SOLUTION: Reading from left to right, a sequence consists of 6 X’s, followed by 24 Y’s, followed
by 96 X’s. After the first n letters, reading from left to right, one letter has occur
Algebra.Com
Question 1136994: Reading from left to right, a sequence consists of 6 X’s, followed by 24 Y’s, followed
by 96 X’s. After the first n letters, reading from left to right, one letter has occurred
twice as many times as the other letter. The sum of the four possible values of n is?
Answer by greenestamps(13214) (Show Source): You can put this solution on YOUR website!
Figure it out on your own -- it's good exercise for the brain:
(1) The first n is when the first 6 X's are twice the total number of Y's.
(2) The second n is when the first 6 X's are half as many as the number of Y's.
(3) The third n is when all 24 Y's are twice the number of X's.
(4) The fourth n is when all 24 Y's are half the total number of X's.
Add the four values of n....
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The thought process, in more detail....
Think about writing out the string of letters, one letter at a time. You are looking for times when the number of X's is twice the number of Y's, or the number of Y's is twice the number of X's.
The first 6 letters are all X's; after that you write a bunch of Y's.
How many Y's do you write before the number of X's is twice the number of Y's?
There are 6 X's, so you need 3 Y's.
So when the total number of letters in the string is 6+3=9, there are twice as many X's as Y's. So the first value of n you are looking for is 9.
Now you continue writing more Y's. The number of X's stays at 6, while the number of Y's increases.
At some point, the number of Y's will be twice the number of X's.
When will that happen? There are 6 X's at the beginning of the string; when the number of Y's reaches 12, there will be twice as many Y's as X's.
So when the total number of letters in the string is 6+12=18, the number of Y's will be twice the number of X's. So the second value of n you are looking for is 6+12 = 18.
Continue adding letters to figure out the last two values of n.
After the first 6 X's in the string, there are 24Y's; then you start writing more X's, with the number of Y's staying at 24.
As you write more X's with the number of Y's staying at 24, you will first reach a point where the number of Y's (24) is twice the total number of X's; then, as you write still more X's, you will reach a point where the number of X's is twice the number of Y's (which is still 24).
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