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 2018-03-08, 09:07 #1 JM Montolio A   Feb 2018 9610 Posts 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.
2018-03-08, 11:35   #2
science_man_88

"Forget I exist"
Jul 2009
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838410 Posts

Quote:
 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.

 2018-03-08, 15:17 #3 CRGreathouse     Aug 2006 176116 Posts 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.)
 2018-03-08, 17:46 #4 JM Montolio A   Feb 2018 9610 Posts Type? Integer, of course. Solutions? The first m.
 2018-03-08, 18:20 #5 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 23·11·23 Posts
 2018-03-08, 19:15 #6 JM Montolio A   Feb 2018 25·3 Posts and you get something with log(), pi, i, etc. Not. Solve: 29(5^m)-11 = 13*D
 2018-03-08, 19:27 #7 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 23·11·23 Posts
 2018-03-08, 19:50 #8 JM Montolio A   Feb 2018 6016 Posts 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.
 2018-03-09, 00:46 #9 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 37508 Posts 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.
2018-03-09, 01:04   #10
a1call

"Rashid Naimi"
Oct 2015
Remote to Here/There

23·11·23 Posts

Quote:
 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.

Last fiddled with by a1call on 2018-03-09 at 01:04

2018-03-09, 01:49   #11
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 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.

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