Sum of three cubes for 42 finally solved

09142019, 03:20 PM
Post: #1




Sum of three cubes for 42 finally solved  
09142019, 05:03 PM
(This post was last modified: 09142019 05:03 PM by Tugdual.)
Post: #2




RE: Sum of three cubes for 42 finally solved
(09142019 03:20 PM)Gerson W. Barbosa Wrote: https://m.phys.org/news/201909sumcube...life.htmlI’m a faithfull Numberfile follower so I saw that a week ago and since have eventually recovered my lost sleep 

09142019, 06:11 PM
Post: #3




RE: Sum of three cubes for 42 finally solved
...and the fun part; the HP/Free/DM42 is not able to calculate it's own self
Esben 28s, 35s, 49G+, 50G, Elektronika MK52 & MK61 

09142019, 09:14 PM
(This post was last modified: 09142019 09:20 PM by Gerson W. Barbosa.)
Post: #4




RE: Sum of three cubes for 42 finally solved
(09142019 06:11 PM)DA74254 Wrote: ...and the fun part; the HP/Free/DM42 is not able to calculate it's own self 34 significant digits are not enough for the task. That would require at least 50. Not a problem for the HP 50g in exact mode, though: '(80538738812075974)^3+80435758145817515^3+12602123297335631^3' EVAL > 42 PS.: Unfortunately there are no solutions for 49 and 50. 

09152019, 06:29 AM
Post: #5




RE: Sum of three cubes for 42 finally solved
Could this problem be converted into a sum of complex products? The 34S e.g. does carry enough digits if so.
I am impressed by the result. Pauli 

09152019, 12:37 PM
Post: #6




RE: Sum of three cubes for 42 finally solved
(09142019 06:11 PM)DA74254 Wrote: ...and the fun part; the HP/Free/DM42 is not able to calculate it's own self DM42 running WP43C works. I did it last week when this problem surfaced. WP43S will work by implication also. Jaco Mostert, Elec Eng 42S, DM42, WP34C/S, 28C, 35S, 32Sii, owned 11C; used 67, 85; iOS: 42s, Free42, WP31S/34S, HCalc; OSX: WP34C. 

09152019, 04:20 PM
(This post was last modified: 09152019 04:39 PM by Albert Chan.)
Post: #7




RE: Sum of three cubes for 42 finally solved
(09142019 09:14 PM)Gerson W. Barbosa Wrote: 34 significant digits are not enough for the task. That would require at least 50. We can split the numbers, like this: a = −80538738812075974 = −80538739e9 + 187924026 = a_{1} + a_{0} b = +80435758145817515 = +80435758e9 + 145817515 = b_{1} + b_{0} c = +12602123297335631 = +12602123e9 + 297335631 = c_{1} + c_{0} a³ + b³ + c³ = (a_{1} + a_{0})³ + (b_{1} + b_{0})³ + (c_{1} + c_{0})³ = (a_{1}³ + b_{1}³ + c_{1}³) + 3*(a_{0}a_{1}² + b_{0}b_{1}² + c_{0}c_{1}²) + 3*(a_{1}a_{0}² + b_{1}b_{0}² + c_{1}c_{0}²) + (a_{0}³ + b_{0}³ + c_{0}³) = 6628842934503040000000000000000000000000000 +6628842934562563667007015000000000000000000 59523703031106040674633000000000 +36024091040674633000000042 = 42 For calculators with precision of 12 decimal digits, we can do mod test: PHP Code: a = {80538738812, 075974}  a^3 ~ 5.2241E+50 lua> modn = {999907,999917,999931,999953,999959,999961,999979,999983,1e6} lua> for _, n in ipairs(modn) do print(modtest(n), '(mod ' .. n .. ')') end 42 (mod 999907) 42 (mod 999917) 42 (mod 999931) 42 (mod 999953) 42 (mod 999959) 42 (mod 999961) 42 (mod 999979) 42 (mod 999983) 42 (mod 1e+006) Except for 1e6, all mods are primes, we get a³ + b³ + c³ ≡ 42 (mod ~ 1e54) Since 1e54 ≫ sum of three cubes , we get a³ + b³ + c³ = 42 

09152019, 07:19 PM
(This post was last modified: 09152019 07:20 PM by DA74254.)
Post: #8




RE: Sum of three cubes for 42 finally solved
(09142019 09:14 PM)Gerson W. Barbosa Wrote:Yes, I too saw that it need 50 significant digits. Though the Prime also in CAS manages the tak.(09142019 06:11 PM)DA74254 Wrote: ...and the fun part; the HP/Free/DM42 is not able to calculate it's own self (09152019 12:37 PM)Jaco@cocooncreations.com Wrote:(09142019 06:11 PM)DA74254 Wrote: ...and the fun part; the HP/Free/DM42 is not able to calculate it's own self Where can I find the 43 program? Esben 28s, 35s, 49G+, 50G, Elektronika MK52 & MK61 

09152019, 11:51 PM
Post: #9




RE: Sum of three cubes for 42 finally solved
(09152019 07:19 PM)DA74254 Wrote:(09142019 09:14 PM)Gerson W. Barbosa Wrote: 34 significant digits are not enough for the task. That would require at least 50. Not a problem for the HP 50g in exact mode, though:Yes, I too saw that it need 50 significant digits. Though the Prime also in CAS manages the tak. WP43C is a keyboard alternative version of WP43S using native DM42 keyboard. WP43C details here: WP43C forum. Here are links to the emulator and DM42 firmware. WP43C code here Gitlab. WP43S can be found here Gitlab, including emulator and a DM42 firmware image in the Gitlab repo. Remember it is a work in progress and not finished. The emulator and firmware are snapshots as is now. The 43S files were updated Sunday night with a major update. The 43C files were update earlier on Sunday before the 43S update. Either way, the functionality you would need for the 42 problem would work on both and is switched on per default, i.e. long integers are input if you enter numbers without a decimal point. Interestingly, to prove the same point, another demo can be done to show the length of integer numbers, i.e. 290 [x!] 1 [+] 290 [x!] [] renders 1. For those who don't read RPN, that is (290! + 1)  290! = 1. Jaco Mostert, Elec Eng 42S, DM42, WP34C/S, 28C, 35S, 32Sii, owned 11C; used 67, 85; iOS: 42s, Free42, WP31S/34S, HCalc; OSX: WP34C. 

09162019, 07:28 AM
(This post was last modified: 09162019 07:29 AM by StephenG1CMZ.)
Post: #10




RE: Sum of three cubes for 42 finally solved
Yes, the Prime can calculate the answer 42 in CAS.
But I guess for the Prime G2 the answer you would be seeking is 2 (1 not being a prime number), not 42. Stephen Lewkowicz (G1CMZ) 

09162019, 05:38 PM
Post: #11




RE: Sum of three cubes for 42 finally solved
I just ran the time command in CAS.
The G2 calculates the answer in 0,000224 sec. Esben 28s, 35s, 49G+, 50G, Elektronika MK52 & MK61 

09272019, 10:52 PM
Post: #12




RE: Sum of three cubes for 42 finally solved
And now there is also a new non trivial solution for 3 Numberphile


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