ANYONE GOOD In Modulo Mathematics

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    introduction to Modulo Mathematics

    All numbers are integers.

    Definition: the modulo operator finds the remainder after the division of one number by another (sometimes called modulus). Given two positive numbers, A (the dividend) and p (the divisor), A modulo p (abbreviated as A mod p) is the remainder of the Euclidean division of A by p. Euclidean division is the process of division of two integers, which produces a quotient and a remainder smaller than the divisor.

    A mod p = C, where A, p, and C are integers, p is the divisor, and C is the remainder. So, we can write

    A = k*p + C; where is k is the quotient, also an integer. We discard k*p

    For example, A=13, p = 3.  If we divide 13 by 3 the remainder is 1 so C =1

    13 = 4*3 + 1 we discard 4*3, the remainder is 1.

    One of the ways to calculate mod would be to use the calculator as follows:

    Divide A by p then discard the fraction. Save the quotient, an integer, say, k. Then calculate

    C = A – k*p.

    For the above example 13/3 is equal to 4.333333, then k= 4

    C = 13 4*3 = 1

    Excel has a built-in mod function which is written as +mod (A, p)

    The function works well when the number of digits is about 14 or less. If the number changes into a decimal form, then the results will not be right.

    Also, some OS such as MS-windows, have a scientific calculator that has a built-in mod function. This function is better than using Excel.

    Mac users, please search a calculator for your OS.

    The following are some identities you can use

    Identity 1 for the sum of two integers:

    (A+B) mod (p) = [A mod(p)+ B mod (p)] mod (p)

    Example:

    Left Hand Side (LHS): (37+41) mod (5) = 78 mod (5) = 3

    Right Hand Side: [37 mod (5) +41 mod (5)] mod (5)

    Substitute: 37 mod (5) = 2 and 41 mod (5) =1

    RHS = [1+2] mod (5) = 3

    Thus LHS = RHS

    Identity 2 for the product of two integers:

    (A*B) mod (p) = [ A mod (p) * B mod (p)] mod (p)

    Example:

    p=42;

    A= 835 = 19*42+37;

    B= 577 = 13*42 +31

    A*B = 835*577 = 481795 = 11471*42+13

    37*31 = 1147 = 27* 42 + 13

    A mod p = 835 mod (42) = 37;

    B mod p = 577 mod (42) = 31

    LHS : (A*B) mod (p) = (835*577) mod (42) = 481795 mod (42) = 13

    RHS: [835 mod (42) * 577 mod (42) ] mod (42)

    (37*31) mod (42) = 1147 mod (42) = 13

    LHS=RHS

    The above identity can be extended to calculate A^n mod(p)

    Suppose we need to calculate 57^ 11 mod (67);

    LHS: Using Windows calculator 57^11 mod (67) = 38

    now 57^11 would be a big number. So, we break the power 11 into say 2*5+1; then first calculate 57^2 mod (67) = 33

    Now we reduced 57 ^ 11 mod (67)

    to calculate [(57^2 mod (67)] ^5 mod (67) * 57 mod (67)] mod (67)

    Using Windows calculator: 57^2 mod (67) = 33

    RHS: [(57^2 mod (67)] ^5mod (67) * 57 mod (67)] mod (67)

    = [33^5 mod (67) * 57] mod (67) {using Window calculator}

    = [23 *57] mod (67)

    = 1311 mod (67);

    =38 :

    For your understanding, you may try different combinations for 11= 3*3+2; if you use excel, avoid getting numbers in e format. If you get in e format, then the calculation would be wrong. Use Windows calculator or Mac equivalent

    For your response:

    Choose two 4-digit numbers A and B and p two-digit divisor and calculate

    A mod(p)  and B mod(p)

    (A+B) mod (p) Calculate directly and using the identity and to prove the identity calculate the mod value of the sum: LHS = RHS

    3        (A*B) mod(p)= [Amod(p)*Bmod(p)] mod(p) to prove the identity calculate the mod value of the product LHS = RHS

    4.       A^5 mod (p) using 5=2+2+1 or any other combination such as 4+1. 

    To prove identities, you need to calculate the left-hand side and the right-hand side independently.  Please let me know if you have any questions or need more clarification

    Please do not choose A or B such that A mod (p) is equal to 0; e.g.

    8888 mod (88) =  0

    Post your results in this forum

    See attached

    Due Fri Feb 7th

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