Seven lamps labeled A through G are arranged in a row. Each lamp has its own switch. Now lamps A, C, E, and G are on and other lamps are off. Ben starts to flip each switch from A to G the following way: if the lamp is on, he turns it off; if the lamp is off, he turns it on. He repeats the pattern until he flips the switches 2011 times. Which lamps are on finally? dont bully me im in 5th grade also this is due in 4 days

O=on

X=off

OXXOXOX  after 2011 flips.

Step-by-step explanation:

There are many ways to get the answer.

A. Actually flip them 2011 times.

One way is to put 7 cards on the table, and name them A to G, with A,E,C,G facing up, and the other facing down.

Now flip the cards in order A to G 2011 times and find the answer.  

Since you have four days to find the answer, you only have to flip about 500 times a day.

B. Simplify procedure A above.

Do the same as above, actually flip the cards, but only 14 times.

You will find that the cards are restored to the original position after 14 flips.  

If we divide 2011 by 14, we get a quotient of 143 with a remainder of 9.

We know we don't have to make 143 sets of 14 flips.  We only need to make 9!

Then again, we know that after 7 flips, the lights will be all opposite.

If O=on, and X=off then the situations are

OXOXOXO  initial

XOXOXOX  after 7 flips

OXXOXOX  after 9 flips, which is also after 2011 flips, and that is the final answer.

the answers c