Understanding the Remainder Problem

The question asks for the remainder when the sum $9^3 + 9^4 + 9^5 + \cdots + 9^{100}$ is divided by 6. This is a problem involving modular arithmetic, specifically finding the value of $(9^3 + 9^4 + 9^5 + \cdots + 9^{100}) \pmod 6$. To find the remainder of a sum, we can find the remainder of each term first, sum those remainders, and then find the remainder of that sum. 

Analyzing Remainders of Powers of 9 Modulo 6

Let's look at the remainder when powers of 9 are divided by 6.

• For $9^1$: $9 = 1 \times 6 + 3$. The remainder is 3. So, $9^1 \equiv 3 \pmod 6$.
• For $9^2$: $9^2 = 81$. $81 = 13 \times 6 + 3$. The remainder is 3. So, $9^2 \equiv 3 \pmod 6$.
• For $9^3$: $9^3 = 729$. $729 = 121 \times 6 + 3$. The remainder is 3. So, $9^3 \equiv 3 \pmod 6$.

It appears that for any positive integer exponent $p \ge 1$, the remainder of $9^p$ when divided by 6 is always 3.

We can prove this using congruences:

Since $9 \equiv 3 \pmod 6$, we can raise both sides to the power $p$:

$\left(9\right)^p \equiv \left(3\right)^p \pmod 6$

So, $9^p \equiv 3^p \pmod 6$. Now we need to find the remainder of $3^p$ when divided by 6 for $p \ge 1$.

• For $p=1$: $3^1 = 3$. $3 \equiv 3 \pmod 6$.
• For $p \ge 2$: $3^p = 3 \times 3^{p-1}$. Since $p \ge 2$, $p-1 \ge 1$, so $3^{p-1}$ is an integer. $3^p$ is a multiple of 3. Also, for $p \ge 2$, $3^{p-1}$ is an odd number (any power of 3 is odd). Let $3^{p-1} = 2k+1$ for some integer $k$.

Then $3^p = 3 \times (2k+1) = 6k + 3$.

This shows that for $p \ge 2$, $3^p$ when divided by 6 leaves a remainder of 3.

Combining the cases for $p=1$ and $p \ge 2$, we conclude that $3^p \equiv 3 \pmod 6$ for all $p \ge 1$.

Since $9^p \equiv 3^p \pmod 6$, it follows that $9^p \equiv 3 \pmod 6$ for all $p \ge 1$.

Calculating the Number of Terms in the Sum

The given sum is $9^3 + 9^4 + 9^5 + \cdots + 9^{100}$. The exponents range from 3 to 100. To find the number of terms, we subtract the starting exponent from the ending exponent and add 1:

Number of terms = $100 - 3 + 1 = 98$.

There are 98 terms in the sum, each of the form $9^p$ where $p \ge 3$. As we established, for each term $9^p$, the remainder when divided by 6 is 3.

Finding the Remainder of the Entire Sum

Let the sum be $S = 9^3 + 9^4 + 9^5 + \cdots + 9^{100}$.

Using the property that the sum of remainders is congruent to the remainder of the sum:

$S \pmod 6 \equiv \left(9^3 \pmod 6 + 9^4 \pmod 6 + \cdots + 9^{100} \pmod 6\right) \pmod 6$

Since $9^p \equiv 3 \pmod 6$ for each term in the sum (where $p$ ranges from 3 to 100):

$S \pmod 6 \equiv \left(3 + 3 + \cdots + 3\right) \pmod 6$

There are 98 terms in the sum, and each term contributes a remainder of 3.

$S \pmod 6 \equiv (98 \times 3) \pmod 6$

Now we calculate $98 \times 3$ and find its remainder when divided by 6.

$98 \times 3 = 294$.

We need to find $294 \pmod 6$. We can perform division:

$294 \div 6$

$294 = 49 \times 6$.

Since 294 is a multiple of 6, the remainder is 0.

Alternatively, using modular arithmetic properties:

$98 \equiv 2 \pmod 6$ (since $98 = 16 \times 6 + 2$).

So, $98 \times 3 \equiv 2 \times 3 \pmod 6$

$98 \times 3 \equiv 6 \pmod 6$

$6 \equiv 0 \pmod 6$.

Thus, the remainder when the sum is divided by 6 is 0.

Step-by-Step Remainder Calculation

Here is a summary of the steps to find the remainder when the sum $9^3 + 9^4 + \cdots + 9^{100}$ is divided by 6:

The remainder when $9^3 + 9^4 + 9^5 + \cdots + 9^{100}$ is divided by 6 is 0.

Revision Table: Key Remainder Concepts

Additional Information: Properties of Modular Arithmetic

Modular arithmetic is fundamental in number theory and has applications in computer science (e.g., cryptography), timekeeping (clocks operate modulo 12 or 24), and more. The congruence relation $a \equiv b \pmod m$ is an equivalence relation, meaning it satisfies reflexivity ($a \equiv a \pmod m$), symmetry ($a \equiv b \pmod m$ implies $b \equiv a \pmod m$), and transitivity ($a \equiv b \pmod m$ and $b \equiv c \pmod m$ implies $a \equiv c \pmod m$).

The properties used in this problem are specific instances of how arithmetic operations behave with congruences:

• If $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then $a+c \equiv b+d \pmod m$.
• If $a \equiv b \pmod m$ and $c \equiv d \pmod m$, then $a \times c \equiv b \times d \pmod m$.
• If $a \equiv b \pmod m$, then $a^p \equiv b^p \pmod m$ for any positive integer $p$.

In our solution, we used the property $9^p \equiv (9 \pmod 6)^p \pmod 6$, and the property that the sum's remainder is the sum of the individual remainders modulo 6. Understanding these basic rules of modular arithmetic is key to solving remainder problems efficiently.