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27/12/2014 Two and Two : nrich.maths.org

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Two and TwoStage: 2 and 3 We received lots of great solutions for this problem, unfortunately too many tomention you all by name! Many of you found that there were up to seven possiblesolutions, by systematically eliminating possibilities well done. Here is an answersubmitted 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 is19 (if you have already carried 1 from the previous column). This also means thatT ≥ 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,

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

If O = 3,

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

If O = 4,

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T W 2+ T W 2− −−−−−−1 2 U R− −−−−−−

T W 3+ T W 3− −−−−−−1 3 U R− −−−−−−1

T W 4+ T W 4− −−−−−−1 4 U R− −−−−−−



then R = 8 and T = 7 and we also know that W < 5 because there can’t beanything 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,

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

then R = 2 and T = 8 and we also know that W < 5 because there can’t beanything 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 = 1672If W = 4, U = 9 which works. 846 + 846 = 1692

If O = 7,

then R = 4 and T = 8 and we also know that W ≥ 5 because there has to be 1carried 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 = 1734W can’t be 7 because O = 7.

T W 5+ T W 5− −−−−−−1 5 U R− −−−−−−1 1

T W 6+ T W 6− −−−−−−1 6 U R− −−−−−−1

T W 7+ T W 7− −−−−−−1 7 U R− −−−−−−1 1



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,

then R = 6 and T = 9 and we also know that W < 5 because there can’t beanything 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,

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=1876928+928=1856867+867=1734846+846=1692836+836=1672765+765=1530734+734=1468

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

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

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

T W 8+ T W 8− −−−−−−1 8 U R− −−−−−−1

T W 9+ T W 9− −−−−−−1 9 U R− −−−−−−1 1



EIGHT

Their solutions are:8523 + 8523 = 170469235 + 9235 = 184709327 + 9327 = 186548652 + 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, forexample, would any of the letters have to be 5 or more and if so, could thecarrying of a digit help achieve another number which has not already been used?They found that because there were no repeated letters between FOUR andEIGHT, then that they would have to use all of the digits between 09 apart from 1which would be remaining. With this they used a process of crossing off anynumbers that they used to see what was left over.

Well done!

Can you think of any more interesting word sums?

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