1st number = 1000a+100b+10c+d; 2nd number = 1000d+100c+10b+a, reverse of 1st number.
1000a+100b+10c+d=4(1000d+100c+10b+a); 3999d+390c=60b+996a; 1333d+130c=20b+332a.
Units: (a,d)=(1,4), (2,8), (3,2), (4,6), (6,4), (7,8), (8,2), (9,6) so d must be one of 2, 4, 6, 8, and d≠0 and a≠5.
Which pair could satisfy 1333d+130c=20b+332a? Clearly a>3 (332a>996).
a d 332a 1333d 1333d-332a (=20b-130c=10(2b-13c))
4 6 1328 7998 6670
6 4 1992 5332 3340
7 8 2324 10664 8340
8 2 2656 2666 10
9 6 2988 7998 5010
From this table, 10(2b-13c)=10 is satisfied when b=7 and c=1, so a=8 and d=2.
The numbers are 8712 and 2178, so the first number is 8712.