20210104, 12:46  #1 
"murat"
May 2020
turkey
3×29 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 20210104 at 12:53 
20210104, 12:48  #2 
Romulan Interpreter
Jun 2011
Thailand
2^{3}·19·61 Posts 
We can not.
If we could, factoring would be a breeze. 
20210104, 13:32  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×3×1,013 Posts 

20210104, 13:36  #4 
Feb 2017
Nowhere
2^{5}·3^{3}·5 Posts 

20210104, 13:45  #5  
"murat"
May 2020
turkey
127_{8} Posts 
Quote:
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 20210104 at 13:59 Reason: Notification of thread closure 

20210106, 11:15  #6 
"murat"
May 2020
turkey
3×29 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 20210107 at 14:29 Reason: As indicated 
20210107, 13:48  #7 
"Jane Sullivan"
Jan 2011
Beckenham, UK
F0_{16} 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. 
20210107, 14:23  #8  
"Viliam Furík"
Jul 2018
Martin, Slovakia
101111100_{2} Posts 
Quote:
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 

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