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We can write $9$ as $9=4+5=2+3+4$. So the question is , Is there a integer $N$ which can be written as a sum of $1990$ consecutive positive integers, and which can be written as a sum of (More that one) consecutive integers in exactly $1990$ ways ???? I Saw this question on net, somewhere I dont remember, but my solution was that there are only 2 such numbers possible, $N=5^{10}×199^{180},N=5^{180}×199^{10}$. I Have a doubt on my answer, so can you please help me to clarify it, also a solution would be appreciated . My Way::: We have $N=\sum_{i=0}^{1989}{(m+i) }=995(2m+1989)$. Therefore, we say that $N$ is odd and is divisible by $995=5×199$. There are exactly $1990$ pairs $(n,k)$ such that $N=\frac{(k+1)(n+2k)}{2}$. therefore $2N$ can be factored in exactly $1990$ ways. Now, we get that $2N$ has exactly $2×11×181$ divisors. Now, we write $2N$ as $2×5^{a_1}×199^{a_2}.....$. So we get $2×11×181=2(a_1+1)(a_2+1)...$therefore $(a_1,a_2)=(10,180)$. Therefore $N=5^{10}×199^{180},N=5^{180}×199^{10}$ are only possible.As all the steps can be reversed, these are the only answers. I want to ask, am I correct. Also can you please provide with another approach for this.

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If you show your answer and the reasoning, we can help with that. Otherwise it is a guessing game. –  Ross Millikan Jun 14 at 3:31
    
OK Sir, I will show my method now,, –  user106079 Jun 14 at 3:40
    
Looks good to me. –  Ross Millikan Jun 14 at 4:22
    
Any other approach for this problem ?? –  user106079 Jun 14 at 5:15

1 Answer 1

Hint: the sum of the numbers from $1$ to $n$ is $\frac 12n(n+1)$. If you want to represent $p$ as the sum of a number of consecutive numbers from $m$ to $n$, you need $p=\frac 12(n(n+1)-m(m-1))$

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Sir, I have added my approach now. Can you please review it –  user106079 Jun 14 at 3:57

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