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**Can you find a sequence of consecutive integers that add up to 1000?**

This puzzle is based on the excellent book A First Step to Mathematical Olympiad Problems – which is full of problems that could be extended to become exploration ideas.

**Step 1 – arithmetic formula**

Our first step is to write out what we want:

a + (a+1) + (a+2) + … (a +n) = 1000

next we notice that the LHS is an arithmetic series with first term a, last term a+n and n+1 terms. Therefore we can use the sum of an arithmetic sequence formula:

S_{n} = 0.5n(u_{1} + u_{n})

S_{n} = 0.5(n+1)(a + a+n) = 1000

S_{n} = (n+1)(2a+n) = 2000

**Step 2 – logic**

However, we currently have 2 unknowns, n and a, and only 1 equation – so we can’t solve this straight away. However we do know that both a and n are integers – and n can be taken as positive.

The next step is to see that one of the brackets (n+1)(2a+n) must be odd and the other even (if n is odd then 2a + n is odd. If n is even then n+1 is odd). Therefore we can look at the odd factors of 2000:

**Step 3 – prime factorisation**

Using prime factorisation: 2000 = 2^{4} x 5³

Therefore any odd factors must solely come from the prime factor combinations of 5 – i.e 5, 25 and 125.

**Step 4 – trial and error**

So we now know that either (n+1) or (2a+n) must be 5, 25, 125. And therefore the other bracket must be 400, 80 or 16 (as 5 x 400 = 2000 etc). Next we can equate the (n+1) bracket to one of these 6 values, find the value of n and hence find a. For example:

If one bracket is 5 then the other bracket is 400.

So if (n+1) = 5 and (2a+n) = 400 then n = 4 and a = 198.

This means that the sequence: 198+199+200+201+202 = 1000.

If (n+1) = 400 and (2a+n) = 5 then n = 399 and a = -197.

This means the sequence: -197 + -196+ -195 … + 201 + 202 = 1000.

We follow this same method for brackets 25, 80 and 125,16. This gives the following other sequences:

28+29+30+…+51+52 = 1000

-54+-53+-52+…+69+70 = 1000

-27+-26+-25+…+51+52 = 1000

55+56+57+…+69+70 = 1000

So with a mixture of mathematical formulae, prime factorisation, logic and trial and error we have our solutions. A good example of how mathematics is often solved in reality!