Wanna dare do this?

It's obvious that the last digit is $5$ else it wouldn't be divisible $5$ or $35$ . It's just left to check how do we arrange the remaining 5 numbers such that it is also divisible by 7.

Of course, the most no-brainer way to is fill in $\overline{ABCDE5}$ with $A,B,C,D,E$ as $5$ or $7$ only. Which is a total of $2^5 = 32$ ways. But that gets a little bit tedious.

We can apply one of divisibility rule of $7$ : "Form the alternating sum of blocks of three from right to left".

We must find $\bigg ( \overline{DE5} - \overline{ABC} \bigg) \bmod 7 = 0$ . Or $\bigg ( (100D + 10E + 5) - (100A+ 10B + C) \bigg) \\ \equiv 2(A-D) + 3(E-B) + (5-C)$ .

Because $A,B,C,D,E$ are only limited to $5,7$ . the possible values are restricted to:

$A - D = 0,2,-2 \\ E-B = 0,2,-2 \\ 5-C = 0, -2 \\$

Or

$2(A-D) = 0,4,-4 \\ 3(E-B) = 0, 6, -6 \\ 5-C = 0, -2 \\$

Taking $S_1, S_2, S_3$ as the possible values of $2(A-D),3(E-B), 5-C$ respectively. We want to match up values of $S_1 + S_2 + S_3$ such that it is a multiple of $7$ .

A simple case by case shows that $(S_1,S_2,S_3) = (0,0,0) , (-4,6,-2)$ only.

Case 1 : If $(S_1,S_2,S_3) = (0,0,0)$ . Then $A=D$ , $B=E$ , $C = 5$ . Then, which means there are $2 \times 2 \times 1 = 4$ solution, namely: $555555,575575,755755,775775$ .

Case 2 : If $(S_1,S_2,S_3) = (-4,6,-2)$ . Then $A =7, D = 5$ , $E = 7, B =5$ , $C = 7$ , the number is only $757575$ .

Adding all these numbers up gives: $(555555 \times 5) + (20020 + 200200 +220220 + 202020) = \boxed{3420235}$ .

The possible numbers are

• 575575
• 757575
• 755755
• 775775
• 555555

Anyone like to prove?

Thanks to one who reported the question for "5 and 7" which is corrected now to "5 or 7"