The unit digit in the decimal expansion of ${7^{25}}$ is
A. 1
B. 3
C. 5
D. 7

Hint: Start with evaluating the first few powers of 7. As we know that unit digit in the increasing power of any positives integers repeats after a fixed period of time so we will notice the same behavior with 7 that the units’ digits in the powers of 7 follow a particular pattern i.e. we find the same digits in the unit’s place in the same order i.e. 7, 9, 3, 1, 7, 9, 3, 1... and so on.

Complete step by step answer:

Here we will take the help of congruence modulo i.e.
 $ \Rightarrow a \equiv b(\bmod \,\,c)$, which means that a or b when divided by c gives the same remainder.
First, we start by evaluating the powers of 7.
 ${7^1} = 7$, the unit’s digit is 7, i.e. ${{\text{7}}^1} \equiv 7{\text{ }}\left( {{\text{mod 10}}} \right)$
 ${7^2} = 49$, the unit’s digit is 9, i.e. ${{\text{7}}^2} \equiv 9{\text{ }}\left( {{\text{mod 10}}} \right)$
 ${7^3} = 343$, the unit’s digit is 3, i.e. ${{\text{7}}^{\text{3}}} \equiv {\text{3 }}\left( {{\text{mod 10}}} \right)$
 ${7^4} = 2401$ , the unit’s digit is 1, i.e. ${{\text{7}}^4} \equiv 1{\text{ }}\left( {{\text{mod 10}}} \right)$
And without evaluating the powers, one can see that the unit’s digit of ${7^5}$ is 7 again. Hence we find the same digits in the unit’s place in the same order i.e. 7, 9, 3, 1.
Now, note that ${7^n}$ has 7 in the units place if $n = 4k + 1$ ; 9 if $n = 4k + 2$ ; 3 if $n = 4k + 3$ , and 1 if $n = 4k$ .
${7^1} \equiv 7{\text{ }}\left( {\bmod {\text{ }}10} \right)$
${7^2} \equiv 9{\text{ }}\left( {\bmod {\text{ }}10} \right)$
${7^3} \equiv 3{\text{ }}\left( {\bmod {\text{ }}10} \right)$
${7^4} \equiv 1{\text{ }}\left( {\bmod {\text{ }}10} \right)$
${{\text{7}}^{{\text{4k + 1}}}} \equiv 7{\text{ }}\left( {{\text{mod 10}}} \right)$
${{\text{7}}^{{\text{4k + 2}}}} \equiv 9{\text{ }}\left( {{\text{mod 10}}} \right)$
${{\text{7}}^{4k + 3}} \equiv {\text{3 }}\left( {{\text{mod 10}}} \right)$
${{\text{7}}^{4k}} \equiv 1{\text{ }}\left( {{\text{mod 10}}} \right)$
Since 25 is of $4k + 1$ form $\left[ {25 = 4 \times \left( 6 \right) + 1} \right]$ , therefore it must have 7 in the unit's place.
Hence, the correct option is (D).


Note: The digit in the unit’s place of any number n is found by taking the mod of that number.
Therefore, if the digit in the unit’s place of ${7^n}$ is d, it can also be written as ${7^n} \equiv {\text{d (mod10)}}$
Now we start with evaluating the first few powers of 7. Note that the units’ digits in the powers of 7 follow a particular pattern i.e. we find the same digits in the unit’s place in the same order i.e. 7, 9, 3, 1, 7, 9, 3, 1...and so on.
Therefore, The pattern has a period of 4.
Therefore,
 ${{\text{7}}^{{\text{4k + 1}}}} \equiv 7{\text{ }}\left( {{\text{mod 10}}} \right)$ ; ${{\text{7}}^{{\text{4k + 2}}}} \equiv 9{\text{ }}\left( {{\text{mod 10}}} \right)$ ; ${{\text{7}}^{4k + 3}}