Key Concept: Fractional Part and Binomial Theorem

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 crucial property for this problem is that if $N$ is an integer and $x$ is a real number, then the fractional part of $(N+x)$ is the same as the fractional part of $x$; i.e., $\{N+x\} = \{x\}$.

The Binomial Theorem states that for any non-negative integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{r=0}^n {^nC_r a^{n-r} b^r} = {^nC_0 a^n b^0} + {^nC_1 a^{n-1} b^1} + \dots + {^nC_n a^0 b^n}$ This theorem is especially powerful when one of the terms in the binomial $(a+b)$ is a multiple of the divisor, as it allows us to isolate a remainder term.

Problem Statement We are asked to find the value of $k$ given that the fractional part of the number $\left\{ {{{{2^{403}}} \over {15}}} \right\}$ is equal to $\frac{k}{15}$.

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Step-by-step Solution

1. Strategic Manipulation of the Numerator

Our goal is to analyze the expression $\frac{2^{403}}{15}$. To effectively use the Binomial Theorem, we want to express the numerator in terms of a base that is close to a multiple of $15$. We observe that $2^4 = 16$, which is precisely $(15+1)$. This makes it an ideal candidate for binomial expansion.

Let's rewrite the exponent $403$: $2^{403} = 2^3 \cdot 2^{400}$ We factored out $2^3$ because $400$ is a multiple of $4$, which allows us to use $2^4 = 16$. Now, we can substitute this back into the original expression: $\frac{2^{403}}{15} = \frac{2^3 \cdot (2^4)^{100}}{15}$ Substitute $2^3 = 8$ and $2^4 = 16$: $= \frac{8 \cdot (16)^{100}}{15}$

Why this step? This initial manipulation is crucial. By rewriting $2^{403}$ as $8 \cdot (16)^{100}$, we've transformed the problem into one where the base of the power $(16)$ is conveniently one greater than the denominator $(15)$. This setup is ideal for applying the Binomial Theorem.

2. Applying the Binomial Theorem to the Power Term

Now we have the expression $\frac{8 \cdot (16)^{100}}{15}$. Let's replace $16$ with $(15+1)$: $= \frac{8 \cdot (15+1)^{100}}{15}$ We now expand $(15+1)^{100}$ using the Binomial Theorem, with $a=15$ and $b=1$: $(15+1)^{100} = {^{100}C_0 (15)^{100} (1)^{0}} + {^{100}C_1 (15)^{99} (1)^{1}} + {^{100}C_2 (15)^{98} (1)^{2}} + \dots + {^{100}C_{99} (15)^{1} (1)^{99}} + {^{100}C_{100} (15)^{0} (1)^{100}}$ Simplifying the terms involving $1^r$ and the combinations: $(15+1)^{100} = {^{100}C_0 \cdot 15^{100}} + {^{100}C_1 \cdot 15^{99}} + {^{100}C_2 \cdot 15^{98}} + \dots + {^{100}C_{99} \cdot 15} + 1$ (Since ${^{100}C_{100}} = 1$ and $15^0=1$).

Why this step? The Binomial Theorem allows us to break down the complex power $(15+1)^{100}$ into a sum of simpler terms. The key insight here is that every term in this expansion, except for the very last one ($1$), contains at least one factor of $15$. This property will make the division by $15$ very straightforward.

3. Isolating the Integer and Fractional Parts

Now, substitute the expanded form of $(15+1)^{100}$ back into our main expression: $\frac{8 \cdot (15+1)^{100}}{15} = \frac{8}{15} \left( {^{100}C_0 \cdot 15^{100}} + {^{100}C_1 \cdot 15^{99}} + \dots + {^{100}C_{99} \cdot 15} + 1 \right)$ Distribute the factor of $\frac{8}{15}$ to each term inside the parenthesis: $= \left( \frac{8}{15} \cdot {^{100}C_0 \cdot 15^{100}} + \frac{8}{15} \cdot {^{100}C_1 \cdot 15^{99}} + \dots + \frac{8}{15} \cdot {^{100}C_{99} \cdot 15} \right) + \left( \frac{8}{15} \cdot 1 \right)$ Let's examine the first group of terms: $\frac{8}{15} \left( {^{100}C_0 \cdot 15^{100}} + {^{100}C_1 \cdot 15^{99}} + \dots + {^{100}C_{99} \cdot 15} \right)$ Each term in this sum has a factor of $15$ that will cancel out with the $15$ in the denominator, resulting in an integer. For example: $\frac{8}{15} ({^{100}C_0 \cdot 15^{100}}) = 8 \cdot {^{100}C_0 \cdot 15^{99}}$ (an integer) $\frac{8}{15} ({^{100}C_{99} \cdot 15}) = 8 \cdot {^{100}C_{99}}$ (an integer) Since each term is an integer, their sum is also an integer. Let this integer sum be $I_{total}$. The second term in our main expression is simply: $\frac{8}{15} \cdot 1 = \frac{8}{15}$ So, the original expression can be written as: $\frac{2^{403}}{15} = I_{total} + \frac{8}{15}$ where $I_{total}$ is an integer.

Why this step? This is where the power of the Binomial Theorem really shines. By factoring out terms divisible by $15$, we clearly separate the entire quantity into an integer part and a potentially fractional part. This makes identifying the fractional part almost immediate.

4. Determining the Fractional Part and the Value of k

The fractional part of $\frac{2^{403}}{15}$ is given by $\left\{ I_{total} + \frac{8}{15} \right\}$. Using the property $\{N+x\} = \{x\}$, where $N = I_{total}$ is an integer: $\left\{ I_{total} + \frac{8}{15} \right\} = \left\{ \frac{8}{15} \right\}$ Since $0 \le \frac{8}{15} < 1$, the fractional part of $\frac{8}{15}$ is simply $\frac{8}{15}$. Thus, the fractional part of $\frac{2^{403}}{15}$ is $\frac{8}{15}$.

The problem states that the fractional part is $\frac{k}{15}$. By equating the two expressions for the fractional part: $\frac{k}{15} = \frac{8}{15}$ Multiplying both sides by $15$, we get: $k = 8$

Why this step? Having successfully isolated the fractional component, the final step involves directly comparing it with the given form $\frac{k}{15}$ to solve for $k$. This aligns with the definition and properties of fractional parts.

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Tips and Common Mistakes:

• Fractional Part Definition: Always remember that the fractional part $\{x\}$ must satisfy $0 \le \{x\} < 1$. If your calculation yields a negative fraction (e.g., $-0.5$) or a fraction $\ge 1$ (e.g., $1.2$), you must adjust it. For instance, $\{-3.5\} = \{-4 + 0.5\} = 0.5$.
• Choosing the Correct Binomial Form: When dividing by $N$, try to express the base of the exponent as $(N+1)$ or $(N-1)$. For example, if dividing by $7$, use $8 = (7+1)$ or $6 = (7-1)$.
• Careful with Initial Factorization: Ensure the initial factorization of the numerator leads to a power with the desired base (e.g., $2^{403} \rightarrow 8 \cdot (2^4)^{100} = 8 \cdot 16^{100}$).

Summary and Key Takeaway: This problem beautifully illustrates how the Binomial Theorem can be leveraged to find the fractional part (or remainder) of a division involving large powers. The core strategy is to rewrite the number being divided ($2^{403}$) such that the base of the exponent ($16$) is in the form of (divisor + 1) or (divisor - 1). Expanding this binomial expression reveals an integer part (composed of terms divisible by the divisor) and a fractional part (from the remainder term). This systematic approach simplifies complex calculations into manageable steps.