What will be the remainder when (265)4081 + 9 is divided by 266?
8
The question asks for the remainder when a specific expression, \((265)^{4081} + 9\), is divided by 266. This is a classic problem involving modular arithmetic, which deals with remainders after division. We need to find the value of \((265)^{4081} + 9 \pmod{266}\).
To solve this efficiently, we can use the properties of modular arithmetic. The key idea is that we can find the remainder of each part of the expression first, and then combine them.
First, let's find the remainder when the base, 265, is divided by 266.
\(265 \div 266\)
Since 265 is less than 266, the remainder is 265. In modular arithmetic, we can also express this using a negative remainder.
\(265 \equiv 265 \pmod{266}\)
Alternatively, we know that \(266 - 1 = 265\). So, 265 is 1 less than a multiple of 266 (which is \(1 \times 266\)).
\(265 \equiv -1 \pmod{266}\)
Using \(-1\) is often helpful when dealing with powers, as it simplifies calculations significantly.
Now we substitute this equivalent value (\(-1\)) back into the original expression \((265)^{4081} + 9 \pmod{266}\).
\((265)^{4081} + 9 \equiv (-1)^{4081} + 9 \pmod{266}\)
Next, we calculate \( (-1)^{4081} \). The rule is:
The exponent is 4081, which is an odd number.
So, \( (-1)^{4081} = -1 \).
Substitute the value of \( (-1)^{4081} \) back into the expression:
\((-1)^{4081} + 9 \equiv -1 + 9 \pmod{266}\)
Now, perform the addition:
\(-1 + 9 = 8\)
So, the expression simplifies to:
\(8 \pmod{266}\)
The remainder when 8 is divided by 266 is simply 8, since 8 is less than 266.
Here is a quick summary of the steps taken:
The remainder when \((265)^{4081} + 9\) is divided by 266 is 8.
| Operation | Calculation | Modular Equivalent (mod 266) |
|---|---|---|
| Base modulo 266 | \(265 \div 266\) | \(265 \equiv -1 \pmod{266}\) |
| Base raised to power | \((265)^{4081}\) | \((-1)^{4081} \equiv -1 \pmod{266}\) |
| Add 9 | \((-1)^{4081} + 9\) | \(-1 + 9 \pmod{266}\) |
| Final Result | \(8\) | \(8 \pmod{266}\) |
Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value—the modulus. Here are some key properties used in solving problems like this:
Using these properties can significantly simplify calculations involving large numbers and exponents when only the remainder is required.
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select the correct answer using the code given below: