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 Register FAQ Search Today's Posts Mark Forums Read 2021-01-04, 12:46 #1 drmurat   "murat" May 2020 turkey 5·23 Posts Calculation C = a x b +a + b İf we know c how can we calculate a and b For example 120 = 10 x 10 + 10 + 10 Last fiddled with by drmurat on 2021-01-04 at 12:53  2021-01-04, 12:48 #2 LaurV Romulan Interpreter   Jun 2011 Thailand 2·43·109 Posts We can not. If we could, factoring would be a breeze.  2021-01-04, 13:32   #3
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

29·211 Posts Quote:
 Originally Posted by drmurat C = a x b +a + b İf we know c how can we calculate a and b For example 120 = 10 x 10 + 10 + 10
The solution is not unique.

e.g. We can always set a=0 and b=C.  2021-01-04, 13:36   #4
Dr Sardonicus

Feb 2017
Nowhere

2·7·11·29 Posts Quote:
 Originally Posted by drmurat C = a x b +a + b İf we know c how can we calculate a and b
By factoring C + 1.

C = a*b + a + b

C + 1 = a*b + a + b + 1

C + 1 = (a + 1)*(b + 1)    2021-01-04, 13:45   #5
drmurat

"murat"
May 2020
turkey

5·23 Posts Quote:
 Originally Posted by Dr Sardonicus By factoring C + 1. C = a*b + a + b C + 1 = a*b + a + b + 1 C + 1 = (a + 1)*(b + 1)  İf

C = 2*a*b + a + b
Or
C = 3*a*b + a + b

MODERATOR NOTE:

Given the equation

C = k*a*b + a + b

there is an obvious algebraic factorization similar to the one I provided above.

Thread closed.

Last fiddled with by Dr Sardonicus on 2021-01-04 at 13:59 Reason: Notification of thread closure  2021-01-06, 11:15 #6 drmurat   "murat" May 2020 turkey 5·23 Posts Any algebria book suggestion Okay I Must solve a equation Think that the equation a>0 b>0 a, b integer 10 * a * b + a + b = c İn this forum I learned a x b + a + b + 1 = c + 1 ( a + 1 ) x ( b + 1 ) = c + 1 Now the equation is 9 x a x b + ( a + 1) x ( b + 1 ) = c + 1 İt is still hard Also I know (5 x a + ... ) x ( 2 x b + ... ) =c + ... Any book sugestion ? MODERATOR NOTE: Thread merged with earlier "Calculation" thread. Last fiddled with by Dr Sardonicus on 2021-01-07 at 14:29 Reason: As indicated  2021-01-07, 13:48 #7 BudgieJane   "Jane Sullivan" Jan 2011 Beckenham, UK 241 Posts I wouldn't call this an equation, because if you reverse it it becomes an instruction for calculating c given a and b. c = 10ab + a + b so there is an infinite number of "solutions". a=b=0 gives c=0 a=b=1 gives c=12 a=0 b=1 or a=1 b=0 gives c=1 and so on ad infinitum. In my humble opinion, you do not need a book on algebra to work this out.  2021-01-07, 14:23   #8
Viliam Furik

"Viliam Furík"
Jul 2018
Martin, Slovakia

2·223 Posts Quote:
 Originally Posted by BudgieJane I wouldn't call this an equation, because if you reverse it it becomes an instruction for calculating c given a and b.
Why not? Equations can have an infinite number of solutions.

And for drmurat:

k * a * b + a + b = c
k^2 * a * b + k * a + k * b = k * c
k^2 * a * b + k * a + k * b + 1 = k * c + 1
(k * a + 1) * (k * b + 1) = k * c + 1 Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post mPat PrimeNet 2 2017-06-24 18:04 radugafener Math 5 2015-02-12 15:41 FreakyPotato Programming 7 2015-02-06 10:33 tcharron PrimeNet 4 2014-06-27 23:27 kurtulmehtap Math 3 2010-10-11 15:02

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