MHB Finding the Limit of Digit Sums in A

[Albert1]

digits sum

$A\in ${1,2,3,4,----2013}
a number p is picking randomly from A
if $q_1$=the digits sum of p
if $q_2$=the digits sum of $q_1$
------
continue this procedure until the digits sum=1 then stop
How many numbers we can pick from A ,and meet the requirement ?

 

[Albert1]

Re: digits sum

$A\in ${1,2,3,4,----2013}
a number p is picking randomly from A
if $q_1$=the digits sum of p
if $q_2$=the digits sum of $q_1$
------
continue this procedure until the digits sum=1 then stop
How many numbers we can pick from A ,and meet the requirement ?

hint:

Spoiler

p can be:
19,28,37,460,1900 ---etc.
all meet the requirement
of course there are more ---

 

[Albert1]

try to find it logically ,don't use any pc code

there are 224 numbers in the set:

{1,2,3,4-------2013}

now prove or find it .

 

[Opalg]

Re: digits sum

$A\in ${1,2,3,4,----2013}
a number p is picking randomly from A
if $q_1$=the digits sum of p
if $q_2$=the digits sum of $q_1$
------
continue this procedure until the digits sum=1 then stop
How many numbers we can pick from A ,and meet the requirement ?

[sp]Digital sum preserves residue class mod 9. So this set is just the set of all numbers between 1 and 2013 that are 1 more than a multiple of 9. Since $2013 \div 9 = 223\!\frac23$, there are $224$ such numbers.[/sp]

 

[Albert1]

my solution:

Spoiler

in fact this is an AP with :
$a_1=1,a_n=2008,\,\,$ and , $d=9$
from :$a_n=a_1+(n-1)d$
we have:$2008=1+(n-1)\times 9$
$2016=9n$
$\therefore n=224$