Parallelogram 17 • Level 2 • 28 Dec 2023Chrismaths - Part 4

We have used p,q,r,s and t for the numbers in certain of the rectangles as shown in the diagram.

We now repeatedly use the fact that the number in each rectangle in the first four rows is the sum of the numbers in the rectangles immediately below it. This enables us to work out the values of p,q,r,s and t.

Applied to the top rectangle, this gives 105=p+47. It follows that p=105−47=58.

Then, as p=31+q, we have 58=31+q. Therefore q=27. Next, from 47=q+r, we deduce that 47=27+r. This gives r=20.

Next, from r=13+s, we have 20=13+s. Hence s=7.

We also have 13=9+t. Therefore t=4. Finally, s=t+x. Therefore 7=4+x.

We can now conclude that x=3.

The man is now 63 years old and was given the rowing machine for his 50th birthday. So he has had the rowing machine for 13 years and, possibly, a few months. Therefore the average number of miles per year that he has rowed is roughly

2504813≈2600013=2000.

Therefore, 2000 is roughly the number of miles per year that the man has rowed.

The digits that are primes are 2, 3, 5 and 7 (It is important to remember that 1 is not a prime number). The numbers of illuminated bars used to represent them are:

2: 5 bars; 3: 5 bars; 5: 5 bars; 7: 3 bars.

We see that each of them is represented by a prime number of bars. So there are 4 of the digits with the required property.

The positive cubes are the numbers in the sequence 1, 8, 27, 64, 125, 216, 343, ... .

It is straightforward to check that none of the first fives cubes in this sequence is the sum of three smaller positive cubes.

For example, as 27 + 27 + 27 = 81, and 81 < 125, any three cubes with sum 125 must include 64 at least once. The three cubes couldn’t include 64 twice because 64 + 64 > 125. However, if we had p+q+64=125, where p and q are positive cubes which are smaller then 64, then p+q=125−64=61, which is impossible as the only values p and q can take are 1, 8 and 27. So 125 is not the sum of three positive cubes.

However, 216 = 27 + 64 + 125, and so 216 is the sum of three positive cubes and so is smallest cube that can be written as the sum of three positive cubes.

We use the notation ‘abcde’ for the number which is represented by the digits a,b,c,d and e when expressed using the standard base 10. Using this notation we can write a five-digit palindromic number as ‘abcde’, where a,b and c are digits.

Since 45 = 5 × 9, and 5 and 9 have no common factors, the five-digit palindromic number ‘abcde’ is a multiple of 45 if, and only if, it is a multiple of both 5 and 9.

A number is a multiple of 5 if, and only if, its units digit is 0 or 5. Here the units digit a cannot be 0, since otherwise ‘abcde’ would not be a five-digit number. We deduce that a is 5. Thus a five-digit palindromic number which is divisible by 5 is of the form ‘5bcb5’, where band c are digits.

A number is a multiple of 9 if, and only if, the sum of its digits is a multiple of 9.

The smallest five-digit palindromic number that is a multiple of 45 has the form ‘50c05’, corresponding to taking b to be 0. The digit sum of ‘50c05’ is 10+c. For this to be a multiple of 9 we need to take the digit c to be 8. Therefore 50805, with digit sum 18, is the smallest five-digit palindromic number which is a multiple of 45.

The largest five-digit palindromic number that is a multiple of 45 has the form ‘59c95’, corresponding to taking b to be 9. The digit sum of ‘59c95’ is 28+c. For this to be a multiple of 9 we need to take the digitc to be 8. Therefore 59895, with digit sum 36, is the largest five-digit palindromic number which is a multiple of 45.

The difference between these two numbers is 59895−50805=9090.