What is the smallest number which when divided by 16, 18, 20, and 25 leaves a remainder of 4 in each case, but is exactly divisible by 7?

Let the required number be $N$

Given: when divided by 16, 18, 20, 25, remainder = 4

So,
$N - 4$ is divisible by 16, 18, 20, 25

Step 1: Find LCM

LCM of 16, 18, 20, 25 = 3600

So,
$N - 4 = 3600k$
$N = 3600k + 4$

Step 2: Divisible by 7

$3600k + 4 \equiv 0 \ (\text{mod } 7)$

$3600 \equiv 2 \ (\text{mod } 7)$

So,
$2k + 4 \equiv 0$
$2k \equiv 3 \ (\text{mod } 7)$

Multiply by inverse of 2 (which is 4):
$k \equiv 12 \equiv 5 \ (\text{mod } 7)$

Smallest $k = 5$

Step 3: Find $N$

$N = 3600 \times 5 + 4 = 18004$

Final Answer: 18004