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# Shifting Times Tables

For this problem, lots of people submitted their methods for finding the times table and the shift, and some people sent in answers to the questions in the problem.

Here are your methods for level 1 and level 2, followed by your answers to the questions, followed by your methods for level 3 and level 4.

Van Anh from British Vietnamese International School in Vietnam sent in a useful insight for finding the times table and shift at level 1 and level 2:

In the times table, even if the times table is shifted, the distance between the two numbers will be the same. Let's say our 2 times table, which has a distance of 2, when shifted by 101, will become 103,105,... which still has a distance of 2. You can use this to help when come dealing with these kinds of problems.

Roshni from Tanglin Trust School in Singapore, EDSR from Priestlands School in the UK and Oren from Sunnynook Primary School in New Zealand used the same method for finding the times table and shift at levels 1 and 2. This is Roshni's work:

In this level, there is a common difference(d) between consecutive terms*(this is the distance that Van Anh was talking about)*. To work out the common difference(d), you simply find the difference between any two adjacent numbers. For example in the sequence 14,22,30,38,46, I could work out the common difference by doing: 22-14 or 30-22 or 38-30 or 46-38,
all of which give the same answer of 8. Once you have found the common difference, you can then subtract (common difference $\times$ n), so you are subtracting 8,16,24,32,40 from the sequence which results in 6,6,6,6,6

Therefore, the shift is up 6. If the difference between the (common difference $\times$ n) and the sequence is negative, then it is a shift down.

Sequence: 14, 22, 30, 38, 46

Common difference $\times$ n: 8, 16, 24, 32, 40

[Result of] subtraction: 6, 6, 6, 6, 6

So, for the example used above the table is 8, it is shifted up by 6.

**What can you say if the numbers are all odd?**

Anh Minh from British Vietnamese International School Hanoi said: the times table you are looking for will always be 2, it will shift up/down 1.

Anh Minh is right that shifting the 2 times table (all even numbers) up or down by 1 will give only odd numbers, but this is not the only way that the numbers could all be odd.

Alex said: That means all the numbers are from an even number's time table and that the shift is an odd number. This is because any number times any even number will always give an even number, and when even numbers are shifted by an odd number they always give an odd number.

**What about if they are all even?**

The Marshmallows and EDSR correctly said: the multiplication table is bound to be even.

John and Jack from Trinity Grammar in Australia gave a fuller explanation: It must be an even number [times table] shifted up by an even number.

**Or a mixture of odd and even?**

Alex said: This means the time table is an odd number as they're the only ones that alternate between odd and even numbers. The shift can be anything because it may turn odds into evens and viceversa but there will still be both odds and evens.

**What can you say if the units digits are all identical?**

The Marshmallows said: the times table is bound to end in zero, which Alex and John and Jack described as a multiple of ten.

**What if there are only two different units digits?**

EDSR said: The table's units digit will have to be a five, which Alex and John and Jack described as a multiple of five - although it must be a multiple of five but not a multiple of ten.

**What can you say if the difference between two numbers is prime?**

EDSR identified that: The prime number will be the multiplication table.

Alex was more precise: Those 2 numbers are consecutive terms of a prime number's time table. The shift can be anything as it won't change the difference.

**What can you say if the difference between two numbers is composite (not prime)?**

EDSR and The Marshmallows both explained the same idea. The Marshmallows said: The times table could be composite but it could have numerous times tables i.e. 4 times table is also in the 2 times table.

Again, Alex was more precise: The numbers are either multiples of a non-prime times table or non-consecutive terms of a prime times table.

**Can you explain how you worked out the table and shift each time, and why your method will always work?**

The Marshmallows explained the steps which form the foundations of all of the methods which were sent in:

First you must find the difference between all the numbers. If they are all the same then you know the times table then add (or subtract) to the original table then you have the shifted answer.

If the difference between the numbers is not the same throughout, then you need to find a common times table before working out the shift.

**How can you find the common times table?**

Hayley from Tarremah Steiner School in Australia, Charlotte from Ryde School in the UK and EDSR all assumed that the common times table would be the smallest of the differences. This is Hayley's example:

My numbers were:

180 438 481 610 653

First you have to find out how much there is between each number, for instance, the difference between 481 & 438 is 43 & the difference between 653 & 610 is also 43.

Then you get the lowest number you have (180) and divide it by your difference (43)

180$\div$43 is approximately 4, then you go 4 $\times$ 43 = 172, then you take 172 from 180 which is 8

So you end up with your times table (43) and the number that you've shifted it by (8).

It is almost like a linear equation $y=mx+c$ .

So you have to find your multiple (43) and your constant (8) so you end up with $y=43x+8$

Charlotte gave the steps of this method in more detail. Click here to see Charlotte's work.

Alex explained why this method doesn't always work:

The highest common factor of the differences will be the times table. At first I thought it would just be smallest difference but then I came across this:

given that the 5 numbers are 9 18 24 30 39

the differences would be 9 6 6 9

so you could think the time table is 6, but it's actually the HCF of 6 and 9, which is 3. This is true because using the 6 times table you can go from 18 to 24, but not from 9 to 18. So the largest time table that will satisfy both will be 3.

Ms Rusnock from Stuart Country Day School of the Sacred Heart in the USA, Roshni and Oren used a more robust method. This is Roshni's work:

You should first reorder the sequence given to you to increasing order. For example, you are given 128,213,60,264,179. The new version would be 60,128,179,213,264.

First, work out the difference between any two numbers next to each other - for example do 179-128 which is 51.

Then work out the difference between another pair of numbers that are next to each other (one of the numbers may have been used for the previous subtraction), so for example do 213 and 179 which is 34.

Noticeably, the differences are different - 51 and 34. So you should find the Highest Common Factor of [all of] the differences, which is 17 in this case, which will give you the table.

Then, for each number in the sequence, find the closest [multiple of 17] that is less than that number. So for the example it is 51, 119, 170, 204, 255. Then, calculate the difference between the closest [multiple of 17] and the ordered sequence which is 9,9,9,9,9.

Sequence: 129,213,60, 264,179

Ordered Sequence(OS): 60, 128,179,213,264

Difference: 68, 51, 34, 51

Highest Common factor for difference: 17

Table closest to OS: 51, 119,170,204,255

Diff between tab and OS: 9, 9, 9, 9, 9

So, for the example used above the table is 17, shifted up by 9

Alex wrote a program to find the times table and the shift. This is Alex's description of the program:

I wrote a python script to which you input all 5 numbers one at a time followed by an enter, they are all put into a list which is then sorted by ascending order. Then the difference between each pair of 2 consecutive terms is then calculated (So the difference between the 1st and 2nd, 2nd and 3rd, 3rd and 4th, etc) listed and sorted in the same way as previously.

The highest common factor of the differences will be the times table. I work out the HCF using the Euclidean algorithm, which takes in only 2 values at a time, however we need to calculate the HCF of the whole thing, to do so we use the following property:

HCF(w,x,y,z) = HCF(HCF(w,x),HCF(y,z))

So this way it arrives at a single term which is the times table.

I used the Euclidean algorithm for the HCF as it makes the code shorter and more efficient, but it could also be computed with a classical list of factors and then picking the largest one.

The shift is calculated using the MOD function (the remainder of a division). When any of the numbers is divided by the times table the remainder will be the shift upwards.

So if the number is 57 and the times table is 20, so MOD(57, 20)= 17 (in python MOD is written as %) so here the shift would be 17 upwards.

The times table and the shift are then outputted.

Click here to see Alex's code, or here to see it as a PDF.## You may also like

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For this problem, lots of people submitted their methods for finding the times table and the shift, and some people sent in answers to the questions in the problem.

Here are your methods for level 1 and level 2, followed by your answers to the questions, followed by your methods for level 3 and level 4.

Van Anh from British Vietnamese International School in Vietnam sent in a useful insight for finding the times table and shift at level 1 and level 2:

In the times table, even if the times table is shifted, the distance between the two numbers will be the same. Let's say our 2 times table, which has a distance of 2, when shifted by 101, will become 103,105,... which still has a distance of 2. You can use this to help when come dealing with these kinds of problems.

Roshni from Tanglin Trust School in Singapore, EDSR from Priestlands School in the UK and Oren from Sunnynook Primary School in New Zealand used the same method for finding the times table and shift at levels 1 and 2. This is Roshni's work:

In this level, there is a common difference(d) between consecutive terms

Therefore, the shift is up 6. If the difference between the (common difference $\times$ n) and the sequence is negative, then it is a shift down.

Sequence: 14, 22, 30, 38, 46

Common difference $\times$ n: 8, 16, 24, 32, 40

[Result of] subtraction: 6, 6, 6, 6, 6

So, for the example used above the table is 8, it is shifted up by 6.

Anh Minh from British Vietnamese International School Hanoi said: the times table you are looking for will always be 2, it will shift up/down 1.

Anh Minh is right that shifting the 2 times table (all even numbers) up or down by 1 will give only odd numbers, but this is not the only way that the numbers could all be odd.

Alex said: That means all the numbers are from an even number's time table and that the shift is an odd number. This is because any number times any even number will always give an even number, and when even numbers are shifted by an odd number they always give an odd number.

The Marshmallows and EDSR correctly said: the multiplication table is bound to be even.

John and Jack from Trinity Grammar in Australia gave a fuller explanation: It must be an even number [times table] shifted up by an even number.

Alex said: This means the time table is an odd number as they're the only ones that alternate between odd and even numbers. The shift can be anything because it may turn odds into evens and viceversa but there will still be both odds and evens.

The Marshmallows said: the times table is bound to end in zero, which Alex and John and Jack described as a multiple of ten.

EDSR said: The table's units digit will have to be a five, which Alex and John and Jack described as a multiple of five - although it must be a multiple of five but not a multiple of ten.

EDSR identified that: The prime number will be the multiplication table.

Alex was more precise: Those 2 numbers are consecutive terms of a prime number's time table. The shift can be anything as it won't change the difference.

EDSR and The Marshmallows both explained the same idea. The Marshmallows said: The times table could be composite but it could have numerous times tables i.e. 4 times table is also in the 2 times table.

Again, Alex was more precise: The numbers are either multiples of a non-prime times table or non-consecutive terms of a prime times table.

The Marshmallows explained the steps which form the foundations of all of the methods which were sent in:

First you must find the difference between all the numbers. If they are all the same then you know the times table then add (or subtract) to the original table then you have the shifted answer.

If the difference between the numbers is not the same throughout, then you need to find a common times table before working out the shift.

Hayley from Tarremah Steiner School in Australia, Charlotte from Ryde School in the UK and EDSR all assumed that the common times table would be the smallest of the differences. This is Hayley's example:

My numbers were:

180 438 481 610 653

First you have to find out how much there is between each number, for instance, the difference between 481 & 438 is 43 & the difference between 653 & 610 is also 43.

Then you get the lowest number you have (180) and divide it by your difference (43)

180$\div$43 is approximately 4, then you go 4 $\times$ 43 = 172, then you take 172 from 180 which is 8

So you end up with your times table (43) and the number that you've shifted it by (8).

It is almost like a linear equation $y=mx+c$ .

So you have to find your multiple (43) and your constant (8) so you end up with $y=43x+8$

Charlotte gave the steps of this method in more detail. Click here to see Charlotte's work.

Alex explained why this method doesn't always work:

The highest common factor of the differences will be the times table. At first I thought it would just be smallest difference but then I came across this:

given that the 5 numbers are 9 18 24 30 39

the differences would be 9 6 6 9

so you could think the time table is 6, but it's actually the HCF of 6 and 9, which is 3. This is true because using the 6 times table you can go from 18 to 24, but not from 9 to 18. So the largest time table that will satisfy both will be 3.

Ms Rusnock from Stuart Country Day School of the Sacred Heart in the USA, Roshni and Oren used a more robust method. This is Roshni's work:

You should first reorder the sequence given to you to increasing order. For example, you are given 128,213,60,264,179. The new version would be 60,128,179,213,264.

First, work out the difference between any two numbers next to each other - for example do 179-128 which is 51.

Then work out the difference between another pair of numbers that are next to each other (one of the numbers may have been used for the previous subtraction), so for example do 213 and 179 which is 34.

Noticeably, the differences are different - 51 and 34. So you should find the Highest Common Factor of [all of] the differences, which is 17 in this case, which will give you the table.

Then, for each number in the sequence, find the closest [multiple of 17] that is less than that number. So for the example it is 51, 119, 170, 204, 255. Then, calculate the difference between the closest [multiple of 17] and the ordered sequence which is 9,9,9,9,9.

Sequence: 129,213,60, 264,179

Ordered Sequence(OS): 60, 128,179,213,264

Difference: 68, 51, 34, 51

Highest Common factor for difference: 17

Table closest to OS: 51, 119,170,204,255

Diff between tab and OS: 9, 9, 9, 9, 9

So, for the example used above the table is 17, shifted up by 9

Alex wrote a program to find the times table and the shift. This is Alex's description of the program:

I wrote a python script to which you input all 5 numbers one at a time followed by an enter, they are all put into a list which is then sorted by ascending order. Then the difference between each pair of 2 consecutive terms is then calculated (So the difference between the 1st and 2nd, 2nd and 3rd, 3rd and 4th, etc) listed and sorted in the same way as previously.

The highest common factor of the differences will be the times table. I work out the HCF using the Euclidean algorithm, which takes in only 2 values at a time, however we need to calculate the HCF of the whole thing, to do so we use the following property:

HCF(w,x,y,z) = HCF(HCF(w,x),HCF(y,z))

So this way it arrives at a single term which is the times table.

I used the Euclidean algorithm for the HCF as it makes the code shorter and more efficient, but it could also be computed with a classical list of factors and then picking the largest one.

The shift is calculated using the MOD function (the remainder of a division). When any of the numbers is divided by the times table the remainder will be the shift upwards.

So if the number is 57 and the times table is 20, so MOD(57, 20)= 17 (in python MOD is written as %) so here the shift would be 17 upwards.

The times table and the shift are then outputted.

Click here to see Alex's code, or here to see it as a PDF.

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