0
$\begingroup$

I know the first term is 5 It's continues till 9 And then there's a break in the series and then resumes at 50 and till 90. But can I get a more formal way?

  • 1
    So, $5-9$ work, then $50-99$, then $500-999$...do you see a pattern? – 2017-01-05
  • 0
    Yeah I know of this method....Wanted a better approach...Well if this is the on!Y one then ok.well thank u – 2017-01-05

1 Answers 1

1

How many numbers with exactly $n$ digits that satisfy the property are there?

$$(10^n-1)-(10^n/2-1)=10^n/2$$

How many numbers with at most $n$ digits that satisfy the property are there?

$$f(n)=\sum_{k=1}^n\frac{10^n}2=\frac{10^{n+1}-10}{18}=\frac59\cdot \underbrace{99\ldots9}_{n\text{ nines}}=\underbrace{55\ldots5}_{n\text{ fives}}$$

Since $f(3)=555$ and $f(4)=5555$, the last of those $2017$ numbers has four digits, so it is $$4999+2017-555=6461$$

  • 0
    Ummm....I did not understand that last step – 2017-01-05
  • 0
    Please tell me! – 2017-01-05
  • 0
    The $556$th number is $5000=4999+556-555$. The $557$th number is $5001=4999+557-555$, etc. – 2017-01-05
  • 0
    How can u tell that the 556th no. Is the given expression...ik it's a dumb question but please tell – 2017-01-05
  • 0
    And how did u get that first expression?Guess or theorem? – 2017-01-05
  • 0
    Follow it through for three digits. The lowest three digit number that carries is $500$, the last is $999$. Those are the two terms on the left. That accounts for $999-500+1=500$ numbers. – 2017-01-05