# Prove that #1^99+2^99+3^99+4^99+5^99# is divisble by 5.?

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To prove that (1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99}) is divisible by 5, we can use the fact that for any integer (n), (n^5 - n) is divisible by 5.

Now, consider the expression modulo 5: [ \begin{align*} 1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99} &\equiv 1^{99} + 2^{99} + 3^{99} + 4^{99} + 0^{99} \pmod{5} \ &\equiv 1^5 + 2^5 + 3^5 + 4^5 + 0 \pmod{5} \ &\equiv 1 + 32 + 243 + 1024 + 0 \pmod{5} \ &\equiv 1 + 2 + 3 + 4 + 0 \pmod{5} \ &\equiv 10 \pmod{5} \ &\equiv 0 \pmod{5} \end{align*} ]

Therefore, (1^{99} + 2^{99} + 3^{99} + 4^{99} + 5^{99}) is divisible by 5.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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