SOLUTION: A centurion in the ancient roman army commanded his 100 troops to stand facing him in two groups: 50 men to his left and the other 50 to his right, numbered 1 to 100, from left to

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Question 1199748: A centurion in the ancient roman army commanded his 100 troops to stand facing him in two groups: 50 men to his left and the other 50 to his right, numbered 1 to 100, from left to right. To choose a few men for an operation he told his troops to face backwards. Then every man with an even number was to face the front. Next, every man with a number divisible by 3 was to change his orientation (front to back or back to front). Next, every man with a number divisible by 4 was to change his orientation, and so on up to every man with a number divisible by 50 was to change his orientation. Finally, he told the 50 men on the right to all change their orientation. Those facing backwards were to go on the mission. How many was that?
Found 2 solutions by math_tutor2020, greenestamps:
Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Answer: 54 soldiers

Explanation:

Let's look at the numbers from 1 to 10. Specifically we'll look at the factors.
NumberFactors
11
21,2
31,3
41,2,4
51,5
61,2,3,6
71,7
81,2,4,8
91,3,9
101,2,5,10
Now what do you notice?
If you were to look at it long enough, you might notice that the perfect squares (1,4,9) have an odd number of factors.
If we ignore "1" from each factor list, then we drop to an even number of factors for each perfect square.

Having an even number of factors means the soldier will end up facing backward since they started in that direction.
2 flips: backward --> forward --> backward
4 flips: backward --> forward --> backward --> forward --> backward
and so on. Each arrow symbol represents a flip.

Something like soldier #6 will have an odd number of factors when ignoring the factor "1". Those factors being {2,3,6}; we can see that soldier #6 will do an odd number of flips, and therefore end up facing forward.

The perfect squares from 1 to 100 are:
1,4,9,16,25,36,49,64,81,100
These soldiers will be the only ones facing backward when following the instructions mentioned, and this is before the "he told the 50 men on the right to all change their orientation" portion is carried out.

The perfect squares between 51 and 100 are:
64, 81, 100
There are 3 such values. These 3 soldiers will turn to face forward after the commander instructs his 50 right-hand-side troops to flip direction. The remaining 50-3 = 47 will turn to face backward.

We have 7 perfect squares on the left that face backward, and 47 non-perfect squares on the right facing backward as well.
7+47 = 54 soldiers total will face backward.
This is the final answer.

A similar concept is discussed in these links
https://www.youtube.com/watch?v=-UBDRX6bk-A
https://math.hmc.edu/funfacts/toggling-light-switches/
Answer by greenestamps(13195)   (Show Source): You can put this solution on YOUR website!


At the start, all 100 men are facing backwards.

The men are then told to change their orientation each time their number is divisible by a number from 2 to 50. So the number of times each man changes orientation is the number of positive divisors greater than 1 that his number has.

An integer has an odd number of positive divisors if and only if it is a perfect square; all other integers have an even number of positive divisors.

However, 1 is a positive divisor of every integer; and the directions for changing orientation only start with being divisible by 2. So, before the final instruction (for all of the men in the second 50 to change orientation), the men whose numbers are perfect squares are the ones that are facing backwards.

In the group of men on the left, numbered 1 to 50, there are 7 men with numbers that are perfect squares: 1, 4, 9, 16, 25, 36, and 49. So among the first 50 men, 43 finish facing forwards and 7 finish facing backwards.

Since the men that go on the mission are the ones who finish facing backwards, only 7 men from the first 50 go on the mission.

In the group of men on the right, numbered 51 to 100, there are 3 with numbers that are perfect squares: 64, 81, and 100. So among the second group of 50 men, 47 end up facing forwards and 3 end up facing backwards. But then every one of those 50 men is instructed to change orientation one more time, leaving only 3 facing forwards and 47 facing backwards.

ANSWER: The number of men facing backwards at the end, and therefore going on the mission, was 7 + 47 = 54.
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