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Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers
Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers
This Question has 1 answers.
To prove that the given expressions are composite numbers, we show that they have factors other than 1 and themselves.
### First Expression: $7 \times 11 \times 13 + 13$
Factor out the common term $13$:
$ 7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) $
Simplify inside the parentheses:
$ = 13 \times (77 + 1) $
$ = 13 \times 78 $
Since $13$ and $78$ are both greater than 1, the expression is composite.
### Second Expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$
Rewrite as:
$ 7! + 5 $
Factor out $5$:
$ 7! + 5 = 5 \times \left(\frac{7!}{5} + 1 \right) $
Since $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ is divisible by $5$, the term inside the parentheses is an integer greater than 1.
Thus, the expression is composite.
Final Answer:
Both given expressions are composite numbers.
### First Expression: $7 \times 11 \times 13 + 13$
Factor out the common term $13$:
$ 7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) $
Simplify inside the parentheses:
$ = 13 \times (77 + 1) $
$ = 13 \times 78 $
Since $13$ and $78$ are both greater than 1, the expression is composite.
### Second Expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$
Rewrite as:
$ 7! + 5 $
Factor out $5$:
$ 7! + 5 = 5 \times \left(\frac{7!}{5} + 1 \right) $
Since $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ is divisible by $5$, the term inside the parentheses is an integer greater than 1.
Thus, the expression is composite.
Final Answer:
Both given expressions are composite numbers.
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