1. Understanding the Concept of Fractional Part

The fractional part of a number $p$, denoted by $\{p\}$, is defined as $p - \lfloor p \rfloor$, where $\lfloor p \rfloor$ is the greatest integer less than or equal to $p$. In simpler terms, if a number $p$ can be written as $I + f$, where $I$ is an integer and $0 \le f < 1$, then its fractional part $\{p\}$ is $f$.

For a fraction $\frac{N}{D}$, where $N$ and $D$ are integers, we can write $N = qD + r$, where $q$ is the quotient (an integer) and $r$ is the remainder ($0 \le r < D$). Then, $\frac{N}{D} = q + \frac{r}{D}$. Since $q$ is an integer and $0 \le \frac{r}{D} < 1$, the fractional part $\left\{ \frac{N}{D} \right\}$ is equal to $\frac{r}{D}$. Therefore, the problem boils down to finding the remainder when $3^{200}$ is divided by 8.

2. Problem Analysis and Strategy

We need to evaluate $\left\{ {{{{3^{200}}} \over 8}} \right\}$. Our goal is to express the numerator $3^{200}$ in the form $8q + r$, where $q$ is an integer and $r$ is the remainder ($0 \le r < 8$). The key strategy here is to transform the base $3$ into a number that is either $1$ or $-1$ modulo 8, or can be easily expressed as $(1 + \text{multiple of } 8)$ or $(-1 + \text{multiple of } 8)$. Notice that $3^2 = 9$. And $9 = 1 + 8$. This is a perfect candidate for binomial expansion.

3. Step-by-Step Solution

Let's find the fractional part:

• Step 1: Rewrite the base to leverage the divisor. We have $3^{200}$. Since our divisor is 8, we look for powers of 3 that are close to a multiple of 8. $3^1 = 3$ $3^2 = 9$ We observe that $9 = 1 + 8$. This allows us to express $3^{200}$ in a form suitable for the Binomial Theorem. $\left\{ {{{{3^{200}}} \over 8}} \right\} = \left\{ {{{{{\left( {{3^2}} \right)}^{100}}} \over 8}} \right\}$ Explanation: We rewrite $3^{200}$ as $(3^2)^{100}$ because $3^2 = 9$, which can be conveniently written as $(1+8)$. This transformation is crucial for applying the Binomial Theorem effectively.

• Step 2: Substitute and prepare for Binomial Expansion. Now, substitute $3^2 = (1+8)$ into the expression: $= \left\{ {{{{{\left( {1 + 8} \right)}^{100}}} \over 8}} \right\}$ Explanation: We've replaced $9$ with $(1+8)$. This form $(1+x)^n$ is directly applicable to the Binomial Theorem.

• Step 3: Apply the Binomial Theorem. Recall the Binomial Theorem: (a+b)^n = {^n C_0 a^n b^0 + {^n C_1 a^{n-1} b^1 + {^n C_2 a^{n-2} b^2 + \dots + {^n C_n a^0 b^n}}}. Here, $a=1$, $b=8$, and $n=100$. \left( {1 + 8} \right)^{100} = {^{100}{C_0}(1)^{100}(8)^0 + {^{100}{C_1}(1)^{99}(8)^1 + {^{100}{C_2}(1)^{98}(8)^2 + \dots + {^{100}{C_{100}}(1)^0(8)^{100}}}} $= 1 \cdot 1 \cdot 1 + {^{100}{C_1} \cdot 8 + {^{100}{C_2} \cdot 8^2 + \dots + {^{100}{C_{100}} \cdot 8^{100}}}}$ $= 1 + {^{100}{C_1} \cdot 8 + {^{100}{C_2} \cdot 8^2 + \dots + {^{100}{C_{100}} \cdot 8^{100}}}}$ Now substitute this back into the fractional part expression: $= \left\{ {{{1 + {}^{100}{C_1}.8 + {}^{100}{C_2}{{.8}^2} + \dots + {}^{100}{C_{100}}{{.8}^{100}}} \over 8}} \right\}$ Explanation: We expand $(1+8)^{100}$ using the Binomial Theorem. This explicitly separates the terms into a constant term (which is not a multiple of 8) and terms that are clearly multiples of 8.

• Step 4: Isolate the remainder term. Observe that every term from ${^{100}{C_1} \cdot 8}$ onwards has a factor of 8. We can factor out 8 from these terms: ${^{100}{C_1} \cdot 8 + {^{100}{C_2} \cdot 8^2 + \dots + {^{100}{C_{100}} \cdot 8^{100}}} = 8 \left( {^{100}{C_1} + {^{100}{C_2} \cdot 8 + \dots + {^{100}{C_{100}} \cdot 8^{99}}}} \right)}$ Let $K = {^{100}{C_1} + {^{100}{C_2} \cdot 8 + \dots + {^{100}{C_{100}} \cdot 8^{99}}}}$. Since all binomial coefficients are integers and 8 is an integer, $K$ must also be an integer. So, the expression becomes: $= \left\{ {{{1 + 8K} \over 8}} \right\}$ Explanation: We group all terms that are multiples of 8. The sum of integers multiplied by 8 will itself be a multiple of 8. We denote this integer multiple of 8 as $8K$. This helps in simplifying the expression to easily identify the remainder.

• Step 5: Separate the integer and fractional parts. $= \left\{ {{1 \over 8} + K} \right\}$ Since $K$ is an integer and $0 \le \frac{1}{8} < 1$, by the definition of the fractional part, $\left\{ {{1 \over 8} + K} \right\} = {1 \over 8}$ Explanation: We separate the expression into an integer part ($K$) and a fractional part ($\frac{1}{8}$). By definition, if $I$ is an integer and $0 \le f < 1$, then $\{I+f\} = f$. Here, $K$ is the integer part and $\frac{1}{8}$ is the fractional part.

Thus, the value of $\left\{ {{{{3^{200}}} \over 8}} \right\}$ is $\frac{1}{8}$.

4. Important Tips and Alternative Approaches

• Modular Arithmetic (Congruence Relation): This problem can be solved very efficiently using modular arithmetic. We need to find $3^{200} \pmod 8$. $3^1 \equiv 3 \pmod 8$ $3^2 = 9 \equiv 1 \pmod 8$ Now, $3^{200} = (3^2)^{100} \equiv (1)^{100} \pmod 8$ $3^{200} \equiv 1 \pmod 8$ This means $3^{200}$ can be written as $8q + 1$ for some integer $q$. Therefore, $\frac{3^{200}}{8} = \frac{8q+1}{8} = q + \frac{1}{8}$. The fractional part is $\left\{ q + \frac{1}{8} \right\} = \frac{1}{8}$. This method is generally faster for such problems once you are comfortable with modular arithmetic.

• Common Mistakes:

  • Forgetting that the fractional part must be non-negative and less than 1. For example, if you get $\{-1/8\}$, the answer is not $-1/8$, but $7/8$ (since $\{-1/8\} = \{-1 + 7/8\} = 7/8$). In our case, the remainder was positive, so this wasn't an issue.
  • Incorrectly applying the Binomial Theorem or making calculation errors with binomial coefficients.
  • Not identifying the correct power of the base to simplify. For instance, if you used $3^1$ directly, the expansion of $(3)^{200}$ would be more complex to relate to modulo 8.

5. Key Takeaway

To find the fractional part $\left\{ \frac{N}{D} \right\}$, the core task is to determine the remainder $r$ when $N$ is divided by $D$. This can often be achieved using:

1. Binomial Theorem: By rewriting $N$ as $(kD \pm 1)^m$ or $(kD \pm a)^m$ where $a$ is small, and expanding it.
2. Modular Arithmetic: Using congruence relations to find $N \pmod D$. Both methods lead to the same result, and understanding both enhances problem-solving flexibility.

The final answer is \boxed{\text{1 \over 8}}.