explain why 7*11*13+13 and 7*6*5*4*3*2*1+5 are composite numbers?

Answer 1

7*11*13+13=(13)(7*11+1)

similarly for the other one, where 5 is a common factor

Answer 2

The number 7*11*13 is a composite number. This number is divible by 13 -- because it is (7*11) * 13.

Adding 13 still leaves it divisible by 13.

(7*11)*13 + (1)*13 = ((7*11)+1)*13 a composite number

The other one can be resolved the same way.

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I just thought of another proof for the first one:

7*11*13 is an odd number. Adding 13 gives an even sum, therefore not prime (even numbers are divible by 2)

This shortcut cannot be used for the second number (7! is even)

Answer 3

7*11*13* + 13 = 13(7*11+1), so 13 is a factor of the
first number. Note also that 7*11+1 = 78 = 3*13*2
So the first number equals 2*3*13²
Next
7*6*5*4*3*2*1+5 = 5045, so 5 is a factor of the
second number. In fact, the second number is 5*1009
and 1009 is a prime.

Answer 4

From Wikipedia: A composite number is a positive integer which has a positive divisor other than one or itself. By definition, every integer greater than one is either a prime number or a composite number.

As to your question, both of the numerical expressions that you stated are composite numbers because they can be divided by 13 and 5, respectively.

Using the first expression as an example,

7*11*13 + 13 is equivalent to (13)(7*11) + (13)(1), or

(13)(7*11 + 1), which is clearly divisible by 13.

Answer 5

7*11*13+13 = (7*11 + 1) * 13
7*6*5*4*3*2*1+ 5 = (7*6*4*3*2*1+1) * 5
They have more than 2 factors.

Answer 6

The key bit is that the number being added is already one of the factors; this means that the result of the sum is still factorable, ergo composite.

Answer 7

it would be a composit number because a composit number consists of two or more factors. and those numbers have more than 2 factors.

Answer 8

They are not prime, they have more factors than just 1 and themselves.