Question

What is the remainder when \[{{74}^{13}}-{{41}^{13}}+{{75}^{13}}-{{42}^{13}}\] is divided by 66?
A. 0
B. 64
C. 1
D. None of these

Answer 1

Hint: Here, first of all we need to solve this question by splitting this problem into two parts then we can use the concept that if it is in the form of \[\dfrac{{{a}^{n}}-{{b}^{n}}}{a-b}\] then remainder will be zero. So, by solving this we get the final answer that the remainder to this question will be zero.

Complete step-by-step solution:
Before solving this problem, we have to first understand the concept of remainder that is
In mathematics, the residual refers to the value that remains after division. When a number (dividend) is not entirely divided by another number (divisor), we are left with a value after division. The residual is the name given to this value.
For example, 7 cannot be completely divided by 2. Because, the nearest value we can get is 2 × 3 = 6, which is 1 less than 7.
Now, we come to the problem,
Here, it is given in the question that we have to find the remainder when \[{{74}^{13}}-{{41}^{13}}+{{75}^{13}}-{{42}^{13}}\]is divided by 66 that means it can be represented by \[\dfrac{{{74}^{13}}-{{41}^{13}}+{{75}^{13}}-{{42}^{13}}}{66}\Rightarrow \,\text{Remainder}\]
So, first of all we need to split these problems into two terms that is
\[\Rightarrow \dfrac{{{74}^{13}}-{{41}^{13}}}{66}+\dfrac{{{75}^{13}}-{{42}^{13}}}{66}\]
Now, here you can see in the denominator that 66 can be written as \[33\times 2\] therefore, substituting this on above equation we get:
\[\Rightarrow \dfrac{{{74}^{13}}-{{41}^{13}}}{33\times 2}+\dfrac{{{75}^{13}}-{{42}^{13}}}{33\times 2}\]
By, rearranging the term we get:
\[\Rightarrow \left( \dfrac{1}{2}\times \dfrac{{{74}^{13}}-{{41}^{13}}}{33} \right)+\left( \dfrac{1}{2}\times \dfrac{{{75}^{13}}-{{42}^{13}}}{33} \right)---(1)\]
Here, for simplicity we can consider the \[M=\left( \dfrac{{{74}^{13}}-{{41}^{13}}}{33} \right)\] and \[N=\left( \dfrac{{{75}^{13}}-{{42}^{13}}}{33} \right)\] hence, the above step becomes
\[\Rightarrow \left( \dfrac{1}{2}\times M \right)+\left( \dfrac{1}{2}\times N \right)\]
Now, we can see that if you subtract 41 from 74 then you get the value in denominator of value of M as 33 and also if you subtract 42 from 75 then also you get the same value in denominator of value of N as 33.
That means \[74-41=33\]and \[75-42=33\]. Substitute this into the values of M and N that is \[M=\left( \dfrac{{{74}^{13}}-{{41}^{13}}}{\left( 74-41 \right)} \right)\] as well as \[N=\left( \dfrac{{{75}^{13}}-{{42}^{13}}}{\left( 75-42 \right)} \right)\]
Above this obtained values substitute on (1) we get:
\[\Rightarrow \left( \dfrac{1}{2}\times \dfrac{{{74}^{13}}-{{41}^{13}}}{\left( 74-41 \right)} \right)+\left( \dfrac{1}{2}\times \dfrac{{{75}^{13}}-{{42}^{13}}}{\left( 75-42 \right)} \right)\]
Now, if you observe the above step carefully then we can notice that it is in the form of \[\dfrac{{{a}^{n}}-{{b}^{n}}}{a-b}\] at that time remainder will be zero. Then we can apply this concept into above step we get:
\[\Rightarrow \left( \dfrac{1}{2}\times 0 \right)+\left( \dfrac{1}{2}\times 0 \right)\]
\[\Rightarrow 0+0=0\]
Therefore, Remainder is zero for this problem that is \[\dfrac{{{74}^{13}}-{{41}^{13}}+{{75}^{13}}-{{42}^{13}}}{66}\Rightarrow 0\]
So, the correct option is “option A”.

Note: Here, you can remember that concept of remainder must be thorough. Students must know about the division algorithm. According to the division algorithm, for every number m and n, there exists a unique r and q such that \[n=mq+r\] where q is called the quotient, r is the remainder and m is the divisor. Because sometimes the question might ask about the remainder by giving the value of dividend quotient and divisor then only you can apply the division algorithm. Some students might make mistakes while substituting the formula. To avoid that we need to solve this particular problem step by step.