Fine the least value of n such that 1+2+3+…+n is divisible by 2013.
Thanks
Simon YAU.
http://ck-math.com
Mathematics
2015-04-05 5:05 am
回答 (3)
2015-04-05 7:24 am
✔ 最佳答案
謝謝知足常樂 ( 知識長 ) 的分析，我們知道︰n^2 + n = n x (n+1) = 4026 x k
We need both n and k is an integer, that makes a quadratic Diophantine equation.
Quadratic Diophantine equation 是一個已經完滿解決的問題，你只要去找一個solver，把coefficient輸進去，就會有答案。
有個solver仲會出step-by-step solution，這個solution不是我的，我就不抄下來了。
查出答案是549。
2015-04-05 7:53 am
1＋2＋3＋⋯⋯＋n＝2013m
==> n(n＋1)/2＝2013m
==> n(n＋1)＝4026m＝61*66m
Consider the multiple of 61 :
61, 122, 183, 244, 305, 366, 427, 488, 549, 610, ...
As 121 has a factor of 11, but both (121, 122) do not have the factor 3.
550 has a factor of 11, and also 549 has the factor 3.
2015-04-04 23:54:12 補充：
So, 549 is the least value of n.
==> n(n＋1)/2＝2013m
==> n(n＋1)＝4026m＝61*66m
Consider the multiple of 61 :
61, 122, 183, 244, 305, 366, 427, 488, 549, 610, ...
As 121 has a factor of 11, but both (121, 122) do not have the factor 3.
550 has a factor of 11, and also 549 has the factor 3.
2015-04-04 23:54:12 補充：
So, 549 is the least value of n.
2015-04-05 5:58 am
Fact 1
2013 = 3 × 11 × 61
Fact 2
1 + 2 + 3 + ... + n = n(n + 1)/2
Consider
n(n + 1)/2 = 2013M
n² + n - 4026M = 0 where n is a positive integer
(n + A)(n - B) = 0
You need A - B = 1 and AB = 4026M and positive integers A and B.
2013 = 3 × 11 × 61
Fact 2
1 + 2 + 3 + ... + n = n(n + 1)/2
Consider
n(n + 1)/2 = 2013M
n² + n - 4026M = 0 where n is a positive integer
(n + A)(n - B) = 0
You need A - B = 1 and AB = 4026M and positive integers A and B.
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