10. The positive integer $n$ is between 1 and 20 . Milly adds up all the integers from 1 to $n$ inclusive. Billy adds up all the integers from $n+1$ to 20 inclusive. Their totals are the same.
What is the value of $n$ ?
A 11
B 12
C 13
D 14
E 15
Solution
D
ComMENTARY
In the solution below we make use of the formula
$$
1+2+\cdots+n=\frac{1}{2} n(n+1)
$$
for the sum of the positive integers from 1 to $n$ inclusive.
If you are not familiar with this formula, see Problems 10.3 and 10.4 below.

Because the total that Milly gets is the same as the total that Billy gets, each of their totals is half the sum of the positive integers from 1 to 20 inclusive. We can obtain this total by putting $n=20$ in the formula $\frac{1}{2} n(n+1)$. This gives $\frac{1}{2} \times 20 \times 21=210$. Therefore the total of the numbers that Milly adds up is half of 210 , that is, 105 . So we need to find the positive integer $n$ such that $1+2+3+\cdots+n=105$

Meтrod 1
We find the value of $n$ by trying each of the given options in turn until we find the correct value. To test whether option $\mathrm{A}$ is correct, we need to see if the sum of the integers from 1 to 11 inclusive is equal to 105 . By putting $n=11$ in the formula $\frac{1}{2} n(n+1)$, we see that
$$
1+2+3+\cdots+11=\frac{1}{2} \times 11 \times 12=66 .
$$

So option A is not the correct answer.
We now test the other options in turn. We have
$$
\begin{array}{l}
1+2+\cdots+12=(1+2+\cdots+11)+12=66+12=78 . \\
1+2+\cdots+13=(1+2+\cdots+12)+13=78+13=91 . \\
1+2+\cdots+14=(1+2+\cdots+13)+14=91+14=105 .
\end{array}
$$

Therefore $n=14$. So option $\mathrm{D}$ is correct.
In the context of the SMC we can stop here.
However, a complete solution should explain why there is no other value of $n$ that works. See Problem 10.1, below

Meтноd 2
To find the smallest positive integer $n$ such that the sum of the integers from 1 to $n$, inclusive, is 105 , we solve the equation $\frac{1}{2} n(n+1)=105$. We have
$$
\begin{aligned}
\frac{1}{2} n(n+1)=105 & \Leftrightarrow n(n+1)=210 \\
& \Leftrightarrow n^{2}+n=210 \\
& \Leftrightarrow n^{2}+n-210=0 \\
& \Leftrightarrow(n+15)(n-14)=0 \\
& \Leftrightarrow n=-15 \text { or } n=14 .
\end{aligned}
$$

Because $n$ is a positive integer, we deduce that $n=14$.
Note that, as in the solution to Question 2, here the symbol $\Leftrightarrow$ stands for if, and only if.
For INVESTIGATION
10.1 Prove that there is exactly one positive integer $n$ that satisfies the equation
$$
\frac{1}{2} n(n+1)=105 \text {. }
$$
10.2 Show that for each positive integer $m$ the equation
$$
\frac{1}{2} n(n+1)=m
$$
has at most one positive integer solution.
10.3 The numbers obtained by adding up all the positive integers from 1 to $n$ inclusive are called the triangular numbers. This is because they correspond to the number of dots in a triangular array. For example, the figure on the right shows an array of dots corresponding to the sum $1+2+3+4+5+6$.
The notation $T_{n}$ is often used for the $n$-th triangular number. That is,
$$
T_{n}=1+2+3+\cdots+n .
$$

In the figure on the right we have put together two triangular arrays of dots corresponding to the sum $1+2+3+4+5+6$. The two triangular arrays together form a rectangle with 6 rows and 7 columns. The rectangle therefore contains $6 \times 7$ dots. So the figure illustrates the fact that $2 T_{6}=6 \times 7$, and hence that $T_{6}=\frac{1}{2}(6 \times 7)$.

Generalize the above to give a proof of the formula $T_{n}=\frac{1}{2} n(n+1)$. for all positive integers $n$.
10.4 We can express the sum of the first $n$ positive integers more succinctly using the sigme notation. That is, we write $\sum_{k=1}^{n} k$ for the sum $1+2+3+\cdots+n$, and, more generally, $\sum_{k=1}^{n} s_{k}$ for the sum $s_{1}+s_{2}+s_{3}+\cdots+s_{n}$. We use this notation to give an algebraic proof of the formula for the triangular numbers.

We begin with the observation that
$$
(k+1)^{2}-k^{2}=\left(k^{2}+2 k+1\right)-k^{2}=2 k+1 .
$$

Therefore summing for $k$ going from 1 to $n$, we obtain
$$
\begin{aligned}
\sum_{k=1}^{n}\left((k+1)^{2}-k^{2}\right) & =\sum_{k=1}^{n}(2 k+1) \\
& =2 \sum_{k=1}^{n} k+\sum_{k=1}^{n} 1 .
\end{aligned}
$$

That is,
$$
\sum_{k=1}^{n}\left((k+1)^{2}-k^{2}\right)=2 T_{n}+n .
$$

Show how the left hand side of the above equation simplifies. Then rearrange the equation to obtain the formula for $T_{n}$.
10.5 The solution to Question 10 shows that the equation $T_{m}=\frac{1}{2} T_{n}$ has the positive integer solution $m=14, n=20$.
Show that if $m=a, n=b$ is a solution of the equation $T_{m}=\frac{1}{2} T_{n}$, then so also is $m=3 a+2 b+2$ and $n=4 a+3 b+3$.
Deduce that the equation $T_{m}=\frac{1}{2} T_{n}$, has infinitely many positive integer solutions. [Note that another way of putting this is to say that there are infinitely many positive integers $T$ such that both $T$ and $2 T$ are triangular numbers.]