Exercise 7.3 — Question 10

Topic: Applications of the Binomial Theorem

This question applies the Binomial Theorem to practical problems including finding remainders, last digits, and divisibility tests for large powers.

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Key Formula: Binomial Theorem

For any positive integer $n$:

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

where $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$.

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Strategy: Remainder Problems

To find the remainder when a large power is divided by a number, rewrite the base so that one part is a multiple of the divisor.

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

$7^{100}=(6+1{)}^{100}={\sum }_{r=0}^{100}(\genfrac{}{}{0ex}{}{100}{r})6^r\cdot 1^{100-r}$

$=1+\left(\genfrac{}{}{0ex}{}{100}{1}\right)6+\left(\genfrac{}{}{0ex}{}{100}{2}\right)6^2+\cdots +6^{100}$

Every term except the first ($r=0$) contains a factor of $6$, so:

$7^{100}=1+6k\text{for some integer }k$

$\therefore \text{Remainder}=\boxed{1}$

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Strategy: Last Digit Problems

To find the last digit (units digit) of a large power, find the remainder when divided by $10$.

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

$3^{100}=(3^4{)}^{25}=81^{25}=(80+1{)}^{25}$

$={\sum }_{r=0}^{25}\left(\genfrac{}{}{0ex}{}{25}{r}\right)80^r\cdot 1^{25-r}$

Every term with $r\ge 1$ contains $80^r$, which is divisible by $10$. So:

$3^{100}=1+10m\text{for some integer }m$

$\therefore \text{Last digit}=\boxed{1}$

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Strategy: Divisibility Tests

To show $N$ is divisible by $d$, write $N=(d\cdot k\pm 1{)}^n$ and expand. All terms except the constant will contain $d$ as a factor.

Example: Show that $6^{100}-1$ is divisible by $5$.

$6^{100}=(5+1{)}^{100}=1+(\genfrac{}{}{0ex}{}{100}{1})5+(\genfrac{}{}{0ex}{}{100}{2})5^2+\cdots$

$6^{100}-1=5\left[\left(\genfrac{}{}{0ex}{}{100}{1}\right)+\left(\genfrac{}{}{0ex}{}{100}{2}\right)5+\cdots \right]=5k$

$\therefore 6^{100}-1\text{ is divisible by }5,\text{ and remainder when }{6}^{100}÷5=\boxed{1}$

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