[superprismatic]

How many sequences of consecutive

positive integers add to 1,000,000?

List them.

[Guest]

5 sequences:

{10⁶/5^m-(5^m-1)/2,...,10⁶/5^m+(5^m-1)/2} for m=0,1,2,3,4.

So the 0th is just {1000000}, the 1st is {19998,...,20002},... and lastly the 4th is {1288,...,1912}. Namely, the 0th has just 1 "consecutive" term, the 1st has 5 consecutive terms,... and the 4th has 5^4=625 consecutive terms.

Had the question not said positive terms, there would be two more sequences when m=5 and when m=6.

[Guest]

There are four sequences not counting the set of 1 member (1,000,000). The logic is:-

1. each sequence must contain an odd number of members with the middle member (M) being a factor of 1,000,000.

2. the number of members of the sequence (N) must be the other factor: N = 1,000,000 / M

3. the first (smallest value) for N is 5 giving rise to the sequence:

200,000 + or - 1 to 2 which is 199,998 199,999 200,000 200,001 200,002

4. the next value for N is 25, giving rise to the sequence 40,000 + or - (N-1)/2 or 40,000 + or - 12

5. it turns out that N = 5^p where p is an integer. Thus the next value for N is 125, giving rise to the sequence 8,000 + or - 62

6. for p = 4, N = 625, M = 1,600 for the sequence 1,600 + or - 312

7. for p = 5, N = 3125, M = 320 which does not work because M must be greater than N/2

So just four sequences

[bushindo]

5 sequences:

{10⁶/5^m-(5^m-1)/2,...,10⁶/5^m+(5^m-1)/2} for m=0,1,2,3,4.

So the 0th is just {1000000}, the 1st is {19998,...,20002},... and lastly the 4th is {1288,...,1912}. Namely, the 0th has just 1 "consecutive" term, the 1st has 5 consecutive terms,... and the 4th has 5^4=625 consecutive terms.

Had the question not said positive terms, there would be two more sequences when m=5 and when m=6.

There's another

that starts at 7749 and goes on for 128 terms.

[Guest]

I got 6 sequences:

1. 1243 to 1882

2. 1288 to 1912

3. 7749 to 7876

4. 7938 to 8062

5. 39988 to 40012

6. 199998 to 200002

[Guest]

Bushindo's right. I failed to address those cases where there are an even number in the sequence. However, there is actually (if I did all of my math correctly)

7 sequences. The five that I listed before, the sixth that bushindo, et. al. posted and a seventh: {1243,...,1882}. (Again, there could be more even-length sequences if it weren't for the stipulation of the terms being positive.

[dark_magician_92]

<p>

bushindo is right as usua,l 6 sequences, mt method was assuming sequence start from k till n,<br />

so sum = n(n+1)/2 - k(k+1)/2=(n-k+1)(n+k)=2000000, so judt factorize it and equated n-k+1 to all odd factors and got 6 sequences

</p>

[dark_magician_92]

sorry my last post shows clear signs of carelessness

[witzar]

Let a be the first number of the sequence, and let n be the number of numbers of the sequence. We have:

(a+(a+n-1))*n/2 = 10^6

or

(*) (2a+n-1)*n = 2^7*5^6

The above equation implies that

(**) n = 2^k*5^l (where k=0,1,...,7 and l=0,1,...6)

We can also observe in (*), that if n is even, then (2a+n-1) is odd and 2^7|n.

So either n is odd or n is a multiplicity of 128.

We can easily check (assuming a=1), that n<1414, so by (**) we know that either

n \in { 5^0, 5^1, 5^2, 5^3, 5^4 }

(when n is odd; higher powers of 5 produce n>=1414) either

n \in { 2^7, 2^7*5}

(when n is even; 2^7*5^l>=1414 for l>1)

Now we can transform (*) into

a = 10^6/n - (n-1)/2

and find a for every possible value of n.

It turns out that all 7 values of n produce a valid sequence.