Parallelogram 14 • Level 2 • 7 Dec 2023Chrismaths - Part 1

Since the numbers are the same in each sum, the largest answer results when the amount subtracted is the smallest. In option A the number 2 is subtracted, in option B it is 3, in option C it is 4, in option D both 3 and 4 are subtracted, and in option E both 2 and 3. So A gives the largest answer.

Alternatively, we can see directly that the sums have the following answers:

• A: 1 − 2 + 3 + 4 = 6
• B: 1 + 2 − 3 + 4 = 4
• C: 1 + 2 + 3 − 4 = 2
• D: 1 + 2 − 3 − 4 = −4
• E: 1 − 2 − 3 + 4 = 0

and therefore that A gives the largest answer.

The angle marked on the diagram as y° and the angle that is 110° are angles on a straight line. Therefore their sum is 180°.

It follows that y°=70°. Therefore, by the Exterior Angle Theorem 120°=x°+y°=x°+70°.

It follows that x=50.

Consider three different prime numbers which include 2, say the prime numbers 2, p and q. Then p and q will both be odd numbers, and therefore 2+p+q will be an even number greater than 2 and so cannot be a prime number. So, if we seek prime numbers that are sums of three different prime numbers, we need only consider sums of three different odd prime numbers.

The three smallest odd prime numbers are 3, 5 and 7, but their sum is 15 which is not prime. If we replace 7 by the next odd prime, 11, we have three odd primes with sum 3 + 5 + 11 = 19, which is a prime number.

We cannot obtain a smaller prime number as a sum using 3 and two other odd prime numbers. If we do not include 3, the smallest sum of three odd prime numbers that we can obtain is 5 + 7 + 11 = 23 which is greater than 19.

We can therefore deduce that 19 is the smallest prime number which is the sum of three different prime numbers.

We let P,Q,R,S and T be the points shown in the diagram. We also let ∠QRP=p° and ∠TRS=q°.

Because it is an angle of a square, ∠PRT=90°. Because they are angles of an equilateral triangle ∠PQR=∠RST=60°.

Because the angles of a triangle have sum 180°, from triangle PQR we have x+p+60=180 and from triangle TRS,y+q+60=180. Therefore x+p=120 and y+q=120.

Because ∠QRP, ∠PRT and ∠TRS are angles on a straight line, p+q+90=180 and therefore p+q=90. It follows that x+y=x+p+y+q−p+q=120+120−90=150.

There is a quick method that it is all right to use in the context of the JMC, but which would not be acceptable if you had to give a full solution with detailed reasons.

We have already shown that p+q=90. Since the question does not give us individual values for p and q, we can assume that the answer is independent of their actual values. So, for simplicity, we assume that p=q=45. Therefore in each of the triangles PQR and RST one of the angles is 60° and one is 45°. Therefore, because the sum of the angles in a triangle is 180°, both x and y are equal to 180−60−45=75. We conclude that x+y=75+75=150.

Let the three gaps between the pictures each be g mm wide.

Since each of the four pictures is 420mm wide and the wall is 4800mm wide:
4×420+3g=4800

and therefore:
3g=4800−4×420=4800−1680=3120

It follows that:
g=13×3120=1040

The distance between the centres of the middle two pictures is equal to the width of one picture and the width of the gap, that is, in mm, 420+g=420+1040=1460.

The distance between the centre of one of these pictures and the centre line is half this distance. Therefore the required distance is, in mm, 12×1460=730.