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Old 2018-03-08, 09:07   #1
JM Montolio A
 
Feb 2018

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Smile 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.
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Old 2018-03-08, 11:35   #2
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Quote:
Originally Posted by JM Montolio A View Post
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.
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Old 2018-03-08, 15:17   #3
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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.)
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Old 2018-03-08, 17:46   #4
JM Montolio A
 
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Type? Integer, of course.
Solutions? The first m.
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Old 2018-03-08, 18:20   #5
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http://m.wolframalpha.com/input/?i=s...cCompTime=true
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Old 2018-03-08, 19:15   #6
JM Montolio A
 
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and you get something with log(), pi, i, etc.
Not.

Solve: 29(5^m)-11 = 13*D
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Old 2018-03-08, 19:27   #7
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http://m.wolframalpha.com/input/?i=s...%3D+13*D+for+m
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Old 2018-03-08, 19:50   #8
JM Montolio A
 
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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.
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Old 2018-03-09, 00:46   #9
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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.
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Old 2018-03-09, 01:04   #10
a1call
 
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Quote:
Originally Posted by a1call View Post
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.

Last fiddled with by a1call on 2018-03-09 at 01:04
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Old 2018-03-09, 01:49   #11
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Quote:
Originally Posted by JM Montolio A View Post

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.
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