Question

CHALLENGE question:

We can find the correct option by scrutinising consecutive pairs of digits in each
of the options, eliminating those which are not multiples of 3: B (637 542) has
consecutive digits 37, C (721 543) has 43, D (751 263) has 26, while E (7 563 421)
has 56 and 34. Option A (75 421) satisfies the conditions of the question, since 75,
54, 42 and 21 are each multiple of 3. The notes below explain why 75 421 is in fact
the largest possible number to work here.


CHALLENGE 2:

Since the mean of Katherine’s prime numbers is 8, their total is 8×3 = 24. The sum
of three odd numbers is an odd number, so at least one of Katherine’s numbers is
even; however, the only even prime number is 2. Hence, the other two numbers
have a total of 22. There are three possibilities where both numbers are prime:
3 + 19, 5 + 17 and 11 + 11. The only possibility where the difference between the
smallest two primes is itself prime is 2, 5 and 17.


LAST ONE IF YOU DARE:

The rectangle in the bottom right with area of 5 cm2 has an area 1
4
times greater
than the rectangle with area of 4 cm2
. They share the same base, and so the lower
rectangle must have a height 1
4
greater than the upper one. Now the same can
be said for the rectangles on the left, and so the area of the unshaded rectangle is
1
4
greater than 3 cm2
, that is 3 × 1.25 = 3.75 cm2
. Hence the total area in square
centimetres of the four rectangles is 3 + 4 + 5 + 3.75, that is 15.75 cm2

Answer 1

Answer:

hii DJ DH DH DH DH DJ DJ DJ hu

Answer 2

Answer:

Step-by-step explanation:do it urself