how many 5-digit numbers satisfy the following conditions How many five-digit numbers divisible by 11 have the sum of their digits equal to 30?
I am able to get the 5-digit numbers divisible by 11
and
I am also able to get the five-digit numbers whose sum of their digits equal to 30.
But i am not able to get how i can get the count of 5 digit numbers satisying both the condition.
Thanks in advance.
Thanks in advance.
combinatorics permutations

A number is divisible by $11$ if the sum of the digits in the tens and thousands place minus the sum of the digits in the ones, hundreds, and ten thousands place is divisible by 11.
So take a number:
$a b c d e$
$(b+d)-(a+c+e)$ is divisible by $11$ (it may be $0$).
$a+b+c+d+e=30$.
$a,b,c,d,e$ are all integers that may be $0$ through $9$.  $a\not=0$.
Combining the two equations above, $2(b+d)-30$ is divisible by $11$, but is also even (since $b$ and $d$ are integers), so it may be $0$, $22$, or $-22$.
If it is $22$, then $(b+d)=26$, which is a contradiction.
If it is $0$, then $(b+d)=(a+c+e)=15$ (condition 1)
If it is $-22$, then $(b+d)=4$ and $(a+c+e)=26$ (condition 2)
Take all such $a,b,c,d,e$ that satisfy the above statements.
If condition 1 holds, then there are $4$ different ways to choose $b$ and $d$ and $4+5+6+7+8+9+9+8+7+6=69$ ways to choose $a,c,e$ that satisfy $a+c+e=15$.  Thus, for condition 1, there are $276$ distinct numbers that satisfy condition 1.
If condition 2 holds, then there are $4$ different ways to choose $b$ and $d$ and $3$ ways to choose $a,c,e$.  So there are $12$ distinct numbers that satisfy condition 2.
Thus, there are $288$ distinct 5-digit numbers that satisfy the desired condition.
