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Shamgar vs the Phillistines


Posted (edited)

Shamgar saved Israel by striking down 600 Philistines with an oxgoad.

Assume the Philistine army were marching downhill in symmetry as follows:

In The Leading Rows

* one soldier in the first row

* two soldiers in the second row

* each row having one extra soldier until row n

The Main Block

* x rows of n + 1 soldiers

The Trailing Rows

* the leading rows in reverse: n soldiers in the first row, n - 1 in the next, down to a final row of 1 (the soldier whom Shamgar strikes directly with his goad)

Thus, Shamgar struck the last man, who toppled the next row, who toppled the next row etc in a domino effect.

What numbers of leading/trailing rows are possible?

I'll give a couple of example possibilities:

1. zero: the main block would be 600 rows of 1 soldier

2. one: the main block would be 299 rows of 2 soldiers

3. two: the main block would be 198 rows of 3 soldiers

Edited by BMAD

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2 answers to this question

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3: 147 rows of 4

4: 116 rows of 5

5: 95 rows of 6

7: 68 rows of 7

9: 51 rows of 10

11: 39 rows of 12

14: 26 rows of 15

19: 11 rows of 20

23: 2 rows of 24

24: no main block


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Let t=trailer or header rows. Then we have

2*t(t+1)/2 + x(t+1) = 600

=> x = -t + 600/(t+1).

so t can only assume such values which make t+1 a factor of 600. Since 600 = 23*31*52, it has (3+1)*(1+1)*(2+1) = 24 factors.

As x>= 0, solving for t we get t<=24

using this upper bound for t, t+1 = 25. Since root of 600 is between 24 and 25 (576 and 625), 25 must be the 13th factor (as out of 24 factors, 12 must be above and 12 below the sqroot)

Hence 13 values are possible for t+1. Therefore 13 trailer/header rows are possible.


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