LCM Problem

[gcooper]

A 1890 feet bridge has a light every 30 feet, including two end sides. In order to save the electricity, the lights will be moved every 42 feet. How many lights will not be moved?

my way of solving it:

So, first, we find the distance at which the first light(that will be moved), is placed. Use LCM on 30 and 42, and you get 210.
Given that all lights are placed at a fixed distance from each other, and that every light that will be moved is at a fixed distance, we can say that each light that will be moved, is at a distance of 210, 420, 630 etc.

So, 1890 / 210 is the amount of lights that will be moved(9).

Now, the problem asks how many lights will NOT be moved, so we have to find the total:
1890 / 30 = 63 lights, add the 2 for the end points, 65 lights.

65 - 9 = 56 (lights that won't be moved).

But that, is apparently very wrong.

The book gives the answer 10, with a HINT: 1890 / LCM + 1

That is almost what I did for finding the lights that WILL BE moved(except for adding 1, since the end light won't be moved).

So, someone can explain this a little bit?

 

[stapel]

A 1890 feet bridge has a light every 30 feet, including two end sides. In order to save the electricity, the lights will be moved every 42 feet. How many lights will not be moved?

my way of solving it:

So, first, we find the distance at which the first light(that will be moved), is placed. Use LCM on 30 and 42, and you get 210.
Given that all lights are placed at a fixed distance from each other, and that every light that will be moved is at a fixed distance, we can say that each light that will be moved, is at a distance of 210, 420, 630 etc.

Try doing the actual computations, or drawing a picture. You should find that you have (1) worked with the lights at the ends, which cannot be moved and (2) have found information related to lights which will not be moved, as they fit either spacing scheme.

 

[HallsofIvy]

With a 30 foot separation, the distance from the end of the bridge to the "k"th light will be 30k. With a 42 foot separation, the distance from the end of the bridge to the "j"th light will be 42j. That is, the distance will be a multiple of 30 and a multiple of 42. Do you see why this is called a "LCM problem"?

 

[gcooper]

Wait just a sec, as english is my second language, maybe i got it wrong: In order to save the electricity, the lights will be moved every 42 feet. How many lights will not be moved?

Doesn't that mean that every light(if it exists) at a position which is a multiple of 42, will be eliminated?

Also, I'm not sure how to picture this: You should find that you have (1) worked with the lights at the ends

LCM of 30 and 42, doesn't that result in the the first light that will be "moved out" ?

 

[Ishuda]

Wait just a sec, as english is my second language, maybe i got it wrong: In order to save the electricity, the lights will be moved every 42 feet. How many lights will not be moved?

Doesn't that mean that every light(if it exists) at a position which is a multiple of 42, will be eliminated?

Also, I'm not sure how to picture this: You should find that you have (1) worked with the lights at the ends

LCM of 30 and 42, doesn't that result in the the first light that will be "moved out" ?

In the new placement you will have a light at
0 ft (the end light at one end), 42 ft, 84 ft, ..., 210 ft, 252 ft, ..., 420 ft, 462 ft, ...,1806 ft, 1848 ft, 1890 ft (the light at the other end)

Right now you have a light at
0 ft, 30 ft, 60 ft, ..., 210 ft, 240 ft, ..., 420 ft, 450 ft, ..., 1830 ft, 1860 ft, 1890 ft.

Which ones will stay in place? Which ones will you have to move?

 

[gcooper]

Right, so all lights are going to be moved, except those who are at the desired position already. To find them, use the LCM on the current distance & desired distance, and you get the distance at which the first light that won't be moved exists. Divide the main distance by that, you get the lights that won't be moved + add the one light at the end = 10.