Ashley claims that the sum of four multiples of 6 will be a multiple of 12. Is Ashley's conjecture reasonable? If not provide a counter-example.

Prove that the perimeter will be even for any rectangle with natural number side lengths.

for the perimeter, since

p = 2(length+width)
p is even, since it is a multiple of 2.

To determine whether Ashley's claim is reasonable, we can analyze the properties of multiples of 6 and multiples of 12.

A multiple of 6 is any number that can be evenly divided by 6, such as 6, 12, 18, 24, etc. A multiple of 12, on the other hand, is any number that can be evenly divided by 12, like 12, 24, 36, 48, etc.

If we take four multiples of 6, such as 6 + 12 + 18 + 24, the sum is 60. This number is not a multiple of 12, because it cannot be divided evenly by 12. Therefore, Ashley's conjecture is not reasonable.

To provide a counter-example, we can take four multiples of 6 that do not add up to a multiple of 12. For instance, let's consider the numbers 6, 12, 18, and 30. The sum of these multiples of 6 is 6 + 12 + 18 + 30 = 66. As 66 is not divisible by 12, this demonstrates that Ashley's claim is incorrect.

Now, let's prove that the perimeter of any rectangle with natural number side lengths is always even.

A rectangle has two pairs of parallel sides. Let's denote the lengths of the sides as a and b, where a and b are natural numbers. The perimeter of the rectangle is given by the formula P = 2a + 2b.

Since a and b are natural numbers, 2a and 2b will be even numbers since any number multiplied by 2 is always even. Additionally, the sum of two even numbers is always even. Therefore, 2a + 2b will always be an even number, regardless of the values of a and b.

Hence, we have proved that the perimeter of any rectangle with natural number side lengths is always even.

False!

Counter-example

4 possible multiples of 6 are
6, 12, 30, and 42
their sum is 90
but 90 is not divisible evenly be 12, so the statement is false