Alan G. answered • 07/14/16
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William,
It isn't hard, but I do not know what you mean by the first year number. Do you mean the smallest?
Alan G.
Okay, thanks. I'll be sending you help later today or early tomorrow. Do you need the complete solution or just some good hints?
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07/14/16
William H.
Just a lot of good hints thanks!
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07/14/16
Alan G.
William,
Good for you!
Since the number 2016 begins with a 2 and also contains 1 and 0, the smaller numbers will start with 2.
There are six unused digits left over: 3, 4, 5, 7, 8, 9. These can occur in any sequence before or after the string of digits "2016."
If you start listing the first six numbers with a single digit appended, you will start with 20163 and end with 20169.
You can then list the next 36 numbers ending in two digits from the set {3, 4, 5, 7, 8, 9}. Now you have a total of 42 numbers in increasing order.
After creating strings of seven, eight, and more digits in the same way, you will be producing a lot more numbers.
The number of strings thus created is counted by the sum
6 + 62 + 63 + 64 + … + 6n,
where n is the number of additional digits appended after "2016."
Here is what you must do to finish this problem.
Find the smallest value of n for which this sum equals or exceeds 2016, then locate the number in the list with the n digits appended after 2016 which corresponds to the 2016th number in this list.
This is a little time consuming, but it is much faster than trying to list the numbers in order one at a time, and that is the wonderful power of using math to answer a question like this.
Let me know if you need more help, but you should be able to get this after my generous hints.
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07/15/16
Garima J.
Hi Alan,
I think you have missed all the numbers starting with x where x <> 2,0,1 or 6. These numbers will come after 20169 I.e. 32016, 42016..... Then there be 2
36 combinations ending in 2 digits, 36 more for '2016' year numbers starting with 2 digits, other 36 combinations with 1 digit in front and 1 in end. So what will be the final nth sequence?
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07/17/16
William H.
I managed to find the answer anyway. It wasn't to dissimilar, I just had to add a lot more values which took a bit more time.
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07/17/16
Alan G.
In response to Garima's comment. You are correct, but I was only providing hints and did not try to solve the problem for William. Hints are free to use or ignore according to their value to the problem solver.
I hope this did not cause anyone concern. Solving math problems is not usually a linear process. This is just a simple fact about solving problems in any subject, and life as well.
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07/17/16
William H.
07/14/16