A school has invited 42 Mathematics teachers, 56 Physics teachers and 70 Chemistry teachers to attend a Science workshop. Find the minimum number of tables required, if the same number of teachers are to sit at a table and each table is occupied by teachers of the same subject.

Step 1: Understand the problem

We are given the number of teachers for each subject: 42 Mathematics teachers, 56 Physics teachers, and 70 Chemistry teachers. The conditions are that the same number of teachers must sit at each table, and each table must be occupied by teachers of the same subject. We need to find the minimum number of tables required.

Step 2: Find the maximum number of teachers per table

To minimize the number of tables, we need to maximize the number of teachers sitting at each table. Since the number of teachers per table must be the same for all subjects and must divide the total number of teachers for each subject, we need to find the Highest Common Factor (HCF) of 42, 56, and 70.

Step 3: Calculate the HCF

First, we find the prime factorization of each number: $42 = 2 \times 3 \times 7$, $56 = 2^3 \times 7$, and $70 = 2 \times 5 \times 7$. The HCF is found by taking the product of the lowest powers of the common prime factors. In this case, the common prime factors are 2 and 7. The lowest power of 2 is $2^1$ and the lowest power of 7 is $7^1$. Therefore, the HCF is $2 \times 7 = 14$. This means 14 teachers will sit at each table.

Step 4: Calculate the number of tables for each subject

Now that we know 14 teachers will sit at each table, we can calculate the number of tables required for each subject. For Mathematics teachers, $42 \div 14 = 3$ tables. For Physics teachers, $56 \div 14 = 4$ tables. For Chemistry teachers, $70 \div 14 = 5$ tables.

Step 5: Calculate the total minimum number of tables

Finally, we sum the number of tables required for each subject to find the total minimum number of tables needed for the workshop. $3 + 4 + 5 = 12$ tables.

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