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

[JM Montolio A]

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

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

[science_man_88]

Originally Posted by JM Montolio A

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

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

You can limit things modularly at very least.

[CRGreathouse]

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]

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

[a1call]

http://m.wolframalpha.com/input/?i=s...cCompTime=true

[JM Montolio A]

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

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

[a1call]

http://m.wolframalpha.com/input/?i=s...%3D+13*D+for+m

[JM Montolio A]

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.

Answer: 29(5^3)-11=13*278.
And 3 is the min value.

Is the same thing im posting all the month.

[a1call]

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.

Thanks in advance.

[a1call]

Originally Posted by a1call

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.

Thanks in advance.

In particular the underlined 1

1+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
added to the end?

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

Thank you for the reply.

[science_man_88]

Originally Posted by JM Montolio A

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

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

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

Anyways.