### Calendar Capers

Choose any three by three square of dates on a calendar page...

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!

### Rotating Triangle

What happens to the perimeter of triangle ABC as the two smaller circles change size and roll around inside the bigger circle?

# Top-heavy Pyramids

##### Stage: 3 Challenge Level:

Answers with limited explanations of your solutions were sent in by a number of you, including club members from Fen Ditton School, Madras College, Waverley Christian College and King Edward VII School. Trial and improvement (as you can see) is a very acceptable approach. It requires you to look at your findings at each stage of your working and use them, in combination with your mathematical knowledge, to move forward. It is this explanation of why you did what you did next that we want to see. Thank you for your answers.

Correct solutions were received from many of you including Alex, Stephanie, Charlotte Brant, Rosie, Hannah, Priscilla and Francesca of The Mount School and Shray of Wilsons Grammar School. All gave solutions based on the one below which was sent in by Alex, whose solution gives an excellent insight into her/his (sorry I am not sure) thinking.

Solution:
I worked out the order of the numbers using just trial and error. I decided to try the numbers in numerical order first but this did not work, so I tried putting the smallest number first, then the largest, then the next smallest, then the next largest etc...(1,12,3,9,4,8). The apex number was too high so I decided to try swapping two of the numbers so one of the higher numbers was at the end and did not have to be added to two different numbers and only to one.

I decided to use the nine, as my number did not need to go down drastically, which it may have done if I had swapped the 12. So now my sequence was 1,12,3,8,4,9. I had swapped the 9 with the 8. I added all the numbers up and the apex came to 200.

 200 103 97 54 49 48 28 26 23 25 13 15 11 12 13 1 12 3 8 4 9

I could find no other possible combinations to make the apex total 200.

******

Lucy, Rosie, Hannah, Priscilla and Francesca also offered the following observation, noticed by a number of you.

Using Pascal's triangle we discovered

 a+3b+3c+d ... ... a+2b+c b+2c+d c+2d+e d+2e+f a+b b+c c+d d+e e+f a b c d e f

We calculated this formula for the top of the pyramid:
a+5b+10c+10d+5e+f

Then using trial and improvement they found the same pyramid as the one above.

****

Andrei of School 205 Bucharest sent in the following. Thank you Andrei.

Let the numbers from the base be a, b,c, d, e and f. Then, the top-heavy pyramid will be:

 a+5b+10c+10d+5b+e a+4b+6c+4d+e b+4c+6d+4e+f a+3b+3c+d b+3c+3d+e c+3d+3e+f a+2b+c b+2c+d c+2d+e d+2e+f a+b b+c c+d d+e e+f a b c d e f

So, in top of the pyramid will be:

a + 5b + 10c + 10d + 5e + f = 200

Giving 5 as common factor for the 4 central numbers, I observe that I have the condition that

(a + f) is divisible by 5. The only possibilities are:

(1) a = 1; f = 4
(2) a = 1; f = 9
(3) a = 3; f = 12
(4) a = 8; f = 20

I analyse them separately:

(1) a = 1; f =4

I have: 5(b + 2c + 2d + e) = 195, and (b + 2c + 2d + e) = 39
The possibilities for b and c are 3 and 8, 3 and 9, 3 and 12, 8 and 9, 8 and 12, 9 and 12.
So, the possibilities for the sum 2c + 2d, and b + e are:

#### Solution

3, 8
22
17
12, 9
-
3, 9
24
12, 8
-
3, 12
30
8, 9
-
8, 9
34
5
3, 12
-
8, 12
40
< 0
-
-
9, 12
42
< 0
-
-

(2) a = 1; f = 9

Here (b + 2c + 2d + e) is 38.

#### Solution

3, 4
14
24
8, 12
-
3, 8
22
16
4, 12
Yes
3, 12
30
8
4, 8
-
4, 8
24
14
3, 12
-
4, 12
32
6
3, 8
-
8, 12
40
< 0
-
-

(3) a = 3; f = 12

The sum (b + 2c + 2d + e) is 37, and I have:

#### Solution

1, 4
10
27
8, 9
-
1, 8
18
19
4, 9
-
1, 9
20
17
8, 4
-
4, 8
12
25
9, 1
-
4, 9
11
26
8, 1
-
8, 9
34
3
1, 4
-

(4) a = 8; f = 20

This is the last solution I have to analyse. Here (b + 2c + 2d + e) = 36.

#### Solution

1, 3
8
28
4,9
-
1,4
10
26
3,9
-
1, 9
20
16
3, 4
-
3, 4
14
22
1, 9
-
3, 9
24
12
4, 1
-
4, 9
26
10
3, 1
-

The only solution is:

a, f = 1, 9
b, e = 4, 12
c, d = 3, 8.

Therefore, the only other way to arrange the digits in addition to the way drawn above is this:
 200 87 113 38 49 64 20 18 31 33 13 7 11 20 13 9 4 3 8 12 1

Well done to the Red Group from Malvern Parish School who pointed this out.