Amrit explained how to find a formula for the sum of consecutive numbers, and then adapted it to answer the other questions:

Pair the values as follows:

1	2	3	...	n
n	n-1	n-2	...	1

Each of the columns sums to $n + 1$, and there are $n$ columns, so the sum
should be $n(n + 1)$. However, as each column is counted twice (once in each
possible order of the two values), the sum is $\frac{n(n + 1)}{2}$.

$2 + 4 + 6 + ... + 96 + 98 + 100 = 2(1 + 2 + 3 + ... + 48 + 49 + 50)$
which equals $2 \times \frac{50(50 + 1)}{2} = 50(50 + 1) = 50 \times 51 = 2550$.

$$\sum_{k=1}^{20}(4k + 12) = 4(\sum_{k=1}^{20} k) + \sum_{k=1}^{20} 12$$
$$= 4 \times \frac{20(20 + 1)}{2} + 12\times20 = 840 + 240 = 1080$$



For the last part, the minimum possible sum for 7 consecutive numbers is 1
+ 2 + 3 + 4 + 5 + 6 + 7 = 28 and the minimum for 8 is 1 + 2 + 3 + 4 + 5 + 6
+ 7 + 8 = 36. Thus, there must be at most 7 consecutive numbers. We try:

2 consecutive numbers: a + a + 1 = 32 -> 2a = 31 -> no solution.
3 consecutive numbers: a + a + 1 + a + 2 = 32 -> 3a = 29 -> no
solution.
4 consecutive numbers: a + a + 1 + a + 2 + a + 3 = 32 -> 4a = 26 ->
no solution.
5 consecutive numbers: a + a + 1 + a + 2 + a + 3 + a + 4 = 32 -> 5a = 22
-> no solution.
6 consecutive numbers: a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 = 32 ->
6a = 17 -> no solution.
7 consecutive numbers: a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6 =
32 -> 7a = 11 -> no solution.

Thus, there is no set of positive integers that sums to 32.


Josh derived the sum for an arithmetic progression with first term $a$, difference $d$ and $n$ terms, and used it to solve the remaining questions:

To prove the sum formula we can do the following:

$S_n = a + (a + d) + (a + 2d) + ... + (a + [n-2]d) + (a + [n-1]d)$

If we reverse the sum we get:

$S_n =  (a + [n-1]d) +  (a + [n-2]d) + ... + (a + 2d) + (a + d) + a$

Note that there are n terms in both these expressions.  Adding the two
equations together, matching up the terms from the first sequence with
their reversed terms gives us

$2S_n = (2a + [n-1]d) + (2a + [n-1]d) + ... + (2a + [n-1]d)$

Note that there are still n terms in this expression. Since they are all
now the same, we now actually have n(2a + [n-1]d) so all that is left is to
divide by two to give:

$S_n = (\frac{n}{2})(2a + [n-1]d)$.

After how many terms would $17+21+25+”¦$ be greater than $1000$?

We can use the formula backwards to find $n$.  We can
see that $a = 17$ and $d = 4$ and that $S_n > 1000$

Therefore
$\frac{n}{2}(34 + 4[n-1]) > 1000$
$30n + 4n^2 > 2000$
$2n^2 + 15n - 1000 > 0$
$n > ( - 15 + \sqrt{225 + 8000} ) / 4$
$n > 18.922946...$
$n = 19$        (I only took the +ve solution of n since we can't have a negative number of terms)

Can you find the sum of all the integers less than 1000 which are not divisible by 2 or 3?

We have to consider a way to express the numbers which aren't divisible by two or three.

The first few are: $1, 5, 7, 11, 13, 17, 19, 23, 25...$
We can see that we have repeating differences of $4$ and $2$, but if we
separate them up we can get two sequences with differences of $6$:

$, 7, 13, 19, 25...$     and $5, 11, 17, 23...$

For the first sequence, $a = 1, d = 6$ and the $n^{th}$ term is the biggest number
of the form $6m + 1$ that is still less than $1000$.  I found this number to be
$997$.

$1 + 6(n-1) = 997$
$n - 1 = 166$
$n = 167$

So $S_167 = 83.5(2 + 996) = 83333$

For the second sequence I did the same thing, but with $a = 5$ and the $n^{th}$
term being the largest number of the form $6m + 5$ but still less than $1000$,
which is $995$.

$5 + 6(n-1) = 995$
$(n-1) = 165$
$n = 166$

So $S_166 = 83(10 + 990) = 83000$

Therefore the sum of all integers that are not divisible by $2$ or $3$ and are
less than $1000$ is $83000 + 83333 = 166,333$

Can you find a set of consecutive positive integers whose sum is 32?

Consecutive => d = 1

So $\frac{n}{2}(2a + n - 1) = 32$, so $n(2a + n - 1) = 64$

Since n > 0, I decided to work with increasingly large values of n up
until the point where they began to return smaller and smaller decimals or
negatives or any other sign that I wouldn't get any more valid values of
a.

$n = 1$
$2a = 64$
$a = 32$  <- this is only one number though so it doesn't really count!

$n = 2$
$4a + 2 = 64$
$4a = 62$
$a = 15.5$ <- invalid number since the series consists of positive integers

$n = 3$
$6a + 6 = 64$
$6a = 58$
$a = 9.66...$ <- invalid number since the series consists of positive integers

At this point I realised a quicker way to get $a$: rearrange the equation to
make a the subject.

$n(2a + n - 1) = 64$
$2a + n - 1 = 64/n$                        [1]
$2a = 64/n + 1 - n = (64 + n - n^2)/n$
$a = (64 + n - n^2)/2n$

By the line marked [1] I could see that n had to be even since otherwise we
would get a non-integer value for 64/n which would lead to a non-integer
a.

$n = 4$ gives $a = 6.5$    Not valid
$n = 6$ gives $a = 2.8333$ Not valid
$n = 8$ gives $a = 0.5$    Not valid
$n = 10$ gives $a = -1.3$  Not valid

As n -> infinity, the constant term in the numerator and
the n term in the numerator become arbitrarily small compared to the -n^2.
So a therefore approaches (-n^2)/2n = -n/2 and so a will continue getting
more and more negative and so we can stop checking n terms and conclude
that there are no sets of consecutive positive integers whose sum is 32.

Rajeev had a different approach for the penultimate question:

Can you find the sum of all the integers less than 1000 which are not divisible by 2 or 3?

Here we first find the sum of all integers which are divisible by 2 and then find the sum of all integers which are divisible by 3 and add these two up and then subtract the sum of the integers that are divisible by 6

$999 \times 1000 /2 = 499500$

Multiples of 2:
$499 \times 1000 /2 =  249500$

Multiples of 3:
$333 \times 1002/2 =  166833$

As multiples of 2 and 3 have duplications they must be removed by counting the number of multiples of 6:
$166 x 1002/2 =  83166$

$499500 - (249500 +166833 - 83166)= 166333$
The sum of all the integers less than 1000 which are not divisible by 2 or 3 is166333


Well done to Abi from Wilberforce Sixth Form College who sent in a clear solution similar to Josh's and Amrit's.