Good to talk with you all again. Before we get back to HP Prime programming goodness, let's make a side trip into algebra land.
The Modulus Function
Definition:
X MOD Y = R where
R = remainder(X/Y) = frac(X/Y) * Y
A practical way to get the modulus is to get the remainder (in other words, what's left over) of the division X/Y.
For example:
13 MOD 5 = 3
Divide 13 by 5, we get the quotient of 2 and the remainder of 3.
Properties:
Let A and B be integers.
A MOD A = 0. Easily justified since an integer divided by itself has no remainder.
If B > A, A > 0, and B > 0, then A MOD B = A. Any integer divided by a larger integer, the remainder will be an original integer.
Example: 11 MOD 12 = 11. The division of 11/12 has the quotient of 0 and the remainder of 11.
A common way to think of the modulus function is dealing clocks and calendars (MOD 12).
x MOD A = B, A ≥ 0 and B ≥ 0
General solution: x = B + A*n, where n is any integer
Example:
x MOD 4 = 2.
Easily x = 2 is a solution since 2/4 leaves a quotient of 0 and remainder of 2.
Observe that x = 6 also works. 6/4 has a quotient of 1 and a remainder of 2.
Hence the solution is: x = 2 + 4 * n
A MOD x = B, A ≥ 0 and B ≥ 0
A way to find possible solutions is to build a table, using the good old trial and error technique. Stop when we get to A MOD A. (See the Properties section above) This can get very lengthy when A gets large. In trade, you can easily see the solutions of A MOD x = B (if solutions exist).
Example 1: 11 MOD x = 3
11 MOD 1 = 0
11 MOD 2 = 1
11 MOD 3 = 2
11 MOD 4 = 3
11 MOD 5 = 1
11 MOD 6 = 5
11 MOD 7 = 4
11 MOD 8 = 3
11 MOD 9 = 2
11 MOD 10 = 1
11 MOD 11 = 0
From this we find that the solutions are x = 4 and x = 8. A very fortunate occasion.
Example 2: 13 MOD x = 8
13 MOD 1 = 0
13 MOD 2 = 1
13 MOD 3 = 1
13 MOD 4 = 1
13 MOD 5 = 3
13 MOD 6 = 1
13 MOD 7 = 6
13 MOD 8 = 5
13 MOD 9 = 4
13 MOD 10 = 3
13 MOD 11 = 2
13 MOD 12 = 1
13 MOD 13 = 0
Darn it! No solution to 13 MOD x = 8. It happens sometimes.
Hopefully this is helpful. Thank you for your comments and questions. Honestly I can't thank you enough. Have a great day!
Eddie
This blog is property of Edward Shore. 2013
A blog is that is all about mathematics and calculators, two of my passions in life.
Saturday, December 14, 2013
Modular Arithmetic: Basics and Solving x MOD A = B and A MOD x = B (with A ≥ 0 and B ≥ 0)
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