Parallelogram 16 • Level 2 • 21 Dec 2023Chrismaths - Part 3

In the context of the JMC we can answer the question by eliminating the options that cannot be correct without having to do a division sum to evaluate the fraction given in this question.

Since 12,345 and 1 + 2 + 3 + 4 + 5 = 15 are both odd numbers, 123451+2+3+4+5 must also be odd. So only options A, D and E could be correct.

However, it is clear that 1234515≠1 and 1234515≠12359

This rules out options A and E and leaves D as the correct option.

The answer may be found by trying the options in turn. In this way we find that 7 has remainder 1 when divided by 2 and when divided by 3.

Alternatively, we see that all the numbers given as options are odd and so each has remainder 1 when divided by 2. So the correct option will be a number which also gives remainder 1 when divided by 3.

It is easy to see that, of the given options, only 7 meets this requirement.

The figure consists of a regular octagon whose vertices are labelled with the positive integers from 1 to 8 inclusive. There is a line joining each pair of the vertices. It follows that the number of labels which are multiples of 3 is equal to the number of pairs of distinct integers in the range from 1 to 8, inclusive, whose sum is a multiple of 3.

The following table shows all multiples of 3 that can be a label, and for each multiple, the different ways of writing it as a sum of two distinct integers in the range from 1 to 8. We do not need to go beyond 15, as the largest integer that can appear as a label is 7 + 8 = 15.

It follows that the number of labels which are multiples of 3 is 1 + 2 + 4 + 2 + 1 = 10.

We are told in the question that 263718459=17.

Now note that 5274=2×2637 and 36918=2×18459. It follows that, by cancelling the common factor 2 in the numerator and the denominator,

527436918=2×26372×18459=263718459=17.

If you do not spot the quick method used above, there is nothing for it but to try out the options in turn.

We first consider option A. By cross multiplication we have

527436918=18⇔5274×8=36918×1.

We can see that the equation 5274×8=36918×1 cannot be correct just by looking at the units digits on the two sides of the equation. On the left hand side 4 × 8 gives a units digit of 2, but on the right hand side 8 × 1 gives a units digit of 8. So the equation is not correct.

We deduce that

527436918≠18

and so option A is not the correct one.

Next we look at option B. Using cross multiplication again, we have

527436918=17⇔5274×7=36918×1.

It is straightforward to check that the equation 5274×7=36918 is true. Therefore option B is correct.

The squares in the corners of the large rectangle are of sizes 1 cm × 1 cm, 2 cm × 2 cm, 3 cm × 3 cm and 4 cm × 4 cm.

The white rectangle has width 1 cm and area 1.5 cm². It follows that it has height 1.5 cm.

We can now deduce that the large rectangle has height 5.5 cm and width 5 cm, and hence that the lengths are, in cm, as shown in the diagram.

We therefore see that the region shown dark grey is made up of two rectangles, one with width 2 cm and height 1.5 cm, and the other with width 2 cm and height 0.5 cm.

Therefore the area of this region is (2 × 1.5) cm² + (2 × 0.5) cm² = 3 cm² + 1 cm² = 4 cm².

Once we have shown, as above, that the large rectangle has height 5.5 cm and width 5 cm, we can find the area of the overlap without finding the dimensions of the two rectangles that make it up. Instead we can give an argument just in terms of areas, as follows.

The area of the large rectangle is 5.5 cm × 5 cm = 27.5 cm². Since the area not covered by any of the squares is 1.5 cm², the area of the large rectangle covered by the squares is 27.5 cm² − 1.5 cm² = 26 cm².

The total area of the squares is 1 cm² + 4 cm² + 9 cm² + 16 cm² = 30 cm².

The difference between these two areas is accounted for by the overlap. Therefore the area of the overlap is 30 cm² − 26 cm² = 4 cm².