7.4 Q-10

Exercise 7.4 — Q10: Last Digit and Remainder Using Binomial Theorem

This question applies the Binomial Theorem to find:

1. The last digit (units digit) of a large power.
2. The remainder when a large power is divided by a number.

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Key Technique 1: Finding the Remainder

To find the remainder when $(a+b{)}^n$ is divided by $a$:

$(a+b{)}^n={\sum }_{r=0}^n(\genfrac{}{}{0ex}{}{n}{r})a^{n-r}{b}^r$

Every term except the last ($b^n$) contains at least one factor of $a$, so all those terms are divisible by $a$.

$(a+b{)}^n\equiv b^n(\mathrm{mod}a)$

Example: Find the remainder when $7^{100}$ is divided by $6$.

Write $7^{100}=(6+1{)}^{100}$. Every term except $1^{100}$ is divisible by 6.

$7^{100}\equiv 1^{100}=1(\mathrm{mod}6)$

Remainder = 1

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Key Technique 2: Finding the Last Digit (Cyclicity Method)

The units digit of powers of any integer follows a repeating cycle.

Steps:

1. List the units digits of successive powers until the pattern repeats.
2. Note the cycle length $c$.
3. Compute $n÷c$ and find the remainder $r$.
4. The units digit of $a^n$ equals the units digit of $a^r$. If $r=0$, use position $c$.

Example: Find the last digit of $7^{102}$.

Cycle length = 4. Now $102÷4=25$ remainder $2$.

So the last digit of $7^{102}$ = last digit of $7^2=9$.

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Key Technique 3: Binomial Approach to Last Digit

Express the base as $(\text{multiple of 10}+d{)}^n$ where $d$ is the units digit, then only the last term $d^n$ determines the units digit of the full expression.

Example: Find the last digit of $3^{100}$.

Cycle of 3: $3,9,7,1$ (length 4). $100÷4=25$ remainder $0$, so use position 4.

Last digit of $3^{100}$ = last digit of $3^4=81$ → last digit = 1.

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