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# More Number Sandwiches

Thank you to Ashlynn from ISF in Hong Kong, who sent in a full solution. This is Ashlynn's work:

*In a "7- sandwich", how many red squares are covered and how many blue squares are*

covered?

There are 7 red squares covered and 7 blue squares covered.

*If it were possible to make a "6 -sandwich", how many red squares and how many blue*

squares would be covered?

For a 6 -sandwich, total there are 12 squares covered. 6 are red, 6 are blue.

*If you place a 1 on a blue square, on which colour will you place the other 1?*

Blue

*If you place a 2 on a blue square, on which colour will you place the other 2?*

Red

*If you place a 3 on a blue square, on which colour will you place the other 3?...*

Blue

*In general, what can you say about the colours on which you place pairs of numbers?*

If it is an odd number, it covers the same colour of tile. e.g. Red -Red, Blue- Blue

If it is an even number, it covers different colours of tile. e.g. Red -Blue, Blue- Red

*When you try to make a sandwich with the numbers from 1 to 5, or from 1 to 6, what goes wrong?*

For a 6 -sandwich, in total there should be 12 tiles covered. 6 are red tiles, 6 are blue tiles.

We try:

It is impossible because in 6 -sandwich there are 3 even numbers which cover 3R and 3B.

Odd numbers can only cover an even number of R or B. They cannot make another 3R or 3B.

For a 5- sandwich, in total there should be 10 tiles covered. 5 are red tiles, 5 are blue tiles.

We try:

It is impossible because in 5 -sandwich there are 2 even numbers which cover 2R and 2B.

Odd numbers can only cover an even number of R or B. They cannot make another 3R or 3B.

*Which other sandwiches are impossible? How can you be sure?*

Odd numbers can only cover an even number of R or B.

We can see a pattern:

If the number sandwich is an even number. When divided by 2, if the quotient is odd number, then it is impossible to make the number sandwich.

If the number sandwich is an odd number. When [decreased] by 1 and then divided by 2, if the quotient is odd number, then it is impossible to make the number sandwich.## You may also like

### Adding All Nine

### Doodles

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Thank you to Ashlynn from ISF in Hong Kong, who sent in a full solution. This is Ashlynn's work:

covered?

There are 7 red squares covered and 7 blue squares covered.

squares would be covered?

For a 6 -sandwich, total there are 12 squares covered. 6 are red, 6 are blue.

Blue

Red

Blue

If it is an odd number, it covers the same colour of tile. e.g. Red -Red, Blue- Blue

If it is an even number, it covers different colours of tile. e.g. Red -Blue, Blue- Red

For a 6 -sandwich, in total there should be 12 tiles covered. 6 are red tiles, 6 are blue tiles.

We try:

Number | Possible Colours of Tiles Covered |
---|---|

1 | R-R |

2 | R-B |

3 | R-R |

4 | R-B |

5 | B-B |

6 | IMPOSSIBLE! |

It is impossible because in 6 -sandwich there are 3 even numbers which cover 3R and 3B.

Odd numbers can only cover an even number of R or B. They cannot make another 3R or 3B.

For a 5- sandwich, in total there should be 10 tiles covered. 5 are red tiles, 5 are blue tiles.

We try:

Number | Possible Colours of Tiles Covered |
---|---|

1 | R-R |

2 | R-B |

3 | B-B |

4 | R-B |

5 | IMPOSSIBLE! |

It is impossible because in 5 -sandwich there are 2 even numbers which cover 2R and 2B.

Odd numbers can only cover an even number of R or B. They cannot make another 3R or 3B.

Odd numbers can only cover an even number of R or B.

Number Sandwich | Number of Red and Blue Need to be Covered |
Even numbers in the sandwich |
No. of R and B covered by even number |
No. of R and B needed to be covered by odd numbers |
POSSIBLE? |

1 | NO | ||||

2 | 2R, 2B | 2 | 1R, 1B | 1R, 1B | NO |

3 | 3R, 3B | 2 | 1R, 1B | 2R, 2B | YES |

4 | 4R, 4B | 2, 4 | 2R, 2B | 2R, 2B | YES |

5 | 5R, 5B | 2, 4 | 2R, 2B | 3R, 3B | NO |

6 | 6R, 6B | 2, 4, 6 | 3R, 3B | 3R, 3B | NO |

7 | 7R, 7B | 2, 4, 6 | 3R, 3B | 4R, 4B | YES |

8 | 8R, 8B | 2, 4, 6, 8 | 4R, 4B | 4R, 4B | YES |

9 | 9R, 9B | 2, 4, 6, 8 | 4R, 4B | 5R, 5B | NO |

10 | 10R, 10B | 2, 4, 6, 8, 10 | 5R, 5B | 5R, 5B | NO |

11 | 11R, 11B | 2, 4, 6, 8, 10 | 5R, 5B | 6R, 6B | YES |

12 | 12R, 12B | 2, 4, 6, 8, 10, 12 | 6R, 6B | 6R, 6B | YES |

13 | 13R, 13B | 2, 4, 6, 8, 10, 12 | 6R, 6B | 7R, 7B | NO |

14 | 14R, 14B | 2, 4, 6, 8, 10, 12, 14 | 7R, 7B | 7R, 7B | NO |

15 | 15R, 15B | 2, 4, 6, 8, 10, 12, 14 | 7R, 7B | 8R, 8B | YES |

We can see a pattern:

If the number sandwich is an even number. When divided by 2, if the quotient is odd number, then it is impossible to make the number sandwich.

If the number sandwich is an odd number. When [decreased] by 1 and then divided by 2, if the quotient is odd number, then it is impossible to make the number sandwich.

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!

Draw a 'doodle' - a closed intersecting curve drawn without taking pencil from paper. What can you prove about the intersections?