This post was motivated by a long public discussion of learning the times table at UK schools.

### Dropping maths at the age of 16

It seems that a public acceptance of being bad in maths may not be considered as a great shame, even by a finance minister, read BBC magazine: UK Chancellor refused to answer a “times table” question. How could that happen?

One retired maths teacher has told us that many students do not continue studying maths and English at the age of 16-18. Here is an update from BBC education: a long-term possibility might be to require all would-be teachers to study maths to 18.

This long term plan can be compared with the long existing tradition in other countries where maths and native language/literature are *compulsory for all students* up to the university level.

### Why are UK students really lucky?

Here is the most popular comment on BBC: “I’m studying a Masters in Physics at University and STILL don’t know my times tables by heart, I got an A* at GCSE Maths and an A at A-Level.”

Does it imply that UK students may not know the times table? It is probably better not to reflect on conclusions that overseas students can make about UK qualifications and universities.

Here is our reply: you are really lucky to study in the UK, because without the times table you wouldn’t progress to a secondary school in a different country.

In many other educational systems, if children start school at the age of 6-7, then they finish learning the times table by the end of their 1st class at the age of 7-8.

### Why is learning the times table so hard?

As with all other obstacles, learning mathematics becomes unnecessarily hard when pre-requisite concepts are missed. The necessary earlier step is the simpler addition table that has the sum i+j in the intersection of the i-th row and j-th column.

+ |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |

1 |
2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

2 |
3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |

3 |
4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

4 |
5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 |

5 |
6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 |

6 |
7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |

7 |
8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |

8 |
9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |

9 |
10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |

Notice how the table above is beautifully symmetric like many great results in mathematics. When students learn the times table not by rote, but by actually trying to compute products in their head, they naturally start adding smaller numbers.

For instance, to find \(7\times 8\), one can sequentially compute the following partial sums: \(8+8=16\), \(\;16+8=24\), \(\; 24+8=32\), \(\; 32+8=40\), \(\; 40+8=48\), \(\; 48+8=56\). These sums require the addition table above, which should be studied before the times table.

So we get \(7\times 8=56\) after completing 6 easy additions. If we are already good at multiplying only by 2, faster way is to do 3 multiplications: \(7\times 2=14\), \(\; 14\times 2=28\), \(\; 28\times 2=56\).

### Can the times table be learned only by rote?

Let us compare the times table with another basic skill of walking. Children usually start learning to walk earlier than learning the times table.

Toddlers fall (fail) many times before they become confident walkers. After a first fall a baby looks at their parent to understand how to react: if the parent looks afraid then the baby starts crying, if the parent smiles then the baby also smiles.

So almost all children successfully learn to walk, not by rote, but by actually trying to walk, despite numerous falls or even small injuries. If a kid can not walk by the age of 3, it is usually considered as a physical disability.

### A criterion of success in learning

In our opinion, the same powerful principle works for learning any other basic skill. So we wish our readers to follow the excellent example of smart babies, namely more practice and consider every failure (or a fall) as one more step in a right direction.

There is even the internal criterion of success without any comparison with your peers. If people can walk, they walk easily and without making any conscious efforts. Similarly, multiplying single-digit numbers should be as effortless and enjoyable as walking or breathing.

**Riddle 17**: can the square \(n^2\) of an integer have the last digit 8?**How to submit**: to write your full answer, submit a comment.**Hint**: the required knowledge is (a small bit of) the times table.**Warning**: give an example or justify that there is no such number.**Prize**: free 1-year access to one of our interactive web tutorials.**Restriction**: only the first correct public answer will be rewarded.

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Does school stop at 5th grade? Do you have no resources for middle school and beyond? If I am not seeing them, then I apologize. Perhaps you could point me in the right direction.

You are right that the school should go beyond 5th grade. Here is the post about the more advanced Extreme Value Theorem.

n^2 will be in the form: 8+Σa_{k}*10^k from k=1 to i. (@)

Let n= Σb_{k}*10^k from k=0 to j.

then n^2=(Σb_{k}*10^k)^2

=b_{0}^2 +10(2b_{0}b_{1}) + 100(2b_{0}b_{2}+b_{1}^2)+…… ($)

Now equating the powers of ten (digits of n^2) of (@) to ($):

=>b_{0}^2=8

b_{0}=+-2^3/2

As b_{0} is not an integer, there is no possible number n, such that n^2 has a last digit of 8.

Dear Jasturan, thank you for your attempt at this question. The implication “=>b_{0}^2=the last digit of n^2” fails for n=4, where b_0=4 and b_{0}^2=16 isn’t equal to the last digit 6 of n^2=16. The idea of considering n modulo 10 can lead to a correct solution, so you could have another attempt. The shorter way to express the same idea is to consider n in the form n=10m+k, where m is integer and 0<=k<=9 is the last digit. Good luck!

Thank you for the feedback but with the generated series, the idea I had was to show that only the last digit of a number will affect the last digit of the square. I suppose this has been a valuable lesson by rigorously looking through the answer you’ve written and how it’s expressed on the page rather than having the intuitive understanding in your head. However I do like the idea that you could have used 10m+k where m is any integer rather than having terms a_{k} where 0 <a_{k}<=9. Makes the maths much similar. I'll have another go at posting a solution but again, thank you for your reply.

Dear Jasturan, it’s great that you are not giving up, especially because you are very close to a complete solution. Your argument with the last digit b_0 of n works perfectly for b_0 = 0,1,2,3 when b_0^2 is still a single digit number. So it’s essentially remains to consider few other remaining cases. We are looking forward to receiving your extra bit so that you can get your well-deserved prize.

Jasturan seems to have given up, so I’ll have a go.

We enumerate the possible cases mod 10:

1^2=1 2^2=4 3^2=9

4^2=6 5^2=5 6^2=6

7^2=9 8^2=4 9^2=1 and last but not least 0^2=0 (all the equalities are mod 10 of course).

Now since 8 does not appear in any of the RHSs, we conclude a square can never have 8 (nor 2,3 or 7 for that matter) as its last digit.

Dear Carlo, yes. Your solution is completely correct. I’ll just add that RHSs means “right hand sides” of the equations in the solution. You could e-mail your choice of a web tutorial to master.maths.tutor@gmail.com and we’ll enrol you for a year as promised.