### [mathematics] From another point of view...

Assumed knowledge: minimal

Difficulty: 2/5

Tedium: 1/5

Insight: 3/5

The problem presented below is used to highlight an elegant approach to solving a particularly nasty mathematical problem.

a + b + c =100; a, b & c are natural numbers (integers greater than zero)

How many possible solutions are there?

To answer the question, we can try counting the number of solutions.

Fix a = 1:

1 + 1 + 98 = 100

1 + 2 + 97 = 100

1 + 3 + 96 = 100

:

:

1 + 98 + 1 = 100

For a = 1, there are 98 possible solutions. What about for a = 2?

Fix a = 2:

2 + 1 + 97 = 100

2 + 2 + 96 = 100

2 + 3 + 95 = 100

:

:

2 + 97 + 1 = 100

For a = 2, there are 97 possible solutions. What about for a = 3, 4, 5, 6 ... 98?

Without crunching numbers for all possible values of a, the semi-enlightened soul might notice that the number of solutions is

98 + 97 + 96 + … + 2 + 1 = 4851.

There are 4851 solutions.

There is an easier approach.

Consider 100 dots arranged in a line:

o o o o o o o o o o o o … o o

1 + 1 + 98 = 100 (a = 1; b = 1; c = 98) can be represented by grouping the dots into groups of 1, 1 and 98, where the first group is for a, the second for b, and the third for c. The grouping will be done by partitioning the dots:

o | o | o o o o o o o o o o … o o

Similarly, 5 + 2 + 93 = 100 (a = 5; b = 2; c = 93) can be represented by grouping the dots into groups of 5, 1 and 94 as shown:

o o o o o | o o | o o o o o … o o

The problem is now reduced to counting how many ways the 2 partitions can be inserted into the 99 gaps between our hundred dots.

The answer is simply 99 C 2 (read “99 choose 2”) = 4851

and

The volume of a circular pizza of thickness a and radius z is simply pizza.

Difficulty: 2/5

Tedium: 1/5

Insight: 3/5

__Content adapted from “The Beauty and Zen of Mathematics” seminar by the Melbourne University Mathematics and Statistics Society (MUMS).__The problem presented below is used to highlight an elegant approach to solving a particularly nasty mathematical problem.

a + b + c =100; a, b & c are natural numbers (integers greater than zero)

How many possible solutions are there?

To answer the question, we can try counting the number of solutions.

Fix a = 1:

1 + 1 + 98 = 100

1 + 2 + 97 = 100

1 + 3 + 96 = 100

:

:

1 + 98 + 1 = 100

For a = 1, there are 98 possible solutions. What about for a = 2?

Fix a = 2:

2 + 1 + 97 = 100

2 + 2 + 96 = 100

2 + 3 + 95 = 100

:

:

2 + 97 + 1 = 100

For a = 2, there are 97 possible solutions. What about for a = 3, 4, 5, 6 ... 98?

Without crunching numbers for all possible values of a, the semi-enlightened soul might notice that the number of solutions is

98 + 97 + 96 + … + 2 + 1 = 4851.

There are 4851 solutions.

There is an easier approach.

Consider 100 dots arranged in a line:

o o o o o o o o o o o o … o o

-The ellipse shows that the series has been truncated to save on space

1 + 1 + 98 = 100 (a = 1; b = 1; c = 98) can be represented by grouping the dots into groups of 1, 1 and 98, where the first group is for a, the second for b, and the third for c. The grouping will be done by partitioning the dots:

o | o | o o o o o o o o o o … o o

Similarly, 5 + 2 + 93 = 100 (a = 5; b = 2; c = 93) can be represented by grouping the dots into groups of 5, 1 and 94 as shown:

o o o o o | o o | o o o o o … o o

The problem is now reduced to counting how many ways the 2 partitions can be inserted into the 99 gaps between our hundred dots.

The answer is simply 99 C 2 (read “99 choose 2”) = 4851

__Finally, some weird word games from the MUMS magazine:__and

The volume of a circular pizza of thickness a and radius z is simply pizza.

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