Let x be the least number that when divided by 16, 28, 40 and 77 leaves a remainder of 7 in each case, but is divisible by 19. What will be the remainder when x is divided by 219?  

Given:

The least number 'x' leaves a remainder of 7 when divided by 16, 28, 40, and 77.

The number 'x' is divisible by 19.

Formula Used:

If a number N leaves a remainder 'r' when divided by 'd', then N = qd + r, or N ≡ r (mod d).

If a number N leaves a remainder 'r' when divided by multiple numbers d₁, d₂, ..., d_n, then N = LCM(d₁, d₂, ..., d_n) × k + r, where k is an integer.

Calculation:

The LCM of 16, 28, 40, and 77.

Prime factorization of 16 = 2⁴

Prime factorization of 28 = 2² × 7

Prime factorization of 40 = 2³ × 5

Prime factorization of 77 = 7 × 11

LCM(16, 28, 40, 77) = 2⁴ × 5 × 7 × 11

⇒ LCM = 16 × 5 × 7 × 11 = 6160

The number 'x' in terms of the LCM.

Since 'x' leaves a remainder of 7 when divided by 16, 28, 40, and 77:

x = 6160k + 7 (where k is a positive integer)

The condition that 'x' is divisible by 19.

x ≡ 0 (mod 19)

6160k + 7 ≡ 0 (mod 19)

The remainder when 6160 is divided by 19:

6160 = 19 × 324 + 4

So, 6160 ≡ 4 (mod 19)

Substitute this back into the congruence:

4k + 7 ≡ 0 (mod 19)

⇒ 4k ≡ -7 (mod 19)

⇒ 4k ≡ 12 (mod 19) (Since -7 + 19 = 12)

To solve for k, we can find the multiplicative inverse of 4 modulo 19. We look for a number 'm' such that 4m ≡ 1 (mod 19). For m = 5, 4 × 5 = 20 ≡ 1 (mod 19).

Multiply both sides of 4k ≡ 12 (mod 19) by 5:

5 × 4k ≡ 5 × 12 (mod 19)

20k ≡ 60 (mod 19)

Since 20 ≡ 1 (mod 19) and 60 = 3 × 19 + 3, so 60 ≡ 3 (mod 19):

⇒ k ≡ 3 (mod 19)

The least positive integer value for k is 3.

Substitute k = 3 into x = 6160k + 7:

x = 6160 × 3 + 7

x = 18480 + 7

⇒ x = 18487

Divide 18487 by 219:

18487 ÷ 219

18487 = 219 × 84 + 91

The remainder is 91.

∴ The remainder when x is divided by 219 is 91.