1. Divisibility rule for 11

For a 7-digit number with digits $d_1d_2d_3d_4d_5d_6d_7$, the number is divisible by $11$ if

is a multiple of $11$.

Here the digits used are exactly $1,2,3,4,5,7,9$.

Their total sum is

Let

• sum of digits in odd places $= S_o$,
• sum of digits in even places $= S_e$.

Then

Also, divisibility by $11$ requires

Since $S_o-S_e$ must have same parity as $31$ (odd), it must be odd. Also its value lies between $-(2+4+5)$ etc., so possible multiples of $11$ are only $\pm 11$.

Thus either

2. Solve for the required sums

From

we get:

• If $S_o-S_e=11$, then

  $$2S_o=42 \Rightarrow S_o=21,$$

  so $S_e=10$.

• If $S_o-S_e=-11$, then

  $$2S_o=20 \Rightarrow S_o=10,$$

  so $S_e=21$.

So the 4 digits in odd positions must sum to $21$ or $10$.

3. Find subsets of 4 digits with required sum

We need 4 digits chosen from $\{1,2,3,4,5,7,9\}$.

Case A: odd-position digits sum to $21$

We need 4 digits summing to $21$.

Check combinations:

• $1+4+7+9=21$
• $2+3+7+9=21$
• $2+4+5+10$ not possible
• $3+4+5+9=21$

So the valid 4-digit sets are:

Hence there are $3$ such selections.

Case B: odd-position digits sum to $10$

We need 4 digits summing to $10$.

Only possible set is

Hence there is $1$ such selection.

So total valid choices for the digits in odd positions:

4. Arrange digits in positions

There are 4 odd positions: $1,3,5,7$ and 3 even positions: $2,4,6$.

For any valid selection of 4 digits for odd positions:

• arrange them in odd positions in $4!$ ways,
• arrange the remaining 3 digits in even positions in $3!$ ways.

Thus for each valid selection, number of numbers formed is

Since there are $4$ valid selections,

5. Final answer

6. Comparison with stored answer

Stored correct answer = $576$.

My derived answer also is $576$, so they agree.