## h0pe one year ago 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?

1. myininaya

hmm... well n(n+1) is the product two consecutive integers so 2|n(n+1) also n(n+1)(n+2) will be divisible by 3 since it contains 3 consecutive integers like for example look at the following triples of consecutives: 8,9,10 1,2,3 3,4,5 4,5,6 all of these triples contain an integer that is divisible by 3 so we have 6|n(n+1)(n+2) and as you see 6|120 and 6|720 and 6|2520 you can probably keep going with this strategy. so looking at if we just had S(n)=n(n+1)(n+2) and we have the first first terms of this sequence S(n) is S(1)=1(2)(3) S(2)=2(3)(4) S(3)=3(4)(5) S(4)=4(5)(6) S(5)=5(6)(7) S(7)=7(8)(9) gcd(S(1),S(2))=2(3)=6 gcd(6,S(3))=6 So it does appear that gcd(S(1),S(2),S(3),...)=6 if S(n)=n(n+1)(n+2)

2. h0pe

It says 6 is wrong.

3. myininaya

You are right for your sequence.

4. myininaya

I gave you an example.

5. myininaya

S(n)=n(n+1)(n+2) is not the same as S(n)=n(n+1)(n+2)(n+3)(n+4)

6. myininaya

As in you are right 6 is not the gcd of your sequence.

7. h0pe

Oh, must have missed that when I read it over. So S(1)=1(2)(3)(4)(5) S(2)=2(3)(4)(5)(6) S(3)=3(4)(5)(6)(7) S(4)=4(5)(6)(7)(8) S(5)=5(6)(7)(8)(9) S(7)=7(8)(9)(10)(11)

8. myininaya

I think if you continue the thought process I had above you can find the gcd of your sequence.

9. h0pe

Alright, thank you

10. h0pe

Got it!

11. myininaya

like you already know 6|n(n+1)(n+2)(n+3)(n+4) and you should be able to find a big number k such that k|n(n+1)(n+2)(n+3)(n+4) notice for example n,(n+1),(n+2),(n+3) contains 4 consecutive integers like so 2*3*4=6(4)=24 so 24|n(n+1)(n+2)(n+3) like as you see if we had S(n)=n(n+1)(n+2)(n+3) S(1)=1(2)(3)(4) ..24|S(1) S(2)=2(3)(4)(5) ..24|S(2) ... and so on... so you can take this a little further for n(n+1)(n+2)(n+3)(n+4)(n+5)

12. h0pe

We claim that the GCD is 120. Since 120 itself is an element of S, the GCD cannot be greater than 120. The prime factorization of 120 is $$2^3 \cdot 3 \cdot 5$$, so it suffices to show that every element in S is divisible by 8, 3, and 5. Every element in S is of the form $$n(n + 1)(n + 2)(n + 3)(n + 4)$$. The numbers n, n + 1, n + 2, and n + 3, and n + 4 form five consecutive positive integers. Therefore, exactly one of them is divisible by 5, which means that their product is also divisible by 5. Also, among 5 consecutive positive integers, at least one integer is divisible by 3. Therefore, their product is also divisible by 3. However, among 5 consecutive positive integers, there is not necessarily one integer that is divisible by 8, so we must treat this case more carefully. We know that among n, n + 1, n + 2, n + 3, and n + 4, at least one is divisible by 4. This gives us two factors of 2. And apart from the number that is divisible by 4, there must be one other number that is even. This gives us another factor of 2, so the whole product is divisible by 120. We conclude that the GCD of the elements in S is 120.

13. myininaya

right

14. myininaya

that last I wrote should have said "so you can take this a little further for n(n+1)(n+2)(n+3)(n+4)"

15. h0pe

Okay