Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Step-by-Step Solution
Key Concept: A natural number greater than 1 is <b>composite</b> if it can be written as a product of two integers, each greater than 1. Hence, showing a factorisation of the given numbers into such a product proves they are composite.
Given:
1) $N_1 = 7 \times 11 \times 13 + 13$
2) $N_2 = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$
To Show: Both $N_1$ and $N_2$ are composite numbers.
Step 1 – Factor out the common factor in each expression.
- For $N_1$, the common factor is $13$:
$$N_1 = 13\bigl(7 \times 11 + 1\bigr).$$
- For $N_2$, the common factor is $5$:
$$N_2 = 5\bigl(7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1\bigr).$$
Step 2 – Simplify the brackets.
- $7 \times 11 = 77$, therefore
$$N_1 = 13\bigl(77 + 1\bigr) = 13 \times 78.$$
- $7 \times 6 \times 4 \times 3 \times 2 \times 1 = 1008$, therefore
$$N_2 = 5\bigl(1008 + 1\bigr) = 5 \times 1009.$$
Step 3 – Observe that each factor is greater than 1.
- In $N_1 = 13 \times 78$, both $13$ and $78$ are integers $>1$.
- In $N_2 = 5 \times 1009$, both $5$ and $1009$ are integers $>1$.
Step 4 – Conclude using the definition of a composite number.
Since each number can be expressed as a product of two integers each exceeding 1, both $N_1$ and $N_2$ satisfy the definition of a composite number.
Conclusion:
- $7 \times 11 \times 13 + 13 = 13 \times 78$ is composite.
- $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times 1009$ is composite.
Correct Answer: Both numbers are composite because they can be written as products of two integers greater than 1: $7\times11\times13+13 = 13\times78$ and $7\times6\times5\times4\times3\times2\times1+5 = 5\times1009$.