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Full Version: Sum of three cubes for 42 finally solved
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(09-14-2019 03:20 PM)Gerson W. Barbosa Wrote: [ -> ]https://m.phys.org/news/2019-09-sum-cube...-life.html
I’m a faithfull Numberfile follower so I saw that a week ago and since have eventually recovered my lost sleep
...and the fun part; the HP/Free/DM42 is not able to calculate it's own self
(09-14-2019 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.
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
(09-14-2019 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.

(09-14-2019 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 = a1 + a0
b = +80435758145817515 = +80435758e9 + 145817515 = b1 + b0
c = +12602123297335631 = +12602123e9 + 297335631 = c1 + c0

a³ + b³ + c³
= (a1 + a0)³ + (b1 + b0)³ + (c1 + c0
= (a1³ + b1³ + c1³) + 3*(a0a1² + b0b1² + c0c1²) + 3*(a1a0² + b1b0² + c1c0²) + (a0³ + b0³ + c0³)

= -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+50b = { 80435758145,  817515}   -- b^3 ~ +5.2041E+50c = { 12602123297,  335631}   -- c^3 ~ +0.0200E+50function cube(x, n)    x = (x[1]%n * 1e6 + x[2]) % n    return x * x % n * x % nendmodtest = function(n) return (cube(a,n) + cube(b,n) + cube(c,n)) % n end `

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
(09-14-2019 09:14 PM)Gerson W. Barbosa Wrote: [ -> ]
(09-14-2019 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.
Yes, I too saw that it need 50 significant digits. Though the Prime also in CAS manages the tak.

(09-15-2019 12:37 PM)Jaco@cocoon-creations.com Wrote: [ -> ]
(09-14-2019 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.

[img]?[/img]

Where can I find the 43 program?
(09-15-2019 07:19 PM)DA74254 Wrote: [ -> ]
(09-14-2019 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:

'(-80538738812075974)^3+80435758145817515^3+12602123297335631^3'

EVAL ->

42

PS.: Unfortunately there are no solutions for 49 and 50.
Yes, I too saw that it need 50 significant digits. Though the Prime also in CAS manages the tak.

(09-15-2019 12:37 PM)Jaco@cocoon-creations.com Wrote: [ -> ]DM42 running WP43C works. I did it last week when this problem surfaced. WP43S will work by implication also.

[img]?[/img]

Where can I find the 43 program?

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.
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.
I just ran the time command in CAS.
The G2 calculates the answer in 0,000224 sec.
And now there is also a new non trivial solution for 3 Numberphile
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