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Theorem: $n(n^2 +5)$ is divisible by 6 for every integer $n\ge0$ Proof: Let the property $P(n)$ be the sentence "$n(n^2 +5)$ is divisible by 6." For our base case we will show that $P(0)$.
Either one of $n$, $n+1$ and $n+2$ is divisible by $3$, because $n$ is either in the form of $3k$, $3k+1$ or $3k+2$. We also have either one of $n$ or $n+1$ is divisible by $2$ because they are two consecutive natural numbers. Thus, the product of $n$, $n+1$ and $n+2$ is always divisible by $2*3=6$.
Las Vegas. / 36.17500°N 115.13639°W / 36.17500; -115.13639. Las Vegas, [6] often known as Sin City or simply Vegas, is the most populous city in the U.S. state of Nevada and the seat of Clark County. The Las Vegas Valley metropolitan area is the largest within the greater Mojave Desert, and second-largest in the Southwestern United ...
18 Ιαν 2024 · prime factorization calculator. A number is divisible by 6 = 2 × 3 if and only if it is divisible by 2 and by 3. A number is divisible by 12 = 2² × 3 if and only if it is divisible by 4 and by 3. A number is divisible by 15 = 3 × 5 if and only if it is divisible by 3 and 5.
The first very obvious thing to note is that $n(n^2 - 1)(n+2)$ can be written as $n(n - 1)(n+1)(n+2)$ which means that they are four consecutive integers. Obviously one of these must be divisible by $4$. The proof: Any number $n$ is either of the form $2m$ or $2m+1$. Case I: $n = 2m$
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