7 divides a*10^3+bcd which implies 7 divides bcd - a.... also 7 divides 3*bcd+a Therefore 7 divides bcd therefore 7 divides a.. So a is 7. a + bcd is congruent to 10bcd + a mod 37. Hence 37 divides bcd.. So the possible nos are:- 7259 7518 7777

That is unfortunately total garbage. Edited for some omissions. We have $\overline{abcd} = 10^3 a + 10^2 b + 10 c + d \equiv a - 11b + 10 c + d \pmod{37}$, so we need $a - 11b + 10 c + d \equiv d - 11c + 10 b + a \pmod{37}$, i.e. $21(b-c) \equiv 0 \pmod{37}$, thus $b=c$. On the other hand, we have $\overline{abcd} = 10^3 a + 10^2 b + 10 c + d \equiv -a + 2b + 3 c + d \pmod{7}$, so we need $5(b+c) \equiv 0 \pmod{7}$, thus $b=c=7$ or $b=c=0$. Then we need $a \equiv d \pmod{7}$, hence $a=d$, or $\{a,d\} = \{1,8\}$, or $\{a,d\} = \{2,9\}$. Thus the numbers are $1771,2772,\ldots,9779$, $1001,2002,\ldots,9009$,

and

Oh so sorry the solution was really garbage... I misread the question I read dcba as bcda.. Really my eyes fool me these days.. Sorry.

mavropnevma wrote: Thus the numbers are $1771,2772,\ldots,9779$. And also $\overline{n00n},\;1\leq n\leq 9$ since $x\equiv0\pmod 7$ for $0\leq x\leq 9$ means either $x=7$ or $x=0$.

2009 might be an answer they were expecting