NUMBER SYSTEMS PART 4

Number System Tutorial IV:  Remainder

 

Remainders

There are a plethora of remainder related questions you can expect in your exam. So it is a mandatory requirement to create a clear awareness of remainders before going to attend the exam.

Based on above illustration, any number 'N' can be expressed as N=dq+r${\displaystyle N=dq+r}$.

Properties of remainders

• When the dividend is less than divisor, then the remainder is the dividend itself. Eg: when 5 divided by 7, the remainder is 5 itself, and then the quotient of this division is 0.
• Remainders are consistent under basic mathematical operations such as addition, subtraction, multiplication and powers.
  Addition:When N1${\displaystyle N_1}$ and N2${\displaystyle N_2}$ (two distinct dividends) divided by a divisor d${\displaystyle d}$, leaves the respective remainders r1${\displaystyle r_1}$ and r2${\displaystyle r_2}$, then the remainder when N1+N2${\displaystyle N_1+N_2}$ divided by d${\displaystyle d}$ is r1+r2${\displaystyle r_1+r_2}$.

  Eg. Rem[119]=2${\displaystyle \text{Rem}\left[\frac{11}{9}\right]=2}$ and Rem[149]=5${\displaystyle \text{Rem}\left[\frac{14}{9}\right]=5}$
  ∴Rem[11+149]=2+5=7${\displaystyle \therefore \text{Rem}\left[\frac{11+14}{9}\right]=2+5=7}$
  Subtraction:When N1${\displaystyle N_1}$ and N2${\displaystyle N_2}$ (two distinct dividends) divided by a divisor d${\displaystyle d}$, leaves the respective remainders r1${\displaystyle r_1}$ and r2${\displaystyle r_2}$, then the remainder when N1−N2${\displaystyle N_1-N_2}$ divided by d${\displaystyle d}$ is r1−r2${\displaystyle r_1-r_2}$.

  Eg. Rem[1910]=9${\displaystyle \text{Rem}\left[\frac{19}{10}\right]=9}$ and Rem[1510]=5${\displaystyle \text{Rem}\left[\frac{15}{10}\right]=5}$
  ∴Rem[19−1510]=9−5=4${\displaystyle \therefore \text{Rem}\left[\frac{19-15}{10}\right]=9-5=4}$
  Multiplication:When N1${\displaystyle N_1}$ and N2${\displaystyle N_2}$ (two distinct dividends) divided by a divisor d${\displaystyle d}$, leaves the respective remainders r1${\displaystyle r_1}$ and r2${\displaystyle r_2}$, then the remainder when N1×N2${\displaystyle N_1\times N_2}$ divided by d${\displaystyle d}$ is r1×r2${\displaystyle r_1\times r_2}$.

  Eg. Rem[1410]=4${\displaystyle \text{Rem}\left[\frac{14}{10}\right]=4}$ and Rem[1210]=2${\displaystyle \text{Rem}\left[\frac{12}{10}\right]=2}$
  ∴Rem[14×1210]=4×2=8${\displaystyle \therefore \text{Rem}\left[\frac{14\times 12}{10}\right]=4\times 2=8}$
  Powers or exponents:When a dividend 'N' divided by divisor 'd', leaves a remainder 'r', then Nn${\displaystyle N^n}$ leaves a remainder rn${\displaystyle r^n}$ when it is divided by d.

  Eg. Rem[108]=2${\displaystyle \text{Rem}\left[\frac{10}{8}\right]=2}$
  ∴Rem[1028]=22=4

• Concept of Negative remainders: As per Eucledian Lemma any number 'N' can be expressed in the following way; N=dq+r${\displaystyle N=dq+r}$
  In a similar arrangement, 7 can be expresses as; 7=3×2+1${\displaystyle 7=3\times 2+1}$, where 3 is the divisor, 2 is the quotient and 1 is the remainder.
  7 can be expressed in the following way too; 7=3×3−2${\displaystyle 7=3\times 3-2}$, where 3 is the same divisor which we considered above, 3 is the new quotient and '-2' is the remainder as per the change in the quotient. Therefore the remainders 1 and -2 are representing the same division. Hence -2 is the corresponding negative remainder of the original remainder 1, when the divisor is 3.
  Negative remainder=original remainder−divisor${\displaystyle \text{Negative remainder}=\text{original remainder}-\text{divisor}}$
  Therefore original remainder=divisor+negative remainder${\displaystyle \text{Therefore original remainder}=\text{divisor}+\text{negative remainder}}$
  Let's look at one more example to understand concept of negative remainders.
  Rem[188]=2${\displaystyle \text{Rem}\left[\frac{18}{8}\right]=2}$ and Rem[58]=5${\displaystyle \text{Rem}\left[\frac{5}{8}\right]=5}$
  ∴Rem[18−58]=2−5=−3${\displaystyle \therefore \text{Rem}\left[\frac{18-5}{8}\right]=2-5=-3}$
  Hence the original remainder=divisor+negative remainder${\displaystyle \text{Hence the original remainder}=\text{divisor}+\text{negative remainder}}$
  =8+(−3)=5${\displaystyle =8+(-3)=5}$
  ie; Rem[18−58]=Rem[138]=5${\displaystyle \text{Rem}\left[\frac{18-5}{8}\right]=\text{Rem}\left[\frac{13}{8}\right]=5}$
• Pattern of remainders:
  Pattern of remainders is one of the most aesthetical and very important features of remainders. Let us consider the following pattern of remainders;
  • Rem[315]=3${\displaystyle \text{Rem}\left[\frac{3^1}{5}\right]=3}$
    Rem[325]=3×3→4${\displaystyle \text{Rem}\left[\frac{3^2}{5}\right]=3\times 3\to 4}$
    Rem[335]=4×3→2${\displaystyle \text{Rem}\left[\frac{3^3}{5}\right]=4\times 3\to 2}$
    Rem[345]=2×3→1${\displaystyle \text{Rem}\left[\frac{3^4}{5}\right]=2\times 3\to 1}$
    Rem[355]=1×3→3${\displaystyle \text{Rem}\left[\frac{3^5}{5}\right]=1\times 3\to 3}$ and so on...
    The above pattern of remainders will repeat cyclically after the occurrence of 1 and the pattern of remainders is 3,4,2,1, 3,4,2,1... In the given example, there are 4 distinct remainders. If the exponent of 3 is any multiple of 4, then the remainder should be 1; Ie. if the exponent is in the form 4k+1→remainder is 3${\displaystyle 4k+1\to \text{remainder is 3}}$
    4k+2→remainder is 4${\displaystyle 4k+2\to \text{remainder is 4}}$
    4k+3→remainder is 2${\displaystyle 4k+3\to \text{remainder is 2}}$
    4k+0→remainder is 1${\displaystyle 4k+0\to \text{remainder is 1}}$, Where k is any whole number, k = 0, 1, 2, 3...
    Find the remainder when 31729${\displaystyle 3^{1729}}$ divided by 5

    As per the above explanation, there are 4 distinct remainders and those are occurring in the order 3, 4, 2, 1. 1729=4k+1${\displaystyle 1729=4k+1}$ (when 1729 divided by 4, it leaves the remainder 1). If the exponent is in the form 4k+1${\displaystyle 4k+1}$, then the remainder is 3.
    Find the remainder when 13271${\displaystyle {13}^{271}}$ divided by 7

    First of all develop the cyclic pattern of possible remainders. 
    Rem[1317]=6${\displaystyle \text{Rem}\left[\frac{{13}^1}{7}\right]=6}$
    Rem[1327]=6×6→1${\displaystyle \text{Rem}\left[\frac{{13}^2}{7}\right]=6\times 6\to 1}$
    Rem[1337]=1×6→6${\displaystyle \text{Rem}\left[\frac{{13}^3}{7}\right]=1\times 6\to 6}$
    Rem[1347]=6×6→1${\displaystyle \text{Rem}\left[\frac{{13}^4}{7}\right]=6\times 6\to 1}$
    Now we got a pattern of remainders in the cyclic order 6,1,6,1...

    Note: In the pattern of remainders if in any step the remainder occurs as 1, then the pattern will start from there. Hence the operation of finding remainders can stop at that point.

    In the above pattern, there are two distinct remainder, ie. when the exponent is odd, remainder is 6 and when the exponent is even, then the remainder is 1. ∴Rem[132717]=6${\displaystyle \therefore \text{Rem}\left[\frac{{13}^{271}}{7}\right]=6}$, because 271 is an odd exponent.

  Fermat's little theorem

  Let N and p are two relatively prime numbers and p itself is a prime, then Np−1p${\displaystyle \frac{N^{p-1}}{p}}$ leaves a remainder of 1.
  Eg. Rem[1003637]=1${\displaystyle \text{Rem}\left[\frac{{100}^{36}}{37}\right]=1}$
  Find the remainder when 10041${\displaystyle {100}^{41}}$ divided by 41

  As per Fermat's theorem, Np−1p${\displaystyle \frac{N^{p-1}}{p}}$ leaves a remainder of 1, where N and p are two relatively prime numbers and p itself is a prime. In the given question, the divisor 41 is a prime number and 100 and 41 are relatively prime.∴Rem[1004041]=1${\displaystyle \therefore \text{Rem}\left[\frac{{100}^{40}}{41}\right]=1}$.
  So, Rem[1004141]=Rem[10040×10041]=Rem[10041]=18${\displaystyle \text{Rem}\left[\frac{{100}^{41}}{41}\right]=\text{Rem}\left[\frac{{100}^{40}\times 100}{41}\right]=\text{Rem}\left[\frac{100}{41}\right]=18}$

  Wilson's theorem

  For all prime p${\displaystyle p}$, [(p−1)!+1]${\displaystyle [(p-1)!+1]}$ is always divisible by p${\displaystyle p}$.
  OR 
  When (p−1)!${\displaystyle (p-1)!}$ Is divided by p${\displaystyle p}$, the remainder is (p−1)${\displaystyle (p-1)}$.
  Let's look at an example to understand Wilson's theorem.
  Find the remainder when 6! Divided by 7.

  As per Wilson's theorem; 7 is a prime →Rem[(7−1)!7]=7−1=6${\displaystyle \to \text{Rem}\left[\frac{(7-1)!}{7}\right]=7-1=6}$

  Remainder theorem and its numerical application

  If p(x)${\displaystyle p\left(x\right)}$ is a polynomial in x, then p(a)${\displaystyle p\left(a\right)}$ is the remainder when p(x)${\displaystyle p\left(x\right)}$ divided by (x−a)${\displaystyle (x-a)}$.
  Understanding the numerical application of this theorem will help you to solve some complicated type of remainder related questions. Let's take a look at few examples to understand remainder theorem better.
  Find the remainder when 2402${\displaystyle 2^{402}}$ divided by 7.

  Consider 2402${\displaystyle 2^{402}}$ as a polynomial in 2. Also, 7 can be expressed in terms of 2 as: 7=23−1${\displaystyle 7=2^3-1}$, which can be considered as a polynomial on 23${\displaystyle 2^3}$
  ∴Rem[24027]=Rem[(23)13423−1]=1134=1${\displaystyle \therefore \text{Rem}\left[\frac{2^{402}}{7}\right]=\text{Rem}\left[\frac{{\left(2^3\right)}^{134}}{2^3-1}\right]=1^{134}=1}$
  Find the remainder when 16133${\displaystyle {16}^{133}}$ divided by 17.

  17=16+1→${\displaystyle 17=16+1\to }$ it is in the form of x+1${\displaystyle x+1}$, where x=16${\displaystyle x=16}$
  16133${\displaystyle {16}^{133}}$ is in the form of x133${\displaystyle x^{133}}$, where x=16${\displaystyle x=16}$
  `"Rem"[16^133/(16 + 1)] = (-1)^133 = -1
  Here the remainde is negative, so original remainder (positive) is 17 - 1 = 16
  
  
  Find the remainder when 27101${\displaystyle {27}^{101}}$ divided by 82.
  82=34+1${\displaystyle 82=3^4+1}$ 
  27101=(33)101=(34)75⋅33=(34)75⋅27${\displaystyle {27}^{101}={\left(3^3\right)}^{101}={\left(3^4\right)}^{75}\cdot 3^3={\left(3^4\right)}^{75}\cdot 27}$
  Rem[2710182]=Rem[(34)75⋅2734+1]=(−1)75⋅27=−27${\displaystyle \text{Rem}\left[\frac{{27}^{101}}{82}\right]=\text{Rem}\left[\frac{{\left(3^4\right)}^{75}\cdot 27}{3^4+1}\right]={(-1)}^{75}\cdot 27=-27}$
  Here we got negative remainder as -27, so actual remainder is 82 - 27 = 55