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.
\(7 \times 11 \times 13 + 13 = 13(7 \times 11 + 1)\) is composite. \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5040 + 5 = 5045 = 5 \times 1009\) is also composite.
We have to show that each given number is composite. A composite number is a natural number greater than 1 which has at least one factor other than 1 and itself. So our aim is to factorise each expression.
1) Number: \(7 \times 11 \times 13 + 13\)
Start with the expression:
\(7 \times 11 \times 13 + 13\)
We can see that both terms contain the number \(13\). So, we factor \(13\) out as a common factor.
First term: \(7 \times 11 \times 13\)
Second term: \(13\)
Take \(13\) common:
\(7 \times 11 \times 13 + 13 = 13 \big(7 \times 11 + 1\big)\)
Now simplify inside the bracket:
\(7 \times 11 = 77\)
So we get:
\(13 \big(7 \times 11 + 1\big) = 13 \big(77 + 1\big)\)
\(13 \big(77 + 1\big) = 13 \times 78\)
Here, both \(13\) and \(78\) are greater than 1.
So the number \(7 \times 11 \times 13 + 13\) has factors \(13\) and \(78\), apart from 1 and itself.
Therefore, \(7 \times 11 \times 13 + 13\) is a composite number.
2) Number: \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\)
First observe the product part:
\(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
This is the same as \(7!\) (factorial 7), but we will work it out step by step.
Multiply step by step:
\(7 \times 6 = 42\)
\(42 \times 5 = 210\)
\(210 \times 4 = 840\)
\(840 \times 3 = 2520\)
\(2520 \times 2 = 5040\)
\(5040 \times 1 = 5040\)
So the product is:
\(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\)
Now add 5:
\(5040 + 5 = 5045\)
So the number is \(5045\).
Now show that 5045 is composite.
Notice that 5045 ends in the digit 5. Any number ending in 5 is divisible by \(5\).
So 5 is a factor of 5045.
Let us divide 5045 by 5:
\(5045 \div 5 = 1009\)
Therefore, we can write:
\(5045 = 5 \times 1009\)
Here, both \(5\) and \(1009\) are greater than 1.
So, 5045 has at least two factors other than 1 and itself, namely \(5\) and \(1009\).
Hence, \(5045\) is a composite number.
Conclusion
We have factorised each expression as a product of two numbers greater than 1:
\(7 \times 11 \times 13 + 13 = 13 \times 78\)
\(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times 1009\)
Therefore, both given numbers are composite.