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June 28th, 2019, 05:44 AM  #1 
Newbie Joined: Jun 2019 From: Italy Posts: 7 Thanks: 0  Simplify modulus expression
Can someone tell me how to simplify this expression with modulus: $\displaystyle (A/100) mod 10 + (A/10) mod 10 + A mod 10$ where $\displaystyle A$ is an integer, but it can also be not divisible by 100. I tried doing this, but it doesn't work... $\displaystyle (A/100+A/10+A)mod 10$ Thanks 
June 28th, 2019, 08:22 AM  #2 
Senior Member Joined: Jun 2019 From: USA Posts: 223 Thanks: 99 
I thought 111 was the key, but it's not working out so far. Not sure if it makes a difference, but are we taking $\displaystyle a mod b$ to be an integer (truncated mod) or allowing real numbers? 
June 28th, 2019, 11:28 AM  #3 
Newbie Joined: Jun 2019 From: Italy Posts: 7 Thanks: 0 
In this case we are takig all real numbers, so that for example $\displaystyle 6.5987 mod 2$ is congruent to $\displaystyle 0.5987$. We aren't considering truncated mod. What's 111, you've mentioned? 
June 28th, 2019, 11:55 AM  #4 
Senior Member Joined: Jun 2019 From: USA Posts: 223 Thanks: 99 
Well, $\displaystyle (A/100) mod 10$ is the hundreds digit (plus everything after rightshifted two decimal places), $\displaystyle (A/10) mod 10$ is the tens digit, and $\displaystyle (A) mod 10$ is the ones digit. $\displaystyle A=1325$ yields 3.25 + 2.5 + 5 = 10.75. I thought multiplying by 111 (i.e., 100*1 + 10*1 + 1*1) either before or after the modulo might achieve the same result, but so far it doesn't seem to be working. 
June 28th, 2019, 12:14 PM  #5 
Newbie Joined: Jun 2019 From: Italy Posts: 7 Thanks: 0 
Thanks for your answer. Can't we use the proprieties:$\displaystyle (a + b)mod c = (a mod c + b mod c) mod c$? 
June 28th, 2019, 03:20 PM  #6 
Senior Member Joined: Jun 2019 From: USA Posts: 223 Thanks: 99 
Not that I can see, because there's no modulo on the whole thing. $\displaystyle (A*1.11) mod 10$, for example almost works, but it's usually going to be off by a multiple of ten. $\displaystyle (1325*1.11) mod 10 = 1470.75 mod 10 = 0.75$, not 10.75. $\displaystyle (1325 mod 1000)*1.11 = 325*1.11 = 360.75$, again not 10.75. 
June 28th, 2019, 11:20 PM  #7 
Newbie Joined: Jun 2019 From: Italy Posts: 7 Thanks: 0 
So, there is no solution, I can't do any gathtering of $\displaystyle A$ or $\displaystyle mod\,\,10$?

June 29th, 2019, 12:56 AM  #8 
Senior Member Joined: Jun 2019 From: USA Posts: 223 Thanks: 99 
I said I can't see one. To say there isn't one, you'd have to define a form of the simplification you expect and formulate a proof that it cannot equal the original expression. As the username says, I'm an engineer, not a mathematician. 
June 29th, 2019, 02:35 PM  #9 
Newbie Joined: Jun 2019 From: Italy Posts: 7 Thanks: 0 
Thanks DarnItJimImAnEngineer. Others?


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