Where can we visit?

Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?

Age

11 to 14

Challenge level

Secondary

NUMBER

Number Operations and Calculation Methods

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Where Can We Visit? printable sheet

100 squares printable sheet

Here is a 100 square board with a counter on 42:

Using either of the two operations $\times 2$ and $-5$, whereabouts on the 100 square is it possible to visit?

You might start like this: $$42, 37, 32, 27, 22, 17, 12, 7, 14, 9, 18, 13, 26, 52, 47, 42, 84 ...$$Notice that you are allowed to visit numbers more than once.

The board would look like this:

Is it possible to visit every number on the grid?

What if you start on a different number?

Can you explain your results?

Choose pairs of operations of your own and investigate what numbers can be visited.

You might like to print off some 100 squares.

Is there a way to predict which numbers it's possible to visit, for a given starting point and a pair of multiplication/subtraction operations?

This problem is also available in French: Où Irons-nous?

Fantastic responses to this problem, Benjamin and George discovered this:

We started with 42 and using the rule multiply by 2 and subtract by 5.

We discovered when any number with the factor of 2 is doubled (multiplied by 2) it gives a units digit of 4, 8, 6, and back to 2 which we started with. Although you can get round this by subtracting 5 it only then takes you to 7, 3 1 and 9 which with the 10 by 10 grid it becomes impossible to numbers with the factors or multiples of 5 e.g. 5, 10 15...

Lydia and Jess explain their findings:

We can not visit numbers which are multiples of 5 (those ending in a 5/0), 99 or 97. We could not visit multiples of 5, as starting from 42, multiplying by 2 or subtracting by 5 would not get us to a multiple of 5. Multiplying 42 by 2 results in 84, you cannot do this again as the result would be above the number 100; however many times you minus 5 you wold only visit numbers ending in 7 or 2. We could not visit 97 or 99 as they are not even meaning we can not double any number to get there. both of these are closer to 100 than 5 meaning we can not subtract 5 to there.

Lyman Shen explored other combinations of multiplying and subtracting:

Solution for $\times$ 2 and -5:

First I started with 1. By multiplying by two I can go to 2, 4, 8, 16, 32, 64. And I go down 64 from -5 all the way to 9 (4 would do the same thing again). And then I multiply by 2 again, and get 9, 18, 36, 72. I can carry this on to get every number, except I can't get any multiple of 5's. Also I can't make 97, and 99. I can't get them because they are odd numbers and can't get it by -5. I can get 98 from $\times$ 2 from 49, which could be get from -5 form numbers end with 9 and 4.

Solution for $\times$ 3 and -5:

By working similarly to before, I can find all the numbers except for the multiple of 5's because no numbers can make it to 5, 10, unless I start with a multiple of 5. I also can't get 92, 97, 98. I can't get them because they aren't multiple of 3's and can't get it from -5.

Solution for $\times$ 4 and -5:

This is time the numbers that can be reached depend on the numbers started with. When the starting number ends with 1, 4, 6, 9, there are no way to get any number that ends differently ($1 \times 4 = 4, 4 \times 4 = 16, 6 \times 4 = 24, 9 \times 4 = 36$). So we can reach everything ending in 1, 4, 6, 9 except the large numbers 94, 99. I can't get those number because it is not divisible from 4 and I can't get them from -5, which I can do to 96.

When the starting number ends with 2, 3, 7, 8, there are no way to get any number that ends differently ( $2\times 4 = 8, 3\times 4 = 12, 7\times 4 = 28, 8\times 4 = 32$). So I can reach anything ending in these digits, apart from 97. I can't get it because it isn't divisible from 4 and can't get it from -5.

When the start number ends with 5 or 0, there are no way to get any number that ends in anything other than 0 and 5.

Solution for $\times$ 5 and -5:

This reaches every multiple of 5, and the number you start with.

The reason is $5\times x = 5x$, which have to end with 0 or 5. No more exceptions after that.

Solution for $\times$ 6 and -5:

This is another one that is dependant on the starting number.

If it ends with 1 or 6, you can't get any numbers that don't end with 1 or 6 ($1 \times 6 = 6, 6 \times 6 = 36$).

If it ends with 0 or 5, you can't get any numbers that don't ends with 0 or 5 ($0 \times 6 = 0, 5 \times 6 = 30$). The ones you can't reach 95 and 100 ($15 \times 6 = 90$).

If it ends with 4 or 9: The same thing ($4 \times 6 = 24, 9 \times 6 = 54$). There are exceptions 99, 94, 89 ($14 \times 6 = 84$).

If the stating number ends with 3 or 8. $3 \times 6 = 18$, $8 \times 6 = 48$ so again you can reach all numbers ending in 3 or 8, but missing 98, 93, 88, 83 as the highest that can be reached is $13 \times 6 = 78$

Why do this problem?

This problem offers all students opportunities to explore fundamental ideas about number theory in a simple context. They are encouraged to explore, conjecture, generalise and justify. There are opportunities for older students who are familiar with algebraic manipulation or modulo arithmetic to produce rigorous proofs.

Possible approach

You need a set of number cards 1-100, with the multiples of 5 removed. Shuffle the cards in front of the class and hand out one to each student. Do not tell them that the multiples of 5 are missing!

Ask each student to turn to their neighbour and work out how to get from one number to the other and back again, using only these two operations:

$\times 2$ and $-5$

For example if the two numbers are 21 and 54 the chains could be:

21, double, 42, take five, 37, take five, 32, take five, 27, double, 54

and

54, take lots of fives, 9, double, 18, take five, 13, double, 26, take five, 21.

Pairs that are proving difficult to connect could be written on the board and offered as a challenge for the whole class to solve. Everyone should be able to arrive at their partner's number!

Finally challenge the class to get to your number (which should be a carefully chosen multiple of five). You may wish to offer a prize......

Once the class give up, ask them to explain why it is impossible.

Display the 1-100 grid, choose 42 as your starting number and explain that by using the operations above we are going to try to visit all the numbers on the grid.

Demonstrate how the numbers are crossed out as they are visited.

Ask the students to predict what will happen.

Will they be able to visit every number on the grid at least once?

Hand out this 1-100 grid and allow some time for students to work in pairs to check their predictions.

Bring the students together to link their ideas to the findings from the earlier exercise.

What would have happened if they had started on a different number?

Can they explain their results?

Ask if they think they will get the same sort of results with other pairs of operations.

You may wish to suggest families of pairs of operations for them to explore. Eg:

x3 and -5

x4 and -5

x5 and -5...

x5 and -2

x5 and -3

x5 and -4...

or they can try some families of their own choosing.

Hand out plenty of the 1-100 grids and ask students to work in pairs or small groups and make a display of their results.

Can they explain their findings and use these to begin to make predictions about other pairs of operations? Encourage them to justify their predictions.

You might find it useful to see if they can identify the pair of operations that produced the patterns in the three grids below. They can choose from either:

$\times 3$ and $-6$,

$\times 6$ and $-3$

See if they can spot which is which, and if the starting number makes a difference.

Key questions

What happens to multiples of 5 when they are doubled?

What about numbers that are 1 more, 2 more, 3 more and 4 more than a multiple of 5?

What happens to multiples of 5 when 5 is subtracted from them?

What about numbers that are 1 more, 2 more, 3 more and 4 more than a multiple of 5?

Possible support

Most students will have little difficulty with doubling or subtracting 5, but may find it more difficult to spot the patterns involved. Using coloured arrows to represent repeated subtractions of 5 on various 1-100 grids (each starting from a different number) may help them to see the patterns emerge.

Possible extension

Students may be interested in this introductory reading on Modular Arithmetic.

Students could have a go at Take Three from Five which has a similar underlying structure. Can they use their insights from the previous problem to solve this challenge?

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