Prove that for a,b,c belongs to natural numbers, if a+b+c is divisible by 30 so is a^5+b^5+c^5.
To prove that if \(a + b + c\) is divisible by 30, then \(a^5 + b^5 + c^5\) is also divisible by 30, we can use the concept of modular arithmetic.
First, let's write down the given condition: \(a + b + c\) is divisible by 30. We can express this as \(a + b + c \equiv 0 \mod 30\), where \(\mod\) represents the modulus operator.
Now, we need to prove that \(a^5 + b^5 + c^5\) is divisible by 30, i.e., \(a^5 + b^5 + c^5 \equiv 0 \mod 30\).
To do that, we'll use the fact that if \(x \equiv y \mod n\) and \(m\) is any positive integer, then \(x^m \equiv y^m \mod n\). This property is known as Fermat's Little Theorem.
Applying Fermat's Little Theorem, we can raise both sides of \(a + b + c \equiv 0 \mod 30\) to the 5th power, giving us:
\((a + b + c)^5 \equiv 0^5 \mod 30\)
Expanding the left side using the binomial theorem:
\((a + b + c)^5 = a^5 + b^5 + c^5 + \text{other terms}\)
Since all the other terms on the right side will have at least one factor of \(a\), \(b\), or \(c\), they will also be divisible by 30. Therefore, we can rewrite the equation as:
\(a^5 + b^5 + c^5 \equiv 0^5 \mod 30\)
Finally, simplifying:
\(a^5 + b^5 + c^5 \equiv 0 \mod 30\)
Thus, we have proven that if \(a + b + c\) is divisible by 30, then \(a^5 + b^5 + c^5\) is also divisible by 30.