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Sending and Receiving Cards

Stage: 2 Challenge Level: Challenge Level:1

We received many good quality solutions this month. A lovely Christmas present for us! We haven't included all of them, but instead we've chosen a variety of solution types to show you. Thank you to Robert, Sarah and Anna for your correct and well-presented solutions.

From The Juniors: The Christian School (Takeley)

Michael: "To find the number of cards you have to times the number of people by the number of cards they sent.''

Class: "The number of cards sent is one less than the number of people.''

Our Table:

Number of
people
Number of
cards sent
1 0
2 2
3 6
4 12
5 20
6 30
7 42
8 56
9 72
10 90
11 110
12 132

Frances: "They are all even numbers.''

Lawrie: "The units digit goes 0, 2, 6, 2, 0, 0, 2, 6, 2, 0''

Michael: "The units digits are symmetrical in the table''

Class: "Let's carry on the number pattern:''

0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 420, 462, 506, 552, 600, 650,...

Lawrie: "The units digit repeats the pattern every five numbers.0, 2, 6, 2, 0''

David - with help from the class and much amazement:

"Every time there is a six you add one more ten to the tens column.''

30 3+1=4

42 4+1=5

56 5+2=7 You've hit the six so you add one more ten

72 7+2=9

90 9+2=11

110 11+2=13

132 13+2=15

156 15+3=18 You've hit the six so you add one more ten

182 18+3=21

210 21+3=24 etc

It does carry on.

Class: "While we were working on last month's dice problem (a bit late) we saw the pattern of triangular numbers and were playing around with squares and doubling the triangular numbers in order to find some patterns etc when:''

Samuel: "All the numbers in the card problem are double the triangle numbers!''

NB Sam is our Mr Memory Man.

Thank you for such an enjoyable problem. My class think you make things look easy and it's your fault that they get so complicated. Apparently you work some kind of magic Bernard!


Siobhan of West Flegg Middle School sent the following solution:

"4 x 3 = 12
5 x 4 = 20
6 x 5 = 30
7 x 6 = 42
8 x 7 = 56
9 x 8 = 72
10 x 9 = 90
11 x 10 = 110
12 x 11 = 132
13 x 12 = 156
14 x 13 = 172
15 x 14 = 210
16 x 15 = 260
17 x 16 = 272
18 x 17 = 306
19 x 18 = 342
20 x 19 = 380
21 x 20 = 440
22 x 21 = 462
23 x 22 = 506
24 x 23 = 552
25 x 24 = 600
26 x 25 = 650
27 x 26 = 702
28 x 27 = 756
29 x 28 = 812
30 x 29 = 870

I worked out all of this because I wanted to find out if there was any pattern in the last numbers. I found out that there was. The pattern is 2, 6, 2, 0, 0 and it carries on about 6 times. Anyway, I found out that the whole class would send to each other all together 870 cards. I worked out the sum by doing this:

30
x29
---
270
600
---
870
===

But, what if two of the friends fell out with each other?

What if they changed to a new school?

What if nearly all the class started to hate each other?

What if one of the children sent a card to the teacher?

Would any of these ifs change the sequence of the numbers at the end?''


Natasha of West Felff Middle School says:

"Jenny sends 4 cards: 1 to Paula, 1 to Sarah, 1 to Roxy, 1 to Hannah
Paula sends 4 cards: 1 to Jenny, 1 to Sarah, 1 to Roxy, 1 to Hannah
Sarah sends 4 cards: 1 to Jenny, 1 to Paula, 1 to Roxy, 1 to Hannah
Roxy sends 4 cards......
Hannah sends 4 cards....
Therefore with a group of 5 children, 20 cards are sent altogether

Steph sends 5 cards: 1 to Emma, 1 to Laura, 1 to Gemma, 1 to June, 1 to Becky
Emma sends 5 cards: 1 to Steph, 1 to Laura, 1 to Gemma, 1 to June, 1 to Becky
Laura sends 5 cards......
Gemma sends 5 cards......
June sends 5 cards.......
Becky sends 5 cards......
Therefore with a group of 6 children 30 cards are sent. ''

Natasha continued using this method, then followed the pattern all the way up to 30 children, as shown below:

"Here are the final numbers of what I have done above, but up to 30 children:

5 children: 20 cards The first digits
of these numbers
go from 2-5
6 children: 30 cards
7 children: 42 cards
8 children: 56 cards
9 children: 72 cards The first digits go
from 7-15, so they
are jumping in twos
10 children: 90 cards
11 children: 110 cards
12 children: 132 cards
13 children: 156 cards
14 children: 182 cards
15 children: 210 cards
16 children: 240 cards There are two 2s,
three 3s, two4s,
two 5s, two 6s,
two 7s and two 8s
17 children: 272 cards
18 children: 306 cards
19 children: 342 cards
20 children: 380 cards
21 children: 420 cards
22 children: 462 cards Every single number
is even because
they have all got an
even number for
the last digit
23 children: 506 cards
24 children: 552 cards
25 children: 600 cards
26 children: 650 cards
27 children: 702 cards
28 children: 756 cards
29 children: 812 cards
30 children: 870 cards''


This is part of another solution sent in by Rachel from West Flegg Middle School. It shows another way to solve the problem.


"The number of arrows round the persons name represents how many cards they have sent and received, which is always one number less that the group number.''


Chris used a grid diagram to discover more patterns.

3 x 4 = 12 . . . 12 cards sent altogether.



5 x 4 = 20 . . . 20 cards sent altogether.

8 x 7 = 56 . . . 56 cards sent altogether.

15 x 14 = 210 . . . 210 cards sent altogether.

I see a pattern.

On a grid you end up with a diagonal line.
The numbers when multiplied are all even in the totals.
There is a pattern in the last digits of the multiplied numbers. The pattern is 2, 0, 0, 2, 6, 2, 0, 0, 2, 6...

There is also a pattern where you multiply the number of people (first number) sending the cards by the same number less one (second number) because you don't send a card to yourself.eg:

4 x 3 = 12
1 st
=
no.
of
people
2 nd

Class of 30 children; 30 x 29 = 870 cards sent.''

Here are results in a spread sheet form;

Ans