Marian Said:

How can I perform the multiplication and division without using arithmetic operator in c++?

We Answered:

bit shifting and addition and subtraction.

say I want to multiply my integer NUM by 5.

Each time you bitshift to the left by 1 space you multiply by 2. So we bitshift left twice (NUM<<2) to get 4x and add it to itself once.
NUM = (NUM<<2)+NUM; //this is a multiplication by 5 without using the multiplication sign.

If you need to divide you bit shift to the right >>



If you mean without using the actual signs to do the same work you can overload operators like && or something like that.

Christine Said:

How does the ancient egyptians do arithmetic in division?

We Answered:

Considering that ancient Egyptian culture was way back in ancient times, I'd say that's a "did", not a "does"!

A simple search on Yahoo.com for 'ancient Egyptian math division' gives back this site, among others:
http://mathforum.org/library/drmath/view…

Jim Said:

Solving Mathmatical Arithmetic problem? Division?

We Answered:

- multiply numerator and denominator by 10
- now you have 3/25
-in your head divide 3 by 25
- you'll have to multiply 3 by 100, or add two zeros to 3 to solve
-30/25 = 1 and 5 remaining, add zero
=> 50/25 = 2
- so you get 12
-but remember you multiplied 3 by 100 (or added two zeros)
- so for your answer 12, you need to start at 2 and move the decimal point two places to the left
=> 0.12
-now to get the percentage, multiply by 100
=> 0.12 * 100 = 12

Willie Said:

How shift registers can be used for arithmetic manipulations? Give example in case of division?

We Answered:

The operation of a shift register naturally multiplies and divides a binary number by 2 for each shift operation. It multiplies when it shifts left, and divides when it shifts right. For example, consider the binary number

0010 0000,

which is decimal 32. If it is shifted once to the right, it becomes

0001 0000,

Which is decimal 16, which is 32 divided by 2.

Bobbie Said:

Find 213/331 in Z/1000 - modulo arithmetic - division?

We Answered:

Hi, I believe that with Z/1000 you mean Z/1000Z, right? Well, in this case all you have to do is:
- check that 331 is invertible mod 1000, that is that gcd(331, 1000) = 1 (this is true since the only prime factors of 1000 are 2 and 5 and none of them divides 331);
- find the inverse of 331 in Z/1000Z, which can be done using Bezout's identity:
from the division algorithm used to find the gcd between 331 and 1000 you have
1000 = 331*3 + 7;
331 = 7*47 + 2;
7 = 2*3 + 1.
The last equality implies
7 - 2*3 = 1,
the second one
2 = 331 - 7*47,
which placed in the last allows you to write
7 - 3*(331 - 7*47) = 1;
142*7 - 3*331 = 1.
Finally, since the first one implies
7 = 1000 - 3*331
you have that
142*(1000 - 3*331) - 3*331 = 1,
that is
142*1000 - 429*331 = 1.
There you see that the inverse of 331 mod 1000 is given by -429. Adding 1000 to this number you also have that 571 is a representative of the inverse of [331] in Z/1000Z;
- multiply 213 and the inverse of 331 found above, obtaining 213*571 = 121623, which reduced mod 1000 leads you to [623], which is what you were looking for.
I hope this helped you. Bye!