To find the least number that leaves the given remainders when divided by 48, 60, 72, 108, and 140, we can use the concept of the Chinese Remainder Theorem (CRT).

The Chinese Remainder Theorem states that if we have a system of congruences in the form:
x ≡ a1 (mod m1)
x ≡ a2 (mod m2)
...
x ≡ an (mod mn)

where a1, a2, ..., an are the remainders and m1, m2, ..., mn are the divisors (in this case, 48, 60, 72, 108, and 140), then there exists a unique solution for x modulo the least common multiple (LCM) of the moduli.

To find the least number, we can follow these steps:

1. Calculate the LCM of the given divisors: 48, 60, 72, 108, and 140.
LCM(48, 60, 72, 108, 140) = 15120.

2. Set up the congruences using the remainders and the LCM of the moduli:
x ≡ 38 (mod 48)
x ≡ 50 (mod 60)
x ≡ 62 (mod 72)
x ≡ 98 (mod 108)
x ≡ 130 (mod 140)

3. Solve the system of congruences using the Chinese Remainder Theorem.

To simplify the calculations, we can reduce the congruences to their simplest form:
x ≡ 14 (mod 48)
x ≡ 50 (mod 60)
x ≡ 62 (mod 72)
x ≡ 26 (mod 108)
x ≡ 10 (mod 140)

4. Apply the Chinese Remainder Theorem to find the solution:
Let's denote the remainders as a1, a2, a3, a4, and a5, and the moduli as m1, m2, m3, m4, and m5, respectively.

M = m1 * m2 * m3 * m4 * m5 = 48 * 60 * 72 * 108 * 140 = 15120

M1 = M / m1 = 15120 / 48 = 315
M2 = M / m2 = 15120 / 60 = 252
M3 = M / m3 = 15120 / 72 = 210
M4 = M / m4 = 15120 / 108 = 140
M5 = M / m5 = 15120 / 140 = 108

Now, we need to find the inverses of Mi modulo mi for i = 1, 2, 3, 4, 5.

The inverse of M1 modulo m1 is 3 because 3 * 315 ≡ 1 (mod 48)
The inverse of M2 modulo m2 is 7 because 7 * 252 ≡ 1 (mod 60)
The inverse of M3 modulo m3 is 5 because 5 * 210 ≡ 1 (mod 72)