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# MOdular arithmetic help pls!

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Let $S$ be the set of numbers of the form $n(n + 1)(n + 2)(n + 3)(n + 4),$ where $n$ is any positive integer. The first few terms of $S$ are \begin{align*} 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120, \\ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 &= 720, \\ 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 &= 2520, \end{align*} and so on. What is the GCD of the elements of $S$?

May 22, 2019

#2
+23786
+3

Let $S$ be the set of numbers of the form $n(n + 1)(n + 2)(n + 3)(n + 4),$ where $n$ is any positive integer.

The first few terms of $S$ are \begin{align*} 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120, \\ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 &= 720, \\ 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 &= 2520, \end{align*} and so on.

What is the GCD of the elements of $S$?

I assume:

The GCD of the elements of $$S$$ is 120.

\begin{align*} 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120 = 2^3\times3\times5\times7^0\times 11^0\times 13^0\times ... \times p^0 \\ \end{align*}

All other values contain a multiple of 4 and another even number, so at least a $$2^3$$
All other values contain a multiple of 3, so at least a $$3^1$$
All other values contain a multiple of 5, so at least a $$5^1$$, while there are 5 concecutive numbers.
The gcd is the product of all prime numbers with the smallest exponent. The exponent zero gives the factor 1.

So this is the value of the first therm.

May 23, 2019

#1
0

May 22, 2019
#2
+23786
+3

Let $S$ be the set of numbers of the form $n(n + 1)(n + 2)(n + 3)(n + 4),$ where $n$ is any positive integer.

The first few terms of $S$ are \begin{align*} 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120, \\ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 &= 720, \\ 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 &= 2520, \end{align*} and so on.

What is the GCD of the elements of $S$?

I assume:

The GCD of the elements of $$S$$ is 120.

\begin{align*} 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120 = 2^3\times3\times5\times7^0\times 11^0\times 13^0\times ... \times p^0 \\ \end{align*}

All other values contain a multiple of 4 and another even number, so at least a $$2^3$$
All other values contain a multiple of 3, so at least a $$3^1$$
All other values contain a multiple of 5, so at least a $$5^1$$, while there are 5 concecutive numbers.
The gcd is the product of all prime numbers with the smallest exponent. The exponent zero gives the factor 1.

So this is the value of the first therm.

heureka May 23, 2019
#3
+1

Another way:

$$n(n+1)(n+2)(n+3)(n+4)=\frac{(n+4)!}{(n-1)!}=5!*\frac{(n+4)!}{(n-1)!*5!}={n+4\choose 5}*5!$$

.
May 24, 2019