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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.
Given: $7\ \times\ 11\ \times\ 13\ +\ 13$ and $7\ \times\ 6\ \times\ 5\ \times\ 4\ \times\ 3\ \times\ 2\ \times\ 1\ +\ 5$.
To do: Here we have to 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.
Solution:
A composite number is a number that is divisible by another number other than by itself and one. The number 2 is the only even prime number. All other even numbers are composite numbers.
Now,
(i) $7\ \times\ 11\ \times\ 13\ +\ 13$
$=\ 13(7\ \times\ 11\ +\ 1)$
$=\ 13\ \times\ (77\ +\ 1)$
$=\ 13\ \times\ 78$
$=\ 1014$
1014 is divisible by other numbers such as 2, 3, 6, 13, 26, 39, 78, 169, 338 and 507 apart from itself (1014) and 1.
So, it is a composite number.
(ii) $7\ \times\ 6\ \times\ 5\ \times\ 4\ \times\ 3\ \times\ 2\ \times\ 1\ +\ 5$
$=\ 5(7\ \times\ 6\ \times\ 4\ \times\ 3\ \times\ 2\ +\ 1)$
$=\ 5(1008\ +\ 1)$
$=\ 5(1009)$
$=\ 5045$
5045 is divisible by other numbers such as 5 and 1009 apart from itself (5045) and 1.
So, it is a composite number.
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