Find the smallest number that must be added to 1994 such that a remainder of 28 is left when the number is divided by 38, and 57?

To solve this problem, we need to find the least common multiple (LCM) of 38 and 57, and then find the smallest number that when added to 1994 leaves a remainder of 28 when divided by the LCM.

Step 1: Find the LCM of 38 and 57 The LCM of two numbers is the smallest number that is a multiple of both numbers.

The prime factors of 38 are 2 and 19, and the prime factors of 57 are 3 and 19.

So, the LCM of 38 and 57 is 2 * 3 * 19 = 114.

Step 2: Find the smallest number that when added to 1994 leaves a remainder of 28 when divided by the LCM When 1994 is divided by 114, the remainder is 56.

To get a remainder of 28, we need to subtract 28 from 56, which gives us 28.

So, the smallest number that must be added to 1994 such that a remainder of 28 is left when the number is divided by 38, and 57 is 28.