In the sequence 0 5 10 15 20 25 30 35 the third odd digit appears in the number 15 the fifth odd digit in the number 25 the seventh in 35 and so on find the 2009 odd digit and the number in which it?

Answer 1

There is actually a predictable pattern that the odd digits follow. Take a look at the first four numbers:

0, 5, 10, 15

The first number has no odds, the second and third both have one, and the fourth has two. This gives a total of four odds. Now, take a look at the next four numbers:

20, 25, 30, 35

Once again, we have no odds in the first number, one odd in the next two numbers, and two odds in the fourth. Once again, this pattern shows up, giving us four odds.

So, you may be tempted to think that this pattern continues indefinitely, giving us four odd digits for every four numbers in the sequence, but this changes once we reach the hundreds. Notice that when we reach the hundreds, we have:

100, 105, 110, 115

The pattern is similar, but each number in the sequence gives us an extra odd number, that 1 in the hundreds place. So now, we have 8 odds for every 4 numbers. This changes again at the two hundreds:

200, 205, 210, 215

Since 2 is even, we don't count it, which means we are back to our original pattern of 4 odds per four numbers. It's easy to see that we are going to alternate between these two patterns all the way up to 1,000, since the hundreds place will keep alternating between even and odd digits. Now, since there are going to be exactly 20 numbers in the pattern from 0 to 95, and 20 more from 100 to 195, and so on, and 20/4 = 5, we will have 5x4 = 20 odd digits for each of the following ranges: 0-95, 200-295, 400-495, 600-695, and 800-895, and 5x8 = 40 for the five remaining ranges. So, from 0 to 995, we are going to have a total of

5x20 + 5x40 = 300 odds

Once we reach the 1,000's, the pattern continues, but with one difference: the 1 in the thousands place gives us an extra odd for each number in the sequence. It's easy to see that this is the only part of the pattern that changes. So, to figure out how many odds there are between 1,000 and 1,995, we first note that since we are counting by 5s, there are 1,000/5 = 200 numbers in this part of the sequence. So, there will be 200 additional odd digits between 1,000 and 1,995. That leaves us with

300 + 200 = 500 odds.

Altogether, that gives us 300 + 500 = 800 odds from 0 to 1,995.

Once we reach the 2,000's, we no longer have an odd thousands digit, so we don't have to count that any more. That means in the range of the 2,000's, we are back to having 300 odd digits. Then, since 3 is odd, we'll have another 500 odd digits in the range of 3,000's (once again, we have this alternating pattern). So, that gives us another 800 odds, which means between 0 and 3,995, we have 800 + 800 = 1,600 odds.

In the 4,000's, we'll have 300 odds, which brings our total up to 1,900 odd digits all together. Now that we are getting close to the 2,009th digit, we'll have to slow down, or else we'll pass it up. From 5,000 to 5,095, there will be 40 odd digits, bringing the total to 1,940. From 5,100 to 5,195, there are 60 odds, bringing our total to 2,000.

We are now close enough to just count our way to the 2,009th digit. Our answer will be the 9th digit to show up in the following part of the sequence:

5200, 5205, 5210, 5215, 5220, ...

Our answer turns out to be 5220, with the 5 being the 2009th odd digit.

Edit:

Notice that even though Tommy's answer was incorrect, it shouldn't have been ruled out simply because it isn't an odd number. The question clearly states that we are looking for the 2009th odd digit. It says nothing about whether or not this will appear in an odd number. As it turns out, the answer was and even number, but the odd digit we were looking for turned up in the thousands place