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-   -   How to solve: k(b^m)-z=n*d. ? (https://www.mersenneforum.org/showthread.php?t=23142)

 JM Montolio A 2018-03-08 09:07

How to solve: k(b^m)-z=n*d. ?

How to solve: k(b^m)-z=n*d. ?

With know values for k,b,z,n.

 science_man_88 2018-03-08 11:35

[QUOTE=JM Montolio A;481851]How to solve: k(b^m)-z=n*d. ?

With know values for k,b,z,n.[/QUOTE]

You can limit things modularly at very least.

 CRGreathouse 2018-03-08 15:17

Are the variables integers, real numbers, or something else?

Are you looking for one solution or a parametrization of all solutions? (Of course there may be no solutions.)

 JM Montolio A 2018-03-08 17:46

Type? Integer, of course.
Solutions? The first m.

 a1call 2018-03-08 18:20

[url]http://m.wolframalpha.com/input/?i=solve+k*(b%5Em)-z%3Dn*d+for+m&incCompTime=true[/url]

 JM Montolio A 2018-03-08 19:15

and you get something with log(), pi, i, etc.
Not.

Solve: 29(5^m)-11 = 13*D

 a1call 2018-03-08 19:27

[url]http://m.wolframalpha.com/input/?i=solve+29%285%5Em%29-11+%3D+13*D+for+m[/url]

 JM Montolio A 2018-03-08 19:50

Not.

29(5^M)-11=13D.

13 + 11 = (5^0)*24
13 + 24 = (5^0)*37
13 + 37 = (5^2)*2
13 + 2 = (5^1)*3
13 + 3 = (5^0)*16
13 + 16 = (5^0)*29.

exponents: 0,0,2,1,0,0.
M = sum exponents = 3.
D = 1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1= 278.

And 3 is the min value.

Is the same thing im posting all the month.

 a1call 2018-03-09 00:46

How do you get
1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1

in
D = 1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1= 278.

 a1call 2018-03-09 01:04

[QUOTE=a1call;481902]How do you get
1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1

in
D = 1+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1= 278.

In particular the underlined 1

[U]1[/U]+1*1+1*1*1+1*1*1*25+1*1*1*25*5+1*1*1*25*5*1

and why is there no +1*1*1*25*5*1*1

Is the added 1 and missing last addend a general rule or subject to variations?

 science_man_88 2018-03-09 01:49

[QUOTE=JM Montolio A;481890]

Solve: 29(5^m)-11 = 13*D[/QUOTE]

Mod 5:
-1=3*D \\ D is 3 mod 5

Mod 13:
3(5^m)=-2 \\ m is congruent to 3 mod 12

Anyways.

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