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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?

Question

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Solution

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.

This problem has been solved

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