1.1 Problem Solving and Sums

There seems to be a definite winning strategy, some number of stones to take depending on the current pile size. The relevant variables are:

• Starting heap size
• Current heap size
• Whose turn it is
• How many stones you're allowed to take (1, 2, or 3)

• Record what happens in several games and look for patterns.
• Start simpler: what if the heap only had 1 stone? 2? 3? 4?
• Does the winner always leave a specific number for the opponent?

Gauss noticed that if you pair the first and last terms, you always get 101:

$$1 + 100 = 101, \quad 2 + 99 = 101, \quad 3 + 98 = 101, \quad \ldots$$

There are exactly 50 such pairs, so:

$$1 + 2 + \cdots + 100 = 50 \times 101 = 5050$$

Call the sum $S$. Write it twice: once forwards, once backwards, and add them:

$$S = 1 + 2 + 3 + \cdots + 99 + 100$$ $$S = 100 + 99 + 98 + \cdots + 2 + 1$$ $$2S = 101 + 101 + 101 + \cdots + 101 + 101$$

This method forces there to be an even number of terms in the sum. In this case, there are 100 copies of 101, so $2S = 100 \cdot 101 = 10100$, which gives $S = 5050$.

Use $t_n = \frac{n(n+1)}{2}$.

1. The first sum is the 48th triangular number: $$t_{48} = \frac{48 \cdot 49}{2} = \frac{2352}{2} = 1176$$
2. This one is the 399th triangular number: $$t_{399} = \frac{399 \cdot 400}{2} = \frac{159600}{2} = 79800$$

This isn't quite a triangular number since it doesn't start at 1. One trick: add up $1$ through $23$, then subtract $1$ through $4$ (the part we don't want).

$$t_{23} - t_4 = \frac{23 \cdot 24}{2} - \frac{4 \cdot 5}{2} = 276 - 10 = 266$$

Factor a 2 out of every term to find a triangular number hidden in the sum!

$$2(1 + 2 + 3 + \cdots + 50) = 2 \cdot t_{50} = 2 \cdot \frac{50 \cdot 51}{2} = 50 \cdot 51 = 2550$$

Sometimes when presented with a difficult problem it can be useful to just to collect data and present it in an organized way. Try building up a table:

Each time we add a triangle we add 2 matchsticks. The 4th figure has 9 sticks.

For larger figures, notice the pattern: Figure 1 starts with 3 sticks, and each new figure adds 2.

The 10th figure: $1 + 2(10) = 21$ sticks.

The 50th figure: $1 + 2(50) = 101$ sticks.

In general, our nth figure will have $1 + 2n$ sticks.

Picking any row or column should do. Lets pick the first column.

$$16+5+9+4=34$$

The magic number must be 34.

If every row adds up to some magic number $m$, then the total of all three rows is $3m$.

$a+b+c=m$, $d+e+f=m$, and $g+h+i=m$. If we add all of the numbers together, we get $3m$.

But that total is also just the sum of all numbers 1 through 9:

$$t_9 = \frac{9 \cdot 10}{2} = 45$$

So $3m = 45$, which means $\mathbf{m = 15}$.

So we could start placing numbers randomly in the magic square to try and hit the magic number, but there are many possible combinations to check. Trial and error might not be best here. Let's look at all the ways to write 15 as a sum of three distinct numbers from 1–9:

$$9+5+1, \quad 9+4+2, \quad 8+6+1, \quad 8+5+2, \quad 8+4+3$$ $$7+6+2, \quad 7+5+3, \quad 6+5+4$$

Count how many times each number appears across these sums:

5 appears 4 times, the most. The center square participates in a row, a column, and both diagonals (4 sums total), so 5 goes in the center.

2, 4, 6, 8 each appear 3 times. Corner squares participate in a row, a column, and one diagonal (3 sums), so the even numbers go in the corners.

1, 3, 7, 9 appear only twice, they go in the edges.

With 5 in the center, placing 2 in a corner forces the opposite corner to be 8 (since $2 + 5 + 8 = 15$). Placing 3 in an adjacent corner forces 7 on the other side. The remaining numbers fall into place:

You can verify: every row, column, and diagonal sums to 15. This isn't the only magic square of this size and given numbers. How might you make another?