find the least 5-digit number which on dividing by 4,12,20 and 24 leaves remainder 3 in each case

Question

Answer

Problem Statement:

Find the least 5-digit number which on dividing by 4,12,20 and 24 leaves remainder 3 in each case.

Solution:

To find the least 5-digit number which leaves a remainder of 3 when divided by 4, 12, 20, and 24, we need to find the LCM of 4, 12, 20, and 24.

Step 1: Find the LCM of 4, 12, 20, and 24

The prime factorization of 4 is 2 x 2

The prime factorization of 12 is 2 x 2 x 3

The prime factorization of 20 is 2 x 2 x 5

The prime factorization of 24 is 2 x 2 x 2 x 3

Now, we list the prime factors with the highest power:

2^3 x 3 x 5 = 120

Therefore, the LCM of 4, 12, 20, and 24 is 120.

Step 2: Add 3 to the LCM to get the required number

Since the number leaves a remainder of 3 when divided by each of 4, 12, 20, and 24, it means that the number is 3 more than a multiple of each of these numbers.

Therefore, we need to add 3 to the LCM of these numbers to get the required number.

The required number is:

120 + 3 = 123

Step 3: Verify that the number satisfies the conditions

We need to verify that the number 123 leaves a remainder of 3 when divided by each of 4, 12, 20, and 24.

123 ÷ 4 leaves a remainder of 3

123 ÷ 12 leaves a remainder of 3

123 ÷ 20 leaves a remainder of 3

123 ÷ 24 leaves a remainder of 3

Thus, the required number is 123.

Answer

10083

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