so simple, as usual ...
Do you have a reference to the original paper from Lander and Parkin? I would like to know more on how they proceeded.
JF
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Hi, JeanFrançois:
JeanFrançois posted:
"Unfortunatly, I had little time for your challenge, but the first part looked especially interesting for me, trying to reproduce a proof made on 1966 on a large computer with a small handheld of the '80s."
Yes, I missed your usual topquality contributions to my challenges, it's a real pity you couldn't give it a try. And I agree on the fun factor of having a quite petite handheld doing essentially the same as an original mainframe computer which used to fill a whole room:
The awesome CDC 6600 mainframe computer
This CDC 6600 was designed by Seymour Cray himself, of Cray supercomputers fame, and had a 60bit processor capable of 4.58 Mflops, impressive for its time indeed.
"And your solution is so simple, as usual ..."
Thanks, JF, but I insist, my challenges *are* simple, that's why I call them "Short & Sweet". They never require more than a few lines of code, and simple code at that.
"Do you have a reference to the original paper from Lander and Parkin? I would like to know more on how they proceeded."
Reference, yes:
"Lander, L. J. and Parkin, T. R. "A Counterexample to Euler's Sum of Powers Conjecture." Math. Comput. 21, 101103, 1967."
Unfortunately, the paper itself, no. It seems to be available only from nonfree subscription services. As far as I know, their procedure was similar to mine, only more ambitious because they also attacked more general equations, involving powers up to the 6^{th}. It's quite possible for they to have made use of congruences (i.e. particular remainders when using specific divisors, etc) to discard ranges of tentative values for the inner loops.
I did try a few congruences but the program got more complicated and worse, no speed was actually gained. In a CDC 6600 assemblylanguage program or compiled code, using just integer operations, congruences would be very effective. But for the floating point HP71B environment, no gains are made in this case.
Thanks for your interest, and please don't miss my next challenge, come next April 1st.
Bes regards from V.
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Hi Valentin,
Quote:
... and please don't miss my next challenge, come next April 1st
Can we expect an April's fool joke?
Marcus
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Hello JeanFrançois,
And don't miss the Euler.net for interesting stuff on conjectures.
Cordialement.
Etienne
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Thank you for an interesting challenge.
I was disappointed to see that no actual math trick could be used here, that was rather a programmer's challenge.
Please accept my apologies for posting a solution without code, I was just thrilled to 'know' the solution of one of the challenges.
Actually I tried Challenge 1 on my 1BBBB, and it ran about the same as solutions showed her, but with no optimizations and I felt it was too slow and ridiculous to be posted here.
To make for my inappropriate post, here is a small challenge for you : solve 28^x = 19^y + 87^z for x, y and z integers.
There IS some math trickery possible...
See you on the 1st of April.
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Hi GE,
28^x = 19^y + 87^z for x, y and z integers
You didn't give us the solution...
JF
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Hi, JeanFrançois:
JeanFrançois posted:
"Hi GE,
28^x = 19^y + 87^z for x, y and z integers
You didn't give us the solution..."
There is no solution to the equation you mention in your post, so it's no big surprise the person you're addressing would fail to provide one.
Per my posted commitment I read or answer nothing by anonymous posters so I don't quite know what's all about. Some kind of joke, perhaps ?
Best regards from V.
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Valentin et al
I had a go at this myself with no joy
Quote:
See you on the 1st of April.
Could this have been a clue!
Regards
Chris
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Hi, Chris:
Chris posted:
"I had a go at this myself with no joy. Quote:
See you on the 1st of April. Could this have been a clue!"
Can't say. As stated, I didn't read the original post JF mentions, so I don't know what was asked or any further comments about it, though judging from JF's post, it seems that a solution was asked to a diophantine equation which has none.
It that was indeed the case, how this is supposed to be considered enjoyable by people wasting their time and computer resources on an impossible task is beyond me. Perhaps the mere idea of having people indulge in fruitless efforts seemed indeed enjoyable to the original poster, but I can't say for sure, who knows ...
Best regards from V.
