Which is the least number that must be subtracted from 1856 so that the remainder when divided by 7, 12, 16 is 4?

Explanation: Let \(N = 1856\). We are looking for the remainder when the resulting number (\(N - x\)) is divided by 7, 12, and 16, which is always 4. The required number \((N - x)\) must be of the form \(k \cdot (\text{LCM}(7, 12, 16)) + 4\). First, find the LCM of 7, 12, and 16. \(\text{LCM}(7, 12, 16) = \text{LCM}(7, \text{LCM}(12, 16)) = \text{LCM}(7, 48)\). Since 7 and 48 are coprime, \(\text{LCM}(7, 48) = 7 \times 48 = 336\). The required number is of the form \(336k + 4\). We need to find the largest multiple of 336 that is less than or equal to \(1856 - 4 = 1852\). Divide 1856 by 336: \(\frac{1856}{336} \approx 5.52\). The largest integer multiple is \(k=5\). The resulting number after subtraction is: \(336 \times 5 + 4 = 1680 + 4 = 1684\). The least number that must be subtracted (\(x\)) is: \(x = 1856 - 1684 = 172\).