Key Concepts

To find the fractional part of a number $\frac{N}{D}$, where $N$ and $D$ are integers and $D \neq 0$, we rely on the definition of the fractional part and properties of modular arithmetic and the Binomial Theorem.

1. Fractional Part: The fractional part of a real number $x$, denoted by $\{x\}$, is defined as $x - \lfloor x \rfloor$, where $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$. By definition, $0 \le \{x\} < 1$. A key property is that for any integer $k$ and real number $x$, we have $\{x+k\} = \{x\}$. This means adding or subtracting an integer does not change the fractional part.

2. Fractional Part via Remainder: If we divide an integer $N$ by a positive integer $D$, we get $N = QD + R$, where $Q$ is the quotient (an integer) and $R$ is the remainder, with $0 \le R < D$. Then $\frac{N}{D} = Q + \frac{R}{D}$. The fractional part is $\left\{\frac{N}{D}\right\} = \left\{Q + \frac{R}{D}\right\} = \left\{\frac{R}{D}\right\}$. Since $0 \le R < D$, we have $0 \le \frac{R}{D} < 1$, so $\left\{\frac{R}{D}\right\} = \frac{R}{D}$. Thus, finding the fractional part of $\frac{N}{D}$ is equivalent to finding the remainder $R$ when $N$ is divided by $D$, and the fractional part will be $\frac{R}{D}$.

3. Binomial Theorem: The Binomial Theorem states that for any non-negative integer $n$, $(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}a^0 b^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ This theorem is particularly useful when one of the terms in the binomial is a multiple of the divisor.

Step-by-Step Solution

We need to find the fractional part of $\frac{4^{2022}}{15}$, i.e., $\left\{\frac{4^{2022}}{15}\right\}$.

Step 1: Express the numerator in a suitable form using the divisor. Our goal is to find the remainder when $4^{2022}$ is divided by 15. We look for a power of 4 that is close to a multiple of 15, or has a simple remainder when divided by 15. Consider powers of 4: $4^1 = 4$ $4^2 = 16$

Notice that $16$ can be written as $15+1$. This is highly useful, as $16 \equiv 1 \pmod{15}$. We can rewrite $4^{2022}$ using $4^2$: $4^{2022} = (4^2)^{1011}$ Explanation: We rewrite $4^{2022}$ as $(4^2)^{1011}$ because the exponent $2022$ is an even number, allowing us to group $4^2 = 16$. This makes it easier to relate the numerator to the divisor 15, since $16$ is just $1$ more than $15$.

Step 2: Apply the Binomial Theorem. Now substitute $4^2 = 16$ into the expression: $4^{2022} = 16^{1011}$ Next, we express $16$ as $(15+1)$ to leverage the Binomial Theorem. $16^{1011} = (15+1)^{1011}$ Now, expand $(15+1)^{1011}$ using the Binomial Theorem: $(15+1)^{1011} = \binom{1011}{0}15^0 \cdot 1^{1011} + \binom{1011}{1}15^1 \cdot 1^{1010} + \binom{1011}{2}15^2 \cdot 1^{1009} + \dots + \binom{1011}{1011}15^{1011} \cdot 1^0$ $(15+1)^{1011} = 1 + \binom{1011}{1}15 + \binom{1011}{2}15^2 + \dots + 15^{1011}$ Explanation: We apply the Binomial Theorem to $(15+1)^{1011}$. By doing so, we can clearly separate the terms that are multiples of 15 from the constant term. Every term in the expansion, except the very first term ($\binom{1011}{0} \cdot 15^0 \cdot 1^{1011} = 1 \cdot 1 \cdot 1 = 1$), will contain a factor of $15^1$ or higher. This means all terms after the first are multiples of 15.

We can write this as: $(15+1)^{1011} = 1 + 15 \left( \binom{1011}{1} + \binom{1011}{2}15 + \dots + 15^{1010} \right)$ Let $K = \left( \binom{1011}{1} + \binom{1011}{2}15 + \dots + 15^{1010} \right)$. Since all binomial coefficients are integers and 15 is an integer, $K$ must be an integer. So, we have: $4^{2022} = 1 + 15K$ This tells us that when $4^{2022}$ is divided by 15, the remainder is 1.

Step 3: Calculate the fractional part. Now we can substitute this back into the expression for the fractional part: $\left\{\frac{4^{2022}}{15}\right\} = \left\{\frac{1 + 15K}{15}\right\}$ $\left\{\frac{1 + 15K}{15}\right\} = \left\{\frac{1}{15} + \frac{15K}{15}\right\}$ $\left\{\frac{1}{15} + \frac{15K}{15}\right\} = \left\{\frac{1}{15} + K\right\}$ Explanation: We split the fraction into two parts. One part is $\frac{1}{15}$, and the other is $\frac{15K}{15} = K$, which is an integer.

Using the property $\{x+k\} = \{x\}$ where $k=K$ (an integer): $\left\{\frac{1}{15} + K\right\} = \left\{\frac{1}{15}\right\}$ Since $0 \le \frac{1}{15} < 1$, the fractional part of $\frac{1}{15}$ is simply $\frac{1}{15}$. $\left\{\frac{1}{15}\right\} = \frac{1}{15}$

Thus, the fractional part of $\frac{4^{2022}}{15}$ is $\frac{1}{15}$.

The final answer is $\boxed{\text{(C)}}$.

Important Tips and Common Mistakes

• Modular Arithmetic: This type of problem can also be solved efficiently using modular arithmetic. We found $4^{2022} \equiv 1 \pmod{15}$, which directly implies that the remainder is 1, and thus the fractional part is $\frac{1}{15}$. Understanding modulo operations can simplify calculations for larger exponents.
• Choosing the right binomial form: When applying the Binomial Theorem for fractional parts, try to express the base as $(D \pm 1)$ or $(D \pm R)$, where $D$ is the divisor. For example, if the divisor were 17, and the base was 16, you might use $(17-1)$.
• Fractional part is always non-negative: Remember that the fractional part $\{x\}$ must always be $0 \le \{x\} < 1$. If your calculation results in a negative remainder (e.g., $-1 \pmod{15}$), convert it to a positive remainder ($14 \pmod{15}$) before forming the fractional part. For example, $\{-0.2\} = 0.8$, not $-0.2$.

Summary

The problem demonstrates how to effectively use the Binomial Theorem to find the remainder of a large power divided by an integer, and consequently determine its fractional part. The key steps involve rewriting the base of the power to relate it to the divisor, expanding the expression using the Binomial Theorem, identifying the remainder, and then converting that remainder into the fractional part. This method is fundamental for problems involving remainders and fractional parts of powers.