What is the least number that when divided by 44 leaves a remainder 31, when divided by 56 leaves a remainder 43, and when divided by 32 leaves a remainder 19?

Question

Sahil Verma · Accepted Answer

First, let's express our problem in terms of congruences:x &equiv; 31 (mod 44)x &equiv; 43 (mod 56)x &equiv; 19 (mod 32)
We need to find the least positive x that satisfies all these congruences.Let's calculate M, which is the product of all moduli:M = 44 * 56 * 32 = 78,848
Now, we calculate Mi for each congruence:M1 = M / 44 = 1,792M2 = M / 56 = 1,408M3 = M / 32 = 2,464
Next, we need to find the modular multiplicative inverses of each Mi modulo the corresponding modulus:1,792^(-1) &equiv; 29 (mod 44)1,408^(-1) &equiv; 25 (mod 56)2,464^(-1) &equiv; 1 (mod 32)
Now we can calculate x:x = (31 * 1,792 * 29 + 43 * 1,408 * 25 + 19 * 2,464 * 1) mod 78,848= (1,615,616 + 1,512,800 + 46,816) mod 78,848= 3,175,232 mod 78,848= 3,707
Therefore, the least number satisfying all the given conditions is 3,707.