Sum to 1000

If the number of summands $n=2k+1$ is odd, then we get $\begin{aligned} &(m-k)+\cdots+(m-1)+m+(m+1)+\cdots+(m+k)=\\ &nm=1000 \end{aligned}$ so $n|k$ . The odd numbers dividing 1000 are 1,5,25,125. If $n=125$ , then the average term is 8, and so some summands are negative. If $n=25$ , then the average term is 40, and all summands are positive. 25 is the largest in this case.

If the number of summands $n=2k$ is even, then we get $\begin{aligned} &(m-k+1)+\cdots+m+\cdots+(m+k)= \\ &m(n+1)-(m-k)=mn+\dfrac{n}{2}=\dfrac{n}{2}(2m+1)=1000 \end{aligned}$ so $\dfrac{n}{2}|1000$ and $n\not{|}1000$ . The smallest number greater than 25 that is twice a factor of 1000, while not being a factor of 1000, is 80. If $n=80$ , then the average term is $12.5$ , and not all terms are positive.

The answer is $\boxed{25}$ .

• The sum $S$ of a series with initial term $a$ and common difference $d$ is $[2an + n(n - 1)d]/2$ .
• Problem tells us we want $S = 1000$ and $d = 1$ . That gives to solve the following equation.
• $[2an + n(n - 1)]/2 = 1000$ .
• So, we're looking for the value of $a$ that will maximize $n$ . Solving the equation for $a$ , we get
• $a = [2000 - n(n-1)]/ 2n$ . It's kind of hard to see what values of n give rise to integers so we can divide out the difference by 2n and we get
• $a = 1000/n - (n-1)/2$ . So $n$ has to be a divisor of $1000$ and odd since $(n-1)/2$ has to be an integer. Since $a > 0$ we have $1000/n > (n -1) / 2$ . You can solve the equation by completing the square, you get an upper bound for n of 88.9 or so.
• So we know $n < 88$ . Listing the divisors of 1000 you get $2, 5, 10, 20, 25, 40,...$ 25 seems to be the biggest odd number < 88 and there goes the answer.

If there are k summands, starting at n, we want n as small as possible.

n+(n+1)+...+(n+k) = 1000 kn + k(k-1)/2 = 1000 k^2 + (2n-1)k - 2000 = 0 so 4n^2-4n+8001 is a perfect square

If n=28, 4n^2+4n+8001=105^2 and k=25

check: summing the arithmetic sequence starting at 28 for 25 terms,

S25 = 25/2 (28 2+24) = 25/2 80 = 1000

So, it looks like 25 is the maximum number of summands.

Let $a_{1} + ... + a_{k} = 1000$ where k is the max possible value.

Then we have $k \times a_{1} + (0 + 1 + ... + k-1) = 1000$ which is $k \times a_{1} + \frac{k(k-1)}{2} = 1000$

So we want $a_{1} = \frac{1-k}{2} + \frac{1000}{k}$ to be positive integer.

$k$ must be odd divisor of $1000 = 2^{3} \times 5^{3}$ thus $k$ is in {1, 5, 25, 125} and those are all the possibilities.

We already know the solutions for 1 and 5. For 125 we have $a_{1} < 0$ and it's not solution of positive integers (some of them are negative). So the longest sequence is the third solution for $\boxed{k=25}$ which is $28 + 29 + ... + 52 = 1000$ .

It would be arithmetic series with increment of 1 and mean as a factor of 1000, there could be only one mean, thus items must be odd. Odd factors are 1, 5, 25, 125 and corresponding means are 1000, 200, 40, 8. Mean 8 with 125 terms would go into -tive terms. Mean 40 with 25 terms is the next largest, and that is possible. So 25 terms.

A number can be written as sum of consecutive integer if and only if it has odd factor. 1000 = $2^{3}$ x $5^{3}$ . So, it has 4 odd factors namely 1, 5, 25, 125. We can represent 1000 as sum of 1, 5, 25, 125 consecutive integers.

To represent as sum of 125 integers, 1000 = -54 - 53 - 52 - ........- 1+ 0 + 1 + ........... + 70 without negative terms, this is 1000 = 55 + 56 +......+70 (16 terms)

To represent as sum of 25 integers, 1000 = 28 + 29 + ...... + 52 (25 terms) So, finelly it's 25

If there are k summands, starting at n, we want n as small as possible.

$n+(n+1)+...+(n+k) = 1000$

$kn + k(k-1)/2 = 1000$

$k^{2} + (2n-1)k - 2000 = 0$

So,

$4n^{2}-4n+8001$ is a perfect square.

If, $n=28, 4n^{2}+4n+8001=105^{2}$ and $k=25$

Check:

Summing the arithmetic sequence starting at $28$ for $25$ terms,

$S25 = \frac{25}{2} (28\times2+24) = \frac{25}{2}\times80 = 1000$

So, it looks like $\boxed{25}$ is the maximum number of summands.

Its given that the numbers which add up to 1000 are consecutive positive integers. So if the first term is x then the last term will be x + (n-1)d, or x +(n-1), as the difference between each term is 1.(here n is the number of terms). So from the formula of Arithmetic progression : n/2 ( x+x+n-1) , we equate it with 1000. here either n is a odd number or (x+x+n-1) is a odd number as the expression evaluates to an integer. resolving 1000 into factors: 2 5 5 5 2 2. the largest odd factor is 125 but substituting it with either n or (2x+n-1) does not satisfy the equation. But 25 , satisfies the equation which is the next largest odd factor of 1000 and the greatest possible value for n. therefore the largest possible number of summands for 1000 is 25