I have a math question that goes like this: There are 500 light that have on and off switches. One person goes and turns all the lights on. The second person switches the lights from 2,4,6,8...and so on. THe 3rd person switches 3,6,9,12....and so on. This goes on for 500 people. How many lights will be left on in the end.
I do not even know how to start this problem please someone help
Try to find a pattern with even-numbered people and odd-numbered people.
I think that it has something to do with series but Im not sure
I tried to find a pattern and can't
You just look at a particular light nr. n and see how many times is is swithced on and off. Now every number has a factorization in terms of prime numbers. We can formally denote the prime numbers by:
p1, p2, p3 etc, where
p1 = 2
p2 = 3
p4 = 5 etc.
Any number n is of the form:
n = p1^k1 * p2^k2 * p3^k3 * ...
e.g. 14 = 2*7 =
2^1 * 3^0 * 5^0 * 7^1 * 11^0*... =
p1^1* p2^0 * p3^0 * p4^1 * p5^0*...
What happens to light nr. n of the form
n = p1^k1 * p2^k2 * p3^k3 * ...
The only persons that can affect light nr. n are persons that have a number r such that:
r = p1^x1 * p2^x2 * p3^x3 * ...
such that xi <= ki
for all i.
because r must be a divisor of n.
How many values for r are there? Each xi has to be in the range from zero to ki, so you have:
(k1+1)*(k2+1)*(k3+1)*...
on and off switchings.
Now you know that a product of numbers is even if one or more factors in the product are even. The only way a product can be odd is if all the factors are odd. Now, you need to switch the light on and off an odd number of times to get it switched on.
This means that all the ki+1 must be odd for the lights that are left on in the end. So, the ki must be even. The prime factorization for the numbers of the lights that are left on thus contain only even powers of prime numbers, which means that they are squares.
So, al the squares (including 1) will be left on. Since sqrt[500] = 22.36 there are 22 lights that are left on:
1^2, 2^2 = 4, 3^2=9,...,22^2= 484
maybe you should just give up? or ask your teacher? or just miss it out
If you are having problems than you should stay after school and ask your teacher for help. Or you can go before school starts. But if that doesn't work you can ask you teacher to give you extra help! Hope this works! Good Luck!!!!!!
To solve this problem, we need to find a pattern in the switches made by each person and determine which lights are left on in the end.
Let's start by looking at the switches made by the first person. They turn all the lights on, so all lights are initially on.
Next, let's analyze the switches made by the second person. They switch the lights at positions 2, 4, 6, 8, and so on. This means they will turn off all the lights that are at even positions.
Now, let's move on to the switches made by the third person. They switch the lights at positions 3, 6, 9, 12, and so on. Notice that the lights at positions 6, 12, 18, and so on have already been switched by the second person, so the third person will reverse the state of these lights back to on.
We can continue this analysis for each subsequent person. The lights at positions that are divisible by their respective person number will be switched. For example, the fourth person will switch the lights at positions 4, 8, 12, and so on.
Now, let's consider a general position n. How many times will it be switched? Well, n will be switched by every person whose number is a divisor of n. For example, for position 12, it will be switched by persons 1, 2, 3, 4, 6, and 12.
To determine whether the light at position n will be left on or off, we need to count how many times it is switched. If the light is switched an odd number of times, it will end up on. If it is switched an even number of times, it will end up off.
To find the lights that are left on in the end, we need to identify the positions that are switched an odd number of times. This occurs when the position has an odd number of divisors.
Position n has divisors that can be represented in the form of p1^k1 * p2^k2 * p3^k3 *..., where p1, p2, p3, ... are prime numbers. In order for n to have an odd number of divisors, each exponent k must be even.
Therefore, the positions that are left on in the end correspond to perfect squares because their exponents are all even powers.
The largest perfect square less than or equal to 500 is 22^2 = 484. Hence, there will be 22 lights left on in the end.
Therefore, the answer to the question is 22 lights will be left on in the end.
If you are still struggling with this problem, I would recommend seeking additional help from your teacher or a tutor. They can provide further explanations and guide you through the process of solving this type of math problem.