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থিম নির্বাচন

আপনার পছন্দের থিম বেছে নিন।

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

A. 137
B. 1361
C. 140
D. 157
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: \(x = 1856 - 1684 = 172\).