Find the units digit in $7^{295}$.

> [!TLDR] > Units digits of $7^n$ repeat with period 4 as $7,9,3,1$. Since $295 \equiv 3 \pmod 4$, the units digit is that of $7^3$, which is **3**. (Confirmed via $7^{295} = (7^2)^{147}\cdot 7 \equiv 1^{147}\cdot 7 \equiv 7 \cdot 1 \cdots$ by congruences below.) > [!FORMULA] > The units digit of a number is its value $\pmod{10}$. Use $a \equiv b \pmod{10}$ and the fact that powers preserve congruence: if $a \equiv b$, then $a^k \equiv b^k \pmod{10}$. 1. Note $7^2 = 49 \equiv 9 \equiv -1 \pmod{10}$. 2. Write $7^{295} = (7^2)^{147}\cdot 7$. …

Find the units digit in 7²⁹⁵.

Question

7^{295}

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Concept understanding — Modular Arithmetic

Modular Arithmetic — The Arithmetic of Remainders

You already know modular arithmetic. You just don't know you know it.

Think about a clock. When it's 10 AM and you add 5 hours, you get 3 PM — not 15 o'clock. The clock "wraps around" after 12. That wrapping is the entire idea of modular arithmetic. We care only about the remainder after division, not the full number.

The Intuition: "What's left over?"

Take any two whole numbers and divide one by the other. The result is a quotient (how many times it fits) and a remainder (what's left). Modular arithmetic is the study of that remainder.

For example: 17 divided by 5 gives quotient 3 and remainder 2. In modular language, we say:

17 \equiv 2 \pmod{5}

Read this as: "17 is congruent to 2 modulo 5." It means 17 and 2 leave the same remainder when divided by 5.

Note

The word "modulo" comes from Latin — meaning "with respect to the modulus." The modulus is the number you divide by (here, 5).

The Precise Definition

Let $a$ , $b$ , and $m$ be integers, with $m > 0$ . We say:

a \equiv b \pmod{m}

if and only if $m$ divides $(a - b)$ exactly — that is, $a - b = m \cdot k$ for some integer $k$ .

Equivalently: $a$ and $b$ have the same remainder when divided by $m$ .

a \equiv b \pmod{m} \iff m \mid (a - b)

Examples to Lock It In

Statement	Why it's true
$23 \equiv 3 \pmod{5}$	$23 - 3 = 20$ , and $5$ divides $20$
$14 \equiv 2 \pmod{12}$	$14 - 2 = 12$ , and $12$ divides $12$
$100 \equiv 1 \pmod{3}$	$100 - 1 = 99$ , and $3$ divides $99$
$7 \equiv 0 \pmod{7}$	$7 - 0 = 7$ , and $7$ divides $7$

Notice the last one: any number is congruent to 0 modulo itself. That's just saying the remainder when you divide a number by itself is 0.

What Modular Arithmetic Doesn't Care About

Two numbers that are congruent modulo $m$ are considered "the same" for all modular purposes. So:

... , -10, -5, 0, 5, 10, 15, ...

are all the same modulo 5. They form an equivalence class — a set of numbers that all share the same remainder.

Watch out

A common mistake: thinking $a \equiv b \pmod{m}$ means $a$ divided by $m$ gives remainder $b$ . That's only true if $0 \leq b < m$ . The definition is about the difference being divisible by $m$ , not about the remainder itself.

Why It Matters …

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Units digits of $7^n$ repeat with period 4 as $7,9,3,1$ . Since $295 \equiv 3 \pmod 4$ , the units digit is that of $7^3$ , which is 3. (Confirmed via $7^{295} = (7^2)^{147}\cdot 7 \equiv 1^{147}\cdot 7 \equiv 7 \cdot 1 \cdots$ by congruences below.)

\pmod{10}

Note $7^2 = 49 \equiv 9 \equiv -1 \pmod{10}$ .
Write $7^{295} = (7^2)^{147}\cdot 7$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/W1XZY/41 markMCQ
$-41 \bmod 9$
›Reveal solutionSolution
$-41 = 9(-5)+4$ , so the least non-negative remainder is $4$ , giving $-41 \bmod 9 = 4$ .

$a \bmod m = r$

Divide and keep the remainder non-negative: $-41 \div 9$ gives a quotient of $-5$ because $9\times(-5) = -45 \le -41$ . …
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CBSE 2025Set 465/S/WXYZ/41 markMCQ
$Q. The smallest positive integer (mod 11) to which 282 is congruent, is : (A) 3 (B) 7 (C) 9 (D) 17$
›Reveal solutionSolution
Divide 282 by 11 and the remainder, 7, is the smallest positive integer it is congruent to mod 11.

$a \equiv r \pmod m \iff a = mq + r,\ 0 \le r < m$

Find how many times 11 divides 282: $\left\lfloor \dfrac{282}{11} \right\rfloor = 25$ .

Compute the multiple: $11 \times 25 = 275$ . …
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CBSE 2023Set 465/EF1GH/41 markMCQ
$(22)^{12}$
(a) $2$
(b) $4$
(c) $6$
(d) $8$
›Reveal solutionSolution
The unit digit of $2^n$ repeats in the cycle $2,4,8,6$ (period 4); $12$ is a multiple of 4, so $(22)^{12}$ ends in $6$ .

$a^n$

The last digit of $(22)^{12}$ equals the last digit of $2^{12}$ .

Find $12 \bmod 4 = 0$ , i.e. $12$ is an exact multiple of the cycle length $4$ . …
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CBSE 2019Set ANNUAL1 markMCQ
$\omega^2$
(a) $2$
(b) $1$
(c) $4$
(d) $3$
›Reveal solutionSolution
The order of $\omega^2$ in the multiplicative group of cube roots of unity is $3$ .

The cube roots of unity are $1,\omega,\omega^2$ where $\omega^3=1$ and this set forms a group under multiplication.

The order of an element $g$ is the smallest positive integer $m$ such that $g^m=1$ .

Compute powers of $\omega^2$ : $(\omega^2)^1=\omega^2\neq1$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$n^{th}$
(a) $\omega^{\frac{1}{k}}$
(b) $\omega^{-1}$
(c) $\omega^{n-k}$
(d) $\omega^{\frac{n}{k}}$
›Reveal solutionSolution
In the group of $n^{th}$ roots of unity, $(\omega^k)^{-1}=\omega^{n-k}$ .

The $n^{th}$ roots of unity form a cyclic group of order $n$ under multiplication, with identity element $1=\omega^0=\omega^n$ .

For an element $\omega^k$ ( $k<n$ ), we need $m$ such that $\omega^k\cdot\omega^m=\omega^n=1$ , i.e. $k+m\equiv0\pmod n$ , so $m=n-k$ .

Hence the inverse of $\omega^k$ is $\omega^{n-k}$ ; equivalently, since all $n^{th}$ roots of unity lie on the unit circle, $\omega^{n-k}=\overline{\omega^k}=(\omega^k)^{-1}$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$[7]$
(a) 9
(b) 6
(c) 3
(d) 1
›Reveal solutionSolution
In the cyclic group $(\mathbb Z_n,+_n)$ , the order of element $[k]$ equals $n/\gcd(k,n)$ ; here $n=9,k=7$ are coprime, so $[7]$ generates the whole group and has order $9$ .

$(\mathbb Z_9,+_9)=\{[0],[1],\dots,[8]\}$ is the cyclic group of integers modulo 9 under addition, of order 9.

For a cyclic group of order $n$ generated by $[1]$ , the order of the element $[k]$ is $\dfrac{n}{\gcd(k,n)}$ .

Here $k=7$ , $n=9$ . Compute $\gcd(7,9)$ : since 7 is prime and does not divide 9, $\gcd(7,9)=1$ .

So order of $[7] = \dfrac{9}{1}=9$ . …
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CBSE 2017Set ANNUAL1 markMCQ
$\{x \in Z / x = 5k+2,\ k \in Z\}$
(a) $[0]$
(b) $[5]$
(c) $[7]$
(d) $[2]$
›Reveal solutionSolution
The set $\{x=5k+2 : k\in\mathbb Z\}$ is, by definition, precisely all integers congruent to $2$ modulo $5$ , i.e. the equivalence class $[2]$ .

Congruence modulo 5 partitions $\mathbb Z$ into equivalence classes $[0],[1],[2],[3],[4]$ , where $[r]=\{x\in\mathbb Z : x\equiv r\pmod5\} = \{x = 5k+r : k\in\mathbb Z\}$ . …
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CBSE 2016Set ANNUAL1 markMCQ
$-i$
(a) $4$
(b) $3$
(c) $2$
(d) $1$
›Reveal solutionSolution
Direct computation of successive powers of $-i$ shows it first returns to $1$ at the 4th power, so its order is $4$ .

The 4th roots of unity are the solutions of $z^4=1$ : $\{1,\ i,\ -1,\ -i\}$ , forming a group of order $4$ under multiplication.

The order of an element $g$ in a finite group is the smallest positive integer $k$ such that $g^k=e$ (here $e=1$ ).

Compute powers of $-i$ :

$(-i)^1=-i\neq1$ .

$(-i)^2=(-i)(-i)=i^2=-1\neq1$ .

$(-i)^3=(-i)^2\cdot(-i)=(-1)(-i)=i\neq1$ .

$(-i)^4=(-i)^3\cdot(-i)=(i)(-i)=-i^2=1$ .

The smallest $k$ for which $(-i)^k=1$ is $k=4$ , so the order of $-i$ is $4$ . …
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Q.Find the units digit in 72957^{295}7295.

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Q.Find the units digit in $7^{295}$ .

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