### Repeaters

Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.

### Digit Sum

What is the sum of all the digits in all the integers from one to one million?

# Two and Two

##### Stage: 3 Challenge Level:
We received lots of great solutions for this problem, unfortunately too many to mention you all by name! Many of you found that there were up to seven possible solutions, by systematically eliminating possibilities - well done. Here is an answer submitted by Charlie, Constance, Gabriel, two Lucases, Matilda, Rachel, Rama, Ronan, Ruby, Sanjay and Stephanie from Strand on the Green Junior School:

The first thing we noticed was that F has to be 1 because the most T + T can be is 19 (if you have already carried 1 from the previous column). This also means that T ≥ 5. We also noticed that R must be even.

We decided to look at the value of O again.
If O = 0, then R would also be 0 so that doesn’t work and O can’t be 1 because F = 1.

If O = 2,

$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{2}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{2}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{2}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$

then R = 4 and T = 6 and we also know that W < 5 because there can’t be anything carried to the hundreds column. The only possible value of W that hasn’t already been used is 3 but this would mean that U is 6 which is the same as T.

If O = 3,

$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{3}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{3}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{3}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1$

then R = 6 and T = 6 which doesn’t work.

If O = 4,

$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{4}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{4}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{4}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$

then R = 8 and T = 7 and we also know that W < 5 because there can’t be anything carried to the hundreds column. So W could be 0, 2 or 3.

W can’t be 0 because then U would be 0 and it can’t be 2 because U would be 4.
If W = 3, U = 6 which works: 734 + 734 = 1468.

If O = 5,

$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{5}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{5}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{5}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1 \, \: ^1$

then R = 0 and T = 7 and we also know that W ≥ 5 because there has to be 1 carried to the hundreds column.

W can’t be 5 because O = 5.
If W = 6, U = 3 which works:  765 + 765 = 1530.
If W = 7, U = 5 which doesn’t work because O and U are the same.
If W = 8, U = 7 which doesn’t work because  T and U are the same.
If W = 9, U = 9 which doesn’t work because W and U are the same.

If O = 6,

$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{6}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{6}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{6}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1$

then R = 2 and T = 8 and we also know that W < 5 because there can’t be anything carried to the hundreds column. So W could be 0, 3 or 4.

If W = 0, U = 1 which doesn’t work because F and U are the same.
If W = 3, U = 7 which works. 836 + 836 = 1672
If W = 4, U = 9 which works. 846 + 846 = 1692

If O = 7,

$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{7}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{7}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{7}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1 \, \: ^1$

then R = 4 and T = 8 and we also know that W ≥ 5 because there has to be 1 carried to the hundreds column.

If W = 5, U = 1 which doesn’t work because F and U are the same.
If W = 6, U = 3 which works.  867 + 867 = 1734
W can’t be 7 because O = 7.
If W = 8 , U = 7 which doesn’t work because  O and U are the same.
If W = 9, U = 9 which doesn’t work because W and U are the same.

If O = 8,

$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{8}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{8}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{8}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1$

then R = 6 and T = 9 and we also know that W < 5 because there can’t be anything carried to the hundreds column. So W could be 0, 2, 3 or 4.

If W = 0, U = 1 which doesn’t work because F and U are the same.
If W = 2, U = 5 which works: 928 + 928 = 1856.
If W = 3, U = 7 which works: 938 + 938 = 1876.
If W = 4, U = 9 which doesn’t work because T and U are the same.

If O = 9,

$\quad \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{9}$
$\underline{+\, \text{T}\hspace{1mm} \text{W}\hspace{1mm} \text{9}}$
$\underline{\, \, \, \text{1}\hspace{1mm} \text{9}\hspace{1mm} \text{U}\hspace{1mm} \text{R}}$
$\quad \, \, ^1 \, \: ^1$

then R = 8 and T = 9 which doesn’t work because O and T are the same.

So there are seven possible answers:
938+938=1876
928+928=1856
867+867=1734
846+846=1692
836+836=1672
765+765=1530
734+734=1468

The group from Strand on the Green Junior School also sent us a solution for the problem ONE+ONE=TWO using a similar strategy, and found 16 possible answers. Well done on such a thorough solution!

Some of Mr Parkin's Students from Alleyn's School also tried finding some other words sums. They also investigated ONE+ONE=TWO. Ed noticed that the word sum FOUR+FOUR=EIGHT has the interesting property, as 'FOUR' and 'EIGHT' have no letters in common. He could only find two solutions, 5469 + 5469 = 10938 and 8235 + 8235 = 16470 but is not sure how to find more, except by using a computer program.

Students from Angel Road Junior School in Norwich have managed to come up with three more answers to the sum:
FOUR
+ FOUR
EIGHT

Their solutions are:
8523 + 8523 = 17046
9235 + 9235 = 18470
9327 + 9327 = 18654
8652 + 8652 = 17304

Their teacher explained how they went about the problem:

They tell me their process started by trying to find any rules that were evident, for example, would any of the letters have to be 5 or more and if so, could the carrying of a digit help achieve another number which has not already been used?
They found that because there were no repeated letters between FOUR and EIGHT, then that they would have to use all of the digits between 0-9 apart from 1 which would be remaining. With this they used a process of crossing off any numbers that they used to see what was left over.

Well done!

Can you think of any more interesting word sums?