Understanding Remainders when Dividing Powers by 6

The problem asks us to find the remainders when the numbers \(5^{16}\) and \(5^{25}\) are divided by 6. We are given that these remainders are \(R_1\) and \(R_2\) respectively. Once we find \(R_1\) and \(R_2\), we need to calculate the value of the expression \(\frac{{{{\rm{R}}_{\rm{1}}}{\rm{ + }}{{\rm{R}}_{\rm{2}}}}}{{{{\rm{R}}_{\rm{2}}}}}.\)

Finding the Remainder \(R_1\) for \(5^{16}\) divided by 6

To find the remainder when \(5^{16}\) is divided by 6, we can use modular arithmetic. We look at the base number, 5, and find its remainder when divided by 6.

\(5 \div 6\)

The remainder is 5. In modular arithmetic notation, this is \(5 \equiv 5 \pmod 6\).

Alternatively, we can notice that 5 is one less than 6. So, \(5 \equiv -1 \pmod 6\).

Now we can work with the power:

\(5^{16} \equiv (5)^{16} \pmod 6\)

Using the property \(5 \equiv -1 \pmod 6\):

\(5^{16} \equiv (-1)^{16} \pmod 6\)

Since 16 is an even number, \((-1)^{16} = 1\).

\(5^{16} \equiv 1 \pmod 6\)

The remainder when \(5^{16}\) is divided by 6 is 1. So, \(R_1 = 1\).

Finding the Remainder \(R_2\) for \(5^{25}\) divided by 6

Now we find the remainder when \(5^{25}\) is divided by 6. Again, we use modular arithmetic, starting with \(5 \equiv -1 \pmod 6\).

\(5^{25} \equiv (5)^{25} \pmod 6\)

Using \(5 \equiv -1 \pmod 6\):

\(5^{25} \equiv (-1)^{25} \pmod 6\)

Since 25 is an odd number, \((-1)^{25} = -1\).

\(5^{25} \equiv -1 \pmod 6\)

However, remainders must be non-negative when we talk about division in elementary arithmetic. A remainder of -1 modulo 6 is the same as \(6 - 1 = 5\) modulo 6.

\(5^{25} \equiv 5 \pmod 6\)

The remainder when \(5^{25}\) is divided by 6 is 5. So, \(R_2 = 5\).

Calculating the Value of the Expression \(\frac{{{{\rm{R}}_{\rm{1}}}{\rm{ + }}{{\rm{R}}_{\rm{2}}}}}{{{{\rm{R}}_{\rm{2}}}}}}

We have found \(R_1 = 1\) and \(R_2 = 5\). Now we substitute these values into the expression \(\frac{{{{\rm{R}}_{\rm{1}}}{\rm{ + }}{{\rm{R}}_{\rm{2}}}}}{{{{\rm{R}}_{\rm{2}}}}}\):

\(\frac{{{{\rm{R}}_{\rm{1}}}{\rm{ + }}{{\rm{R}}_{\rm{2}}}}}{{{{\rm{R}}_{\rm{2}}}}} = \frac{1 + 5}{5}\)

Simplify the numerator:

\(\frac{1 + 5}{5} = \frac{6}{5}\)

So, the value of the expression is \(\frac{6}{5}\).

Summary of Remainders and Calculation

Calculation of \(\frac{{{{\rm{R}}_{\rm{1}}}{\rm{ + }}{{\rm{R}}_{\rm{2}}}}}{{{{\rm{R}}_{\rm{2}}}}}:\)

\(\frac{{{{\rm{R}}_{\rm{1}}}{\rm{ + }}{{\rm{R}}_{\rm{2}}}}}{{{{\rm{R}}_{\rm{2}}}}} = \frac{1 + 5}{5} = \frac{6}{5}\)

The final answer is \(\frac{6}{5}\).

Revision Table: Remainders and Powers

Additional Information: Properties of Remainders

When working with remainders, especially with powers, understanding the cyclical nature of remainders is key. For example, let's look at powers of 5 divided by 6:

• \(5^1 \div 6\), Remainder = 5
• \(5^2 = 25 \div 6\), Remainder = 1 (since \(25 = 4 \times 6 + 1\))
• \(5^3 = 125 \div 6\), Remainder = 5 (since \(125 = 20 \times 6 + 5\))
• \(5^4 = 625 \div 6\), Remainder = 1 (since \(625 = 104 \times 6 + 1\))

We can see a pattern: the remainders alternate between 5 and 1. Specifically, \(5^n \pmod 6\) is 5 if \(n\) is odd, and 1 if \(n\) is even.

This pattern arises because \(5 \equiv -1 \pmod 6\).

• If \(n\) is even, \(5^n \equiv (-1)^n \equiv 1 \pmod 6\).
• If \(n\) is odd, \(5^n \equiv (-1)^n \equiv -1 \equiv 5 \pmod 6\).

Using this pattern directly:

• For \(5^{16}\), the exponent 16 is even, so the remainder \(R_1\) is 1.
• For \(5^{25}\), the exponent 25 is odd, so the remainder \(R_2\) is 5.

This confirms the remainders we found using the direct calculation with -1.