find the total no. of +ve integerbetween 1000&9999 such that when its digits are reverse,it increases by 2088?

Answer 1

x = 1000a + 100b + 10c + d.
Reverse digits:
1000d + 100c + 10b + a = x + 2088
→ 1000d + 100c + 10b + a = 1000a + 100b + 10c + d + 2088
→ 999d + 90c - 90b - 999a = 2088
→ 111(d-a) + 10(c-b) = 232
→ d-a = 2 and c-b = 1
So a (leading digit) can vary from 1 to 7 (7 possibilities)
And b can vary from 0 to 8 (9 possibilities)
So we have 7*9 = 63 solutions:
1013, 1123, 1233, 1343, 1453, 1563, 1673, 1783, 1893,
2014, 2124, 2234, 2344, 2454, 2564, 2674, 2784, 2894,
3015, 3125, 3235 etc....
7019, 7129, 7239, 7349, 7459, 7569, 7679, 7789, 7899.