It looks like you're seeking an answer to the question of determining the remainder when the expression [tex]\(15^{20} + 24^{15}\)[/tex] is divided by 7. Let's walk through the solution step by step.
First, note the problem requires us to find the remainder, or in mathematical terms, the modulo, of a complex expression:
[tex]\[ U = 15^{20} + 24^{15} \][/tex]
when divided by 7.
1. To begin, let's consider the expression [tex]\(15^{20}\)[/tex] modulo 7:
- [tex]\( 15 \mod 7 \)[/tex] simplifies to [tex]\( 15 \div 7 \)[/tex] which is 2 remainder 1. Therefore, [tex]\( 15 \equiv 1 \mod 7 \)[/tex].
- Thus, [tex]\( 15^{20} \mod 7 \)[/tex] simplifies to [tex]\( 1^{20} \mod 7 \)[/tex]. Since any number to the power of 1 is 1, [tex]\( 1^{20} = 1 \)[/tex].
Therefore:
[tex]\[ 15^{20} \equiv 1 \mod 7 \][/tex]
2. Next, consider the expression [tex]\(24^{15}\)[/tex] modulo 7:
- [tex]\( 24 \mod 7 \)[/tex] simplifies to [tex]\( 24 \div 7 \)[/tex] which is 3 remainder 3. Therefore, [tex]\( 24 \equiv 3 \mod 7 \)[/tex].
- Now we need [tex]\( 3^{15} \mod 7 \)[/tex].
To simplify [tex]\( 3^{15} \mod 7 \)[/tex], we can utilize properties of modular arithmetic and patterns:
- Notice the periodicity:
- [tex]\( 3^1 \equiv 3 \mod 7 \)[/tex]
- [tex]\( 3^2 \equiv 9 \equiv 2 \mod 7 \)[/tex]
- [tex]\( 3^3 \equiv (3^2 \cdot 3) \equiv 2 \cdot 3 \equiv 6 \mod 7 \)[/tex]
- [tex]\( 3^4 \equiv (3^3 \cdot 3) \equiv 6 \cdot 3 \equiv 18 \equiv 4 \mod 7 \)[/tex]
- [tex]\( 3^5 \equiv (3^4 \cdot 3) \equiv 4 \cdot 3 \equiv 12 \equiv 5 \mod 7 \)[/tex]
- [tex]\( 3^6 \equiv (3^5 \cdot 3) \equiv 5 \cdot 3 \equiv 15 \equiv 1 \mod 7 \)[/tex]
- Observing the pattern, [tex]\( 3^6 \equiv 1 \mod 7 \)[/tex], implies the cycle repeats every 6 powers.
Since [tex]\( 3^{15} \equiv 3^{6 \cdot 2 + 3} \equiv (3^6)^2 \cdot 3^3 \equiv 1^2 \cdot 3^3 \equiv 3^3 \equiv 6 \mod 7 \)[/tex].
Therefore:
[tex]\[ 3^{15} \equiv 6 \mod 7 \][/tex]
thus,
[tex]\[ 24^{15} \equiv 6 \mod 7 \][/tex]
3. Combining the results from the two parts:
[tex]\[ 15^{20} + 24^{15} \equiv 1 + 6 \equiv 7 \equiv 0 \mod 7 \][/tex]
Therefore, the remainder when [tex]\( 15^{20} + 24^{15} \)[/tex] is divided by 7 is 0. Maria is correct.
So, the final answer is:
[tex]\[ \boxed{0} \][/tex]
