Which of the following can be used to illustrate that not all prime numbers are odd? 

 

The best approach is to pick a number that satisfies the rules that x and y are greater than 0. So for example you could choose 25 for y.

The value of x before doubling is then 25 - 50/25 = 23

The value after doubling y will be 50 - 50/50 = 49, which is more than double.

We have to rearrange the equation to make R the subject.

Start by cross multiplying by (r + R); V (r + R) = 12R

Multiply out the bracket Vr + VR = 12R

The maximum number of bows will be 4 yards (= 4 x 36 inches) divided by 15 inches.

This gives 9.6. But as a fraction of a bow is no use, we can only make 9 bows.

The sum of three consecutive integers can be written as n + (n + 1) + (n + 2) = 3n + 3

If the sum is 24, we need to solve the equation 3n + 3 = 24;

=> 3n = 21;

=> n = 7

The greatest of the three numbers is therefore 7 + 2 = 9

When the larger cube is cut into smaller cubes, the corner cubes will have paint on three sides. The cubes in the middle of the faces will have paint on only one side, but the cubes cut from the edges will have paint on two sides. In this case, there will be only one cube one each edge (excluding the corners), and since there are 12 edges, there will be 12 cubes with paint on two sides.

You can find the integers which when divided by 5 have a remainder 2 by adding 2 to all multiples of 5. So we have n = 7 , 12, 17, 22 etc.

From this series we can see that n does not have to be odd.

Also n + 1 can be a prime because, for example, 12 + 1 = 13

And (n + 2) / 7 has a remainder 2 in some cases but not all.

Remember the question asks us for what MUST be true, and we see that none of the statements are true in all cases. However, adding 3 to any of the values of n will always give a multiple of 5.

A perfect cube will have prime factors that are in groups of 3; for example 125 has the prime factors 5 x 5 x 5 , and 64 x 125 will also be a cube because its factors will be 4 x 4 x 4 x 5 x 5 x 5

Consider the answer choices in turn.

8 is the cube of 2, and p is a cube, and so the product will also be a cube.

pq will also be a cube as shown above.

pq is a cube and so is 27, but their sum need not be a cube. Consider the case where p =1 and q = 8, the sum of pq and 27 will be 35 which has factors 5 x 7 and is not a cube.

-p will be a cube.

Since the difference between p and q is raised to the power of 6, this expression will be a cube (with cube root = difference squared).

In case B, odd times even will give even