superprismatic 11 Posted September 1, 2011 Report Share Posted September 1, 2011 How many sequences of consecutive positive integers add to 1,000,000? List them. Quote Link to post Share on other sites

0 Guest Posted September 1, 2011 Report Share Posted September 1, 2011 5 sequences: {10^{6}/5^{m}-(5^{m}-1)/2,...,10^{6}/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. Quote Link to post Share on other sites

0 Guest Posted September 1, 2011 Report Share Posted September 1, 2011 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 Quote Link to post Share on other sites

0 bushindo 14 Posted September 1, 2011 Report Share Posted September 1, 2011 5 sequences: {10^{6}/5^{m}-(5^{m}-1)/2,...,10^{6}/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. Quote Link to post Share on other sites

0 Guest Posted September 1, 2011 Report Share Posted September 1, 2011 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 Quote Link to post Share on other sites

0 Guest Posted September 1, 2011 Report Share Posted September 1, 2011 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. Quote Link to post Share on other sites

0 dark_magician_92 4 Posted September 1, 2011 Report Share Posted September 1, 2011 <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> Quote Link to post Share on other sites

0 dark_magician_92 4 Posted September 1, 2011 Report Share Posted September 1, 2011 sorry my last post shows clear signs of carelessness Quote Link to post Share on other sites

0 witzar 18 Posted September 1, 2011 Report Share Posted September 1, 2011 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. Quote Link to post Share on other sites

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## superprismatic 11

How many sequences of consecutive

positive integers add to 1,000,000?

List them.

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