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

Answer 1

Sadie Alexander

See proof below

#1=1\mod 5#

#1^{99}=1^99\mod 5#

#1^{99}=1\mod 5#

Similarly,

#2^{99}=2^{99}\mod 5#

#3^{99}=(-2)^99\mod 5#

#4^{99}=(-1)^99\mod 5#

#5^{99}=0\mod 5#

#\therefore 1^99+2^99+3^99+4^99+5^99#

#=(1+2^99+(-2)^99+(-1)^99+0)\ mod 5#

#=(1+2^99-2^99-1)\ mod 5#

#=0\ mod 5#

hence the given number is divisible by #5#

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Answer 2

Theodore Amidon

We Know by divisibility rule that #a^n+b^n# is divisible by #a+b# when #n# is odd. Since inserting #a=-b# the value of #a^n+b^n= (-b)^n+b^n=0#

For similar reason in our problem

#1^99+4^99# is divisible by #1+4=5#

Again #2^99+3^99# is divisible by #2+3=5#

And #5^99# is divisible by #5#

Hence the sum #1^99+2^99+3^99+4^99+5^99# must be divisible by #5#

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Answer 3

Anna Allen

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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Answer from HIX Tutor

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.