Problem 58 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\\{x\\}\) denotes the fractional part of \(x\), then \(\left\\{\frac{2^{2003}}{17}\right\\}\) is (A) \(\frac{2}{17}\) (B) \(\frac{4}{17}\) (C) \(\frac{8}{17}\) (D) \(\frac{16}{17}\)

Step-by-Step Solution

Answer

The fractional part is \(\frac{8}{17}\), so the answer is (C).

Step 1: Understanding the Fractional Part

The fractional part of a number \(x\), denoted \(\{x\}\), is defined as the difference between the number and the greatest integer less than or equal to \(x\). Therefore, \(\{x\} = x - \lfloor x \rfloor\). Our task is to find the fractional part of \(\frac{2^{2003}}{17}\).

Step 2: Applying Fermat's Little Theorem

Fermat's Little Theorem states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then \(a^{p-1} \equiv 1 \pmod{p}\). For our problem, let \(a = 2\) and \(p = 17\), so \(2^{16} \equiv 1 \pmod{17}\).

Step 3: Simplifying the Exponent Using Moduli

We are dealing with \(2^{2003}\). First, divide the exponent 2003 by 16 (since 2003 modulo 16 will give us the same result as \(2^{2003} \mod 17\)). The calculation is: \(2003 \div 16 = 125\) remainder 3. So, \(2003 \equiv 3 \pmod{16}\).

Step 4: Calculate Residual Power Modulo 17

Using the result from Step 3, compute \(2^{3} \mod 17\). Since \(2^{3} = 8\), and \(8 < 17\), we have \(2^{3} \equiv 8 \pmod{17}\). This implies \(2^{2003} \equiv 8 \pmod{17}\).

Step 5: Calculate the Fractional Part

Having determined that \(2^{2003} \equiv 8 \pmod{17}\), we have \(\frac{2^{2003}}{17} = n + \frac{8}{17}\) for some integer \(n\). Therefore, the fractional part is \(\left\{ \frac{2^{2003}}{17} \right\} = \frac{8}{17}\).

Step 6: Verify the Answer

Reconfirming our calculations using Fermat's Little Theorem and the properties of moduli, we find that each step aligns with our theoretical background. Thus, the correct answer amongst the options is \(\frac{8}{17}\).

Key Concepts

Fermat's Little Theorem

Fermat's Little Theorem is a key concept in number theory that simplifies calculations involving powers and moduli. The theorem states that if we take a prime number \(p\) and any integer \(a\) that is not divisible by \(p\), then \(a^{p-1}\) will always be congruent to 1 when divided by \(p\). In mathematical terms, this is expressed as \(a^{p-1} \equiv 1 \pmod{p}\).
This theorem is incredibly useful for simplifying large exponent calculations, as it allows us to reduce the exponent by recognizing that once it reaches \(p-1\), the power starts repeating. In our problem, with \(p=17\) and \(a=2\), this theorem helps us determine that \(2^{16} \equiv 1 \pmod{17}\).
This means that every time the exponent reaches a multiple of 16, the power of 2 modulo 17 resets to 1, a crucial simplification for handling the large exponent 2003. By applying Fermat's theorem, complex power computations are transformed into more manageable ones.

Modulo Arithmetic

Modulo Arithmetic, often abbreviated as mod, is a system of arithmetic for integers where numbers wrap around upon reaching a certain value, called the modulus. Imagine a clock where after hitting 12, the next hour is 1 again. That is a simple form of modulo arithmetic with modulus 12.
The expression \(a \equiv b \pmod{m}\) informs us that \(a\) and \(b\) leave the same remainder when divided by \(m\). This concept is vital in simplifying math operations and finding equivalency among numbers.
In the context of the original exercise, working with \(2^{2003} \pmod{17}\) requires us to first simplify the huge exponent using modulus. By dividing 2003 by 16 (the power cycle found using Fermat's theorem), we determine that \(2003 \equiv 3 \pmod{16}\). Therefore, rather than calculating directly \(2^{2003}\), we compute \(2^{3} \equiv 8 \pmod{17}\). This approach greatly reduces complexity in calculations.

Exponents

Exponents are a shorthand way to express repeated multiplication of a number by itself. For example, \(2^3\) means \(2 \times 2 \times 2\), resulting in 8.
When working with exponents, especially large ones, specific patterns or rules like Fermat's Little Theorem can simplify the computation.
In mathematical exercises, handling large exponents involves breaking them down with the help of modulo operations. In our case, we have \(2^{2003}\). Direct calculation would be practically unfeasible, so we determine its behavior in modulo setting, reducing it to \(2^3\). Simplification techniques, such as observing repeated cycles in modulo arithmetic, turn daunting exponent tasks into manageable solutions, which is essential in many areas of discrete mathematics.