Key Concepts This problem involves two main mathematical concepts:

1. Geometric Progression (GP): A sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The sum of the first $n$ terms of a GP is given by the formula $S_n = a \frac{r^n - 1}{r - 1}$, where $a$ is the first term and $r$ is the common ratio.
2. Modular Arithmetic: A system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value—the modulus. We use the notation $a \equiv b \pmod m$ to indicate that $a$ and $b$ have the same remainder when divided by $m$. To find the remainder of an expression like $X - Y \pmod m$, we can find $(X \pmod m) - (Y \pmod m) \pmod m$.

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Step 1: Simplify the given sum to find the value of K

The given equation is: ${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$

Explanation: First, we need to identify the pattern of the terms on the left-hand side. This is a sum of a finite series.

Let's list the first few terms: The first term $T_1 = \frac{1}{2 \cdot 3^{10}}$ The second term $T_2 = \frac{1}{2^2 \cdot 3^9}$ The third term $T_3 = \frac{1}{2^3 \cdot 3^8}$ ... The last term $T_{10} = \frac{1}{2^{10} \cdot 3}$

To check if it's a Geometric Progression, we compute the ratio of consecutive terms: $r = \frac{T_2}{T_1} = \frac{\frac{1}{2^2 \cdot 3^9}}{\frac{1}{2 \cdot 3^{10}}} = \frac{2 \cdot 3^{10}}{2^2 \cdot 3^9} = \frac{3}{2}$ Since the ratio is constant, this is indeed a Geometric Progression.

We have:

• First term, $a = \frac{1}{2 \cdot 3^{10}}$
• Common ratio, $r = \frac{3}{2}$
• Number of terms, $n = 10$ (from the power of 2 going from $1$ to $10$, or power of 3 going from $10$ to $1$)

Now, we apply the formula for the sum of a GP, $S_n = a \frac{r^n - 1}{r - 1}$: $S_{10} = \frac{1}{2 \cdot 3^{10}} \cdot \frac{\left(\frac{3}{2}\right)^{10} - 1}{\frac{3}{2} - 1}$

Calculation: $S_{10} = \frac{1}{2 \cdot 3^{10}} \cdot \frac{\frac{3^{10}}{2^{10}} - 1}{\frac{1}{2}}$ $S_{10} = \frac{1}{2 \cdot 3^{10}} \cdot \frac{\frac{3^{10} - 2^{10}}{2^{10}}}{\frac{1}{2}}$ $S_{10} = \frac{1}{2 \cdot 3^{10}} \cdot \frac{3^{10} - 2^{10}}{2^{10}} \cdot 2$ $S_{10} = \frac{3^{10} - 2^{10}}{2^{10} \cdot 3^{10}}$

Now, we equate this sum with the right-hand side of the given equation: $\frac{3^{10} - 2^{10}}{2^{10} \cdot 3^{10}} = \frac{K}{2^{10} \cdot 3^{10}}$

Explanation: By comparing the numerators of both sides, since the denominators are equal, we can directly find $K$.

Result: $K = 3^{10} - 2^{10}$

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Step 2: Find the remainder when K is divided by 6

We need to find $K \pmod 6$. We have $K = 3^{10} - 2^{10}$.

Explanation: To find $K \pmod 6$, we can evaluate $3^{10} \pmod 6$ and $2^{10} \pmod 6$ separately and then subtract their remainders. This is a property of modular arithmetic: if $A \equiv a \pmod m$ and $B \equiv b \pmod m$, then $A - B \equiv a - b \pmod m$.

First, let's find $3^{10} \pmod 6$:

• $3^1 = 3 \equiv 3 \pmod 6$
• $3^2 = 9 \equiv 3 \pmod 6$
• $3^3 = 27 \equiv 3 \pmod 6$ Observation: For any integer $n \ge 1$, $3^n \equiv 3 \pmod 6$. This is because $3^n = 3 \cdot 3^{n-1}$. If $n \ge 1$, then $3^n$ contains a factor of 3. Also, if $n \ge 2$, $3^n$ contains $3 \cdot 3 = 9$, which is not divisible by 2. This observation is key. $3^n$ is always odd. $3^n - 3 = 3(3^{n-1}-1)$. For $n \ge 1$, $3^{n-1}$ is always odd or 1 (for $n=1$). So $3^{n-1}-1$ is always even. Thus $3(3^{n-1}-1)$ is divisible by $3 \times 2 = 6$. Therefore, $3^{10} \equiv 3 \pmod 6$.

Next, let's find $2^{10} \pmod 6$:

• $2^1 = 2 \equiv 2 \pmod 6$
• $2^2 = 4 \equiv 4 \pmod 6$
• $2^3 = 8 \equiv 2 \pmod 6$
• $2^4 = 16 \equiv 4 \pmod 6$ Observation: For $n \ge 1$, the powers of $2 \pmod 6$ follow a cycle of $(2, 4)$. Specifically, $2^n \equiv 2 \pmod 6$ if $n$ is odd, and $2^n \equiv 4 \pmod 6$ if $n$ is even. Since $10$ is an even number, we have: Therefore, $2^{10} \equiv 4 \pmod 6$.

Now, we can find $K \pmod 6$: $K \equiv (3^{10} \pmod 6) - (2^{10} \pmod 6) \pmod 6$ $K \equiv 3 - 4 \pmod 6$ $K \equiv -1 \pmod 6$

Explanation: A remainder must be non-negative. To convert $-1 \pmod 6$ to a positive remainder, we add the modulus (6) to it until it's in the range $[0, 5]$. $K \equiv -1 + 6 \pmod 6$ $K \equiv 5 \pmod 6$

Common Mistake: A common mistake is to try to use binomial expansion for $(3^{10} - 2^{10})$ directly modulo 6, especially $(1+2)^{10} - 2^{10}$. While technically correct if done carefully, working with composite moduli like 6 can lead to errors if the divisibility rules for binomial coefficients are not rigorously applied. The direct evaluation of powers modulo 6 is much simpler and less prone to errors.

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Summary and Key Takeaway

By recognizing the series as a Geometric Progression, we first simplified the expression to find $K = 3^{10} - 2^{10}$. Then, using modular arithmetic properties, we evaluated $3^{10} \pmod 6$ and $2^{10} \pmod 6$ separately. We found that $3^{10} \equiv 3 \pmod 6$ and $2^{10} \equiv 4 \pmod 6$. Subtracting these remainders, we got $K \equiv 3 - 4 \equiv -1 \equiv 5 \pmod 6$.

The remainder when $K$ is divided by 6 is 5.

The final answer is $\boxed{5}$.```json [ {"description": "Identify the given series as a Geometric Progression (GP) and determine its first term, common ratio, and number of terms.", "status": "completed"}, {"description": "Apply the sum of GP formula to simplify the left-hand side of the equation.", "status": "completed"}, {"description": "Equate the simplified sum to the right-hand side to find the expression for K.", "status": "completed"}, {"description": "Determine the value of 3^10 modulo 6.", "status": "completed"}, {"description": "Determine the value of 2^10 modulo 6.", "status": "completed"}, {"description": "Calculate K modulo 6 using the properties of modular arithmetic.", "status": "completed"} ]