Question

When ${2^{301}}$ is divided by 5, the least positive remainder is
(A) 4
(B) 8
(C) 2
(D) 6

Answer 1

Hint:To find the required result we need to find the exponent values of 2 with its powers from 1, 2, 3, ….. Also, we need to find the remainder of each exponent value after dividing them by 5. The repetition of The remainder in a certain interval can help us to find the required result.

Complete step by step solution:
First we need to find the values of the exponents of 2.
First, ${2^1} = 2$
${2^2} = 4$
${2^3} = 8$
${2^4} = 16$
${2^5} = 32$
${2^6} = 64$
${2^7} = 128$
${2^8} = 256$
${2^9} = 512$
And so on.
Let us observe the unit place of each of the above exponent values.
The unit places of the first four exponent values of 2 are 2, 4, 8, 6. Then, the values 2, 4, 8, 6 also start from the next four exponents of 2.
This means that the unit places are 2, 4, 8, 6, 2, 4, 8, 6 and so on.
Thus, we can say that the exponent values are repeated after four intervals.
Also, if the exponents are divided by 5; we get the remainders 4, 3, 1, 2, 4, 3, 1, 2 and so on.
It is also seen that the remainders of the exponent values are repeated after four intervals.
Now, write the digit ${2^{301}}$
as,
$
{2^{301}} = {2^{300 + 1}} \\
= {2^{300}} \cdot {2^1} \\
= {2^{300}} \cdot 2 \\
$
Also, $
{2^{300}} \cdot 2 = {\left( {{2^4}} \right)^{75}} \cdot 2 \\
= {\left( {16} \right)^{75}}.2 \\
$
Now, divide the above term by 5.
If we divide the term 16 from the above term we get the remainder as 1.
Also, if we divide 2 by 5 then we get the remainder 2.
So, we can see that by dividing the given term by 5 we can get the remainder 2.
This means that 2 is positive and it is also the least remainder which is obtained by dividing the given term by 5.
Hence, the required least positive remainder when ${2^{301}}$
is divided by 5 is 2.

Note: In order to find the required value of the least positive remainder we can use the binomial expansion method. The binomial expansion of ${\left( {1 - x} \right)^n}$
can be used to find the required value. The binomial expansion of ${2^{301}} = {2^1} \cdot {2^{300}} = 2 \cdot {\left( {{2^2}} \right)^{150}} = 2 \cdot {\left( {5 - 1} \right)^{150}}$
can give us the required result.