Question

The last digit of ${2137^{753}}$is
A. 9
B. 7
C. 3
D. 1

Answer 1

Hint: The unit place of the powers of a number follows a trend which is repeated after a certain interval. For this question, we need to find that trend for 7. Then we can find the interval and divide the power with it. The unit place will depend on the remainder after division.

Complete step by step Answer:

We have the digit ${2137^{753}}$. Its unit place always only depends on 7 and its power.
So, we can consider the powers of 7
${7^1} = 7$
When the power is 1, the unit place is 7.
${7^2} = 49$
When the power is 2, the unit place is 9.
${7^3} = 343$
When the power is 3, the unit place is 3.
${7^4} = 2401$
When the power is 4, the unit place is 1.
${7^5} = 16807$
When the power is 5, the unit place is again 7.
So, the digit in the unit place will repeat after the 4th power and $4{n^{th}}$powers
We can write 753 as the quotient and remainder of 4.
$753 = 4 \times 188 + 1$
As the remainder of 753 is 1, the last digit of ${2137^{753}}$ will be the same as that of ${2137^1}$.
So, the last digit of ${2137^{753}}$ is 7.

Therefore, the correct answer is option B.

Note: The concept we used for finding the repeating unit place digits of the powers of 7 can be used for any natural number. We can use this concept to check whether a given number is a power of a particular number. For example, all the powers of 7 will end in either of the numbers 7, 9, 3, and 1. The concept of division algorithm is used to write 753 as the product of 4 and its remainder. According to the division algorithm, for every number n and m, there exist q and r such that $n = mq + r$. This helps us to write a number as the multiple of others with a remainder. The unit place of any power of any number is determined by the unit place of the base only. No other digits have any importance in determining the unit’s place.