I have $l$ and $r$ and I have to find the number of such integers $x$ $(l ≤ x ≤ r)$, that the first digit of integer $x$ equals the last one (in decimal notation). For example, such numbers as $101$, $477474$ or $9$ will be included in the answer and $47$, $253$ or $1020$ will not.
Examples:
#$1:$
$l = 2$ and $r = 47$
$answer = 12$
#$2:$
$l = 47$ and $r = 1024$
$answer = 98$
Sorry for my english. if I do any wrong , sorry and please edit it.
I find answer so:
if($first$ $number$ $of$ $n$ $>$ $last$ $number$ $of$ $n$) $k = 1$ else $k = 0$
$answer = n/10 + 9 - k$
answer = (answer for $r$) - (answer for ($l-1$))
I want to know proof of this formula
I saw these:
len -> length of n
| len | count of numbers that are x |
$| 1 | 9 |$
$| 2 | 9 |$
$| 3 | 90 |$
$| 4 | 900 |$
...
| $k$ |$9000...00$ (count of zeros is $k-2$)|
explanation of the above:
$len = 1:$
$1, 2, 3, 4, 5, 6, 7, 8, 9$
$ len = 2: $
$11, 22, 33, 44, 55, 66, 77, 88, 99$
...
answer is sum of all numbers(for len) to $n$. $->$ $9 + 9 + 90 + 900 + 9000 ...$
This is geometric sequence.