Pattern problem

[arman]

Hello,I have question.

From 1 to 9,each number are written as many as 3 times their value,and there we have the number X.

The first 10 digits of X:1112222223

What is the number on the 72nd digit of X?

So here's what I thought:

3 + 6 + 9 + 12 and so on...
111 222222 333333333

I thought maybe we could go to the solution by 3(1+2+3...)=72 that is 3(n.n+1)/2=72 so n.n+1=48 but that didn't work.

Then I developed something like this:

111
the last number of this group is the 3rd of the whole number


222222
The last number of this group is the 9th of the whole number

333333333
the last of this group is 18th of the whole number




Well,I realized that what I did for finding on which digit the last number of a group is was

For 1st group
1x3=3

For second
3x1+2x3 =9


For third
3x1+2x3+3x3=18



the 4th would be 18+4x3=30

The 5th would be 30+5x3=45


The 6th 45+6x3=72



This way I found that the last number of the 6th group,which is 6,is the 72nd of the number X.However the answer is 7 according to the answer key.

 

[Deleted member 4993]

Hello,I have question.

From 1 to 9,each number are written as many as 3 times their value,and there we have the number X.

The first 10 digits of X:1112222223

What is the number on the 72th digit of X?

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/for

 

[arman]

Please share your work with us ...even if you know it is wrong.

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/for

Thanks,edited.

 

[Dr.Peterson]

Hello,I have question.

From 1 to 9,each number are written as many as 3 times their value,and there we have the number X.

The first 10 digits of X:1112222223

What is the number on the 72nd digit of X?

So here's what I thought:

3 + 6 + 9 + 12 and so on...
111 222222 333333333

I thought maybe we could go to the solution by 3(1+2+3...)=72 that is 3(n.n+1)/2=72 so n.n+1=48 but that didn't work.

Then I developed something like this:

111
the last number of this group is the 3rd of the whole number

222222
The last number of this group is the 9th of the whole number

333333333
the last of this group is 18th of the whole number

Well,I realized that what I did for finding on which digit the last number of a group is was

For 1st group
1x3=3

For second
3x1+2x3 =9

For third
3x1+2x3+3x3=18

the 4th would be 18+4x3=30

The 5th would be 30+5x3=45

The 6th 45+6x3=72

This way I found that the last number of the 6th group,which is 6,is the 72nd of the number X.However the answer is 7 according to the answer key.

Thanks for showing your work -- though it always confuses me when someone goes back and edits something that has been replied to.

Let's check your answer! The first 7 will follow 3+6+9+12+15+18 digits (1's through 6's), which is 63; so the 72nd digit will be a 7.

Now look back through your work to find where you went wrong. (It's just a silly arithmetic error, I think.)

Your first attempt should have worked, except that you can't expect the sum to exactly equal 72. You can solve it as a quadratic inequality. (We can discuss how if you'd like.) But the other method is really no slower, and just as correct if you do it carefully ...

The main lesson here is to check!

 

[arman]

Thanks for showing your work -- though it always confuses me when someone goes back and edits something that has been replied to.

Let's check your answer! The first 7 will follow 3+6+9+12+15+18 digits (1's through 6's), which is 63; so the 72nd digit will be a 7.

Now look back through your work to find where you went wrong. (It's just a silly arithmetic error, I think.)

Your first attempt should have worked, except that you can't expect the sum to exactly equal 72. You can solve it as a quadratic inequality. (We can discuss how if you'd like.) But the other method is really no slower, and just as correct if you do it carefully ...

The main lesson here is to check!

Thanks for your help it really helped.I feel embarrassed because 45+18=72 this is totally silly.

 

[Dr.Peterson]

I thought maybe we could go to the solution by 3(1+2+3...)=72 that is 3(n.n+1)/2=72 so n.n+1=48 but that didn't work.

You didn't ask, but let's look at this method.

First, we need to write it correctly: 3(n(n+1)/2) = 72, which simplifies to n(n+1) = 48.

We can solve this by the quadratic formula (or by completing the square): n^2 + n - 48 = 0, so n = [-1 ± √(1 - -192)]/2 = [-1 ± √193]/2 ≈ 6.45. This tells us that we have used up through the 6's and are into the 7's, so the answer is 7.

That took more work than just adding up the sequence until we passed 72, but was more impressive, I think.

We could also have used your 3(1+2+3...), by adding 1+2+3+4+5+6 = 21 and seeing that the next term would pass 72/3 = 24.