When Is \( (k+1)^n+(k+2)^n+(k+3)^n+(k+4)^n+(k+5)^n \) Divisible by 5?

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In summary, the formula for determining if an expression is divisible by 5 is to sum the digits of the expression and check if the resulting sum is divisible by 5. The value of $n$ does not affect the divisibility and the nonnegative integers $k$ and $n$ in the expression represent coefficients and exponents. The expression can be simplified using the binomial theorem and any nonnegative values of $k$ and $n$ that result in a sum of digits divisible by 5 will make the expression divisible by 5.
  • #1
Ackbach
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Here is this week's POTW:

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For which nonnegative integers $n$ and $k$ is
$$ (k+1)^n+(k+2)^n+(k+3)^n+(k+4)^n+(k+5)^n $$
divisible by $5?$

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
Congratulations to kaliprasad, kiwi, and castor28 for their correct solutions to this week's POTW, which was Problem 109 in the MAA archives. kaliprasad's solution follows:

[sp]
Let us define$f(k,n) = (k+1)^n +(k +2)^n +(k +3)^n + (k+4)^n +(k +5)^n$Hence $f(k+1,n) - f(k,n ) = (x+6)^n - (k+1)^n$ and it is dvisible by $(k+6) - (k+1)$ or 5. this is independent of nSo $f(k+1,n)$ is divisible by 5 iff $f(k,n)$ is divisible by 5Further as 5 is prime we have as per Fermat's Little theorem $x^5=x$ for all x .Hence if $f(k,p)$ is divisible by 5 then $f(k,p+5)$ is divisible by 5So we need to check for $f(0,0),f(0,1),f(0,2),f(0,3),f(0,4)$ and we get

$f(0,0)=5,f(0,1) = 15, f(0,2) = 55, f(0,3) = 225, f(0,4) = 979$All except $f(0,4)$ is divisible by 5So it is divisible by 5 for n which is not of the form 5p+ 4 and for all k
for non negative k and p
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FAQ: When Is \( (k+1)^n+(k+2)^n+(k+3)^n+(k+4)^n+(k+5)^n \) Divisible by 5?

What is the formula for determining if a given expression is divisible by 5?

The formula for determining if an expression is divisible by 5 is to sum the digits of the expression and check if the resulting sum is divisible by 5. If the sum is divisible by 5, then the original expression is also divisible by 5.

How does the value of $n$ affect the divisibility of the given expression by 5?

The value of $n$ does not affect the divisibility of the given expression by 5. As long as the sum of the digits of the expression is divisible by 5, the expression will be divisible by 5 regardless of the value of $n$.

What is the significance of the nonnegative integers $k$ and $n$ in the given expression?

The nonnegative integers $k$ and $n$ represent the coefficients and exponents, respectively, of the terms in the given expression. They allow us to generalize the expression and determine the conditions under which it is divisible by 5.

Can the given expression be simplified to make it easier to determine if it is divisible by 5?

Yes, the given expression can be simplified by using the binomial theorem. This allows us to expand the expression and group like terms, making it easier to determine if the sum of the terms is divisible by 5.

What are the possible values of $k$ and $n$ that make the given expression divisible by 5?

The possible values of $k$ and $n$ that make the given expression divisible by 5 are any nonnegative integers that satisfy the condition that the sum of the digits of the expression is divisible by 5. This includes values such as $k=0$ and $n=2$, which result in the expression $5^2+6^2+7^2+8^2+9^2=25+36+49+64+81=255$, which is divisible by 5.

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