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Summing Consecutive Numbers

15 = 7 + 8 and 10 = 1 + 2 + 3 + 4. Can you say which numbers can be expressed as the sum of two or more consecutive integers?

Always the Same

Arrange the numbers 1 to 16 into a 4 by 4 array. Choose a number. Cross out the numbers on the same row and column. Repeat this process. Add up you four numbers. Why do they always add up to 34?


The well known Fibonacci sequence is 1 ,1, 2, 3, 5, 8, 13, 21.... How many Fibonacci sequences can you find containing the number 196 as one of the terms?

The Simple Life

Age 11 to 14
Challenge Level

Lily from Jersey College for Girls found another possibility:
I found one possibilty for this question, and my answer became $3(x+2y) + 2(x+y)$, this would lead to $3x+6y + 2x+2y$, then the answer becoming $5x + 8y$.

Minjoon from Seoul Foreign British School found another one:

(so $a=6$ and $b=-1$)

Brian found a possibility using Alison's algebraic method:
$$a(x+2y)+b(2x+3y)=5x+8y\\\Rightarrow\begin{cases}ax+2bx=5x \\2ay+3by=8y\end{cases}\\
\Rightarrow\begin{cases} a+2b=5\\2a+3b=8\end{cases}\\
\Rightarrow a=1\hspace{4mm}\text{and}\hspace{4mm}b=2$$

Rishika from Nonsuch High School for Girls used her own algebraic method, similar to Alison's, to find all of the possibilities:
I used an algebraic method to find the multiples using the following method:
For example, $$a(x+y)+b(x+2y) = 5x+8y$$
{$a$ and $b$ represent the values of the multiples we need to find}

We can then form two equations if we substitute $x=0$ and $y=0$,
When $x=0$,
$$ay+2by = 8y$$
When $y=0$,
$$ax+bx = 5x$$
Then, we can cancel the like terms in each equation:

Equation 1: $a+2b = 8$

Equation 2: $a+b = 5$

Which we can solve simultaneously:

Equation 1 - Equation 2 (as '$a$' is present in both): $b = 3$

Substitute into Equation 2: $a = 2$
Therefore, $2 (x+y) +3 (x+2y) = 5x+8y$

I repeated this for all other possibilities.
Rishika made some mistakes, but these are the ones that Rishika found correctly:
$$\begin{align}2 (x+y) &+3 (x+2y) \hspace{4mm}\text{{from above}}\\
6 (x+y) &-1 (x-2y)\\
4 (x+y) &+1 (x+4y)\\
1 (x+2y) &+2 (2x+3y)\\

Rishika identified all of the possible pairs of expressions. Can you use Rikisha's method, or any other method, to find multiples of these pairs that add up to $5x+8y$?
$(x+y)$ and $(2x+3y)$
$ (x+2y)$ and $(x-2y)$
$ (x+2y)$ and $(x+4y)$
$ (x-2y)$ and $ (x+4y)$
$(x-2y) $ and $ (2x+3y)$
$(x+4y)$ and $ (2x+3y)$