Puzzles

Polya strikes out

Source: Thinking Mathematically by John Mason, Leone Burton & Kaye Stacey

Write the numbers 1, 2, 3, ... in a row. Strike out every third number beginning with the third. Write down the cumulative sums of what remains:

1, 2, 3, 4, 5, 6, 7, ...

1, 2, 4, 5, 7, ...

1=1; 1+2=3; 1+2+4=7; 1+2+4+5=12; 1+2+4+5+7=19; ...

1, 3, 7, 12, 19, ...

Now strike out every second number beginning with the second. Write down the cumulative sums of what remains. What is the final sequence? Why do you get this sequence?

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1, 3, 7, 12, 19, ...

1, 7, 19, ...

1=1; 1+7=8; 1+7+19=27; ...

1, 8, 27, ...

The final sequence is the cube numbers. To show why, let

n

be an integer and follow through the process.

Cross out every third number:

1, 2, 3, 4, 5, 6, ..., 3n,

3 n + 1

3 n + 2

, ...

1, 2, 4, 5, ...,

3 n + 1

3 n + 2

, ...

Find the cumulative sums:

1 = 1

1 + 2 = 1 + 2 = 3

1 + 2 + 4 = 1 + 2 + 3 + 4 - 3 = 7

1 + 2 + 4 + 5 = 1 + 2 + 3 + 4 + 5 - 3 = 12

. . .

1 + 2 + 4 + 5 + . . . + (3 n + 1) = \sum_{i = 1}^{3 n + 1} - \sum_{i = 1}^{n} 3 i

= \frac{1}{2} (3 n + 1) (3 n + 2) - \frac{3}{2} n (n + 1)

= 3 n^{2} + 3 n + 1

1 + 2 + 4 + 5 + . . . + (3 n + 2) = 3 n^{2} + 3 n + 1 + (3 n + 2)

= 3 n^{2} + 6 n + 3

. . .

1, 3, 7, 12, ...,

3 n^{2} + 3 n + 1

3 n^{2} + 6 n + 3

, ...

Cross out every second number, starting with the second:

1, 3, 7, 12, ...,

3 n^{2} + 3 n + 1

, ~~3n2+6n+3~~, ...

1, 7, ...,

3 n^{2} + 3 n + 1

, ...

Find the cumulative sums. The

m

^th sum is:

\sum_{n = 0}^{m} 3 n^{2} + 3 n + 1

= 3 \sum_{n = 0}^{m} n^{2} + 3 \sum_{n = 0}^{m} n + \sum_{n = 0}^{m} 1

= \frac{3}{6} m (m + 1) (2 m + 1) + \frac{3}{2} m (m + 1) + m + 1

= \frac{1}{2} (m + 1) (m (2 m + 1) + 3 m + 2)

= \frac{1}{2} (m + 1) (2 m^{2} + m + 3 m + 2)

= \frac{1}{2} (m + 1) (2 m^{2} + 4 m + 2)

= (m + 1) (m^{2} + 2 m + 1)

= (m + 1) (m + 1)^{2}

= (m + 1)^{3}

Hence the numbers obtained are the cube numbers.

Extension

What happens if you cross out every third number starting at the second? Or every fifth number starting at the fifth? Or every

n

^th number starting at the

m

^th?

Tags: numbers

If you enjoyed this puzzle, check out Sunday Afternoon Maths XXV,
puzzles about numbers, or a random puzzle.

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Polya strikes out

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