# Two and Two

How many solutions can you find to this sum? Each of the different letters stands for a different number.

*Two and Two printable worksheet*

Each of the different letters below stands for a different number.

**How many solutions can you find to this cryptarithm?**

**How can you be sure you have found them all?**

**Can you create other similar cryptarithms?**

Here are some suggestions to start you off.

**ONE + ONE = TWO**

**ONE + TWO = THREE**

**ONE + THREE = FOUR**

**FOUR + FIVE = NINE**

**Can you make a cryptarithm subtraction?**

*If you enjoyed this problem you may like to take a look at **Cryptarithms*

Try some different values for **O** and see what happens.

What must **F** be?

What does that tell you about **T**?

Well done to everybody who found a solution to this activity. Some of the solutions we received used the same number for more than one letter, so if you think you've found a solution that isn't mentioned here then take another look at it to see if you might have made a mistake with that.

Ferdia from Clifton Hill Primary School in Australia explained how to tell when a solution was impossible:

You can tell that some will be impossible because some of the o's are different

Good idea, Ferdia - because the letter O is in TWO and FOUR, that's a very useful way to check whether or not a solution is going to work.

P A from Lexden Primary in the UK explained their ideas:

We used number cards to systematically work out the problem as part of a mastery group. We found that if we carried into the tens column U had an odd number. We found that the last number needed to be even, but currently we are unsure why.

So we found 734 + 734 = 1468 worked and 836 + 836 = 1672.

Well spotted - if O is large, then U will end up being W+W+1, rather than just W+W, because of the carrying/exchanging. Lots of the solutions said that the last digit in TWO had to be even, but there are actually some solutions where it is odd instead.

Imeth from Pristine Private School in Dubai, UAE found the solution 928 + 928 = 1856 and explained their reasoning:

Since 0+0=0, nothing can be zero, as every number doesn't link to itself. 1234/2 = 617, so TWO>617. Since 2T = FO, F, and therefore O is even. Let's say it's 8. Since 18/2 = 9, T=9, F=1, and O=8. Two eights are 16, carry the one, so R=6. You can test different values of W, but let's say it's 2. 2+2+1(carried)=3: U=3. This is just one solution: O can be any even number except 0. But 928 + 928= 1856 is still one.

I think if there was no carrying/exchanging, then you'd be completely right, Imeth, and the solution you've found is correct. But I think we could have T+T+1=FO, so O doesn't necessarily have to be even. I agree that TWO has to be at least 617 in order for the solution to be a four digit number with four different digits!

Akshobhyan from Ganit Kreeda, Vicharvatika, India had some similar ideas. Their teacher said:

Akshobhyan noticed that two 3-digit number’s sum is 4-digit means hundredths place is generating one carryover. That means

T is greater than 4, or T = 5 /6 /7 /8 /9

And F=1.

O can’t be 0, as 0+0=0 and O and R cannot be both 0.

R can only be even. So, R can be either 2,4,6 or 8.

Well done, Akshobhyan! Your reasoning is all correct here, so we know that O definitely can't be zero, and R has to be even.

Shaunak from Ganit Manthan, Vicharvatika, India found four different solutions: 734+734=1468, 836+836=1672, 928+928=1856 and 867+867=1734. They explained their reasoning in this video:

Thank you, Shaunak, this is very clearly explained.

Youyou from Sha Tin Junior School in Hong Kong found five different solutions:

When we take a closer look at the problem, F in FOUR must equal 1, since no two 3-digit numbers add up to equal or exceed 2000. Also, T in TWO must be higher than 4. No 3-digit number lower than 500 would be multiplied by 2 and equal 1000 or higher.

But trial and improvement shows that T must be higher than 6, because no combination where T equals 5 or 6 satisfies the conditions of TWO + TWO = FOUR. Therefore, the 3-digit numbers in TWO + TWO = FOUR combinations must be higher than 699.

I’ve found 5 possible combinations: 734+734=1468, 765+765=1530, 836+36=1672, 867+867=1734, and 938+938=1876. 969+969=1938 does not work, so those are all the solutions.

I agree with everything you've said here, Youyou, but I think there are a couple more solutions that you haven't spotted. One of the other solutions is one that Imeth and Shaunak found in their solutions above (928+928=1856). I wonder what the last one is?

Arjun from Canadian International School Lakeside, Singapore had a different way of recording their ideas:

So basically the table in the image shows different possibilities of what the letters value could be. So in order to interpret and understand it you have to read it like a column. So if O=4 then R is 8 and then so on below in the column. So what I did was I tried different numbers for O and I realised that the letter T has to be five or above because the alphanumerics/ cryptarithm is 3 digit plus 3 digit is a four digit number. And then F=1 because the maximum number that T plus T could add up to is 19 so F will always be one. And then I tried different numbers for O and then found that the numbers repeat until O=4 where it can fit and then from there onwards until 8 it works because if you try O=9 then O is the same as T so it repeats the same number.

This is an interesting way of looking at it, Arjun! These solutions are all correct but there are two solutions that you haven't found. (Interestingly, these are the same two solutions that Youyou missed out, in their solution above.) When O=6 and when O=8 there are two possible solutions.

Jeremy and Lucas from Sha Tin Junior School in Hong Kong found all of the possibilities for each digit:

F can only be 1.

W can be 2, 3, 4 or 6.

O can be 4, 5, 6, 7 or 8.

U can be 3, 5, 6, 7 or 9.

R can be 4, 6 or 8.

T can be 7, 8 or 9.

Well done - these are all correct, but I wonder how they fit together to form the seven possible solutions? You might like to take a look at Jeremy and Lucas's full solution to see more of their ideas.

Eleanor and Stan from Horfield CEVC Primary School in the UK found every solution and explained their reasoning:

We have found seven solutions to this problem which are:

1. T = 7, W = 3, O = 4, F = 1, U = 6 and R = 8

2. T = 7, W = 6, O = 5, F = 1, U = 3 and R = 0

3. T = 8, W = 3, O = 6, F = 1, U = 7 and R = 2

4. T = 8, W = 4, O = 6, F = 1, U = 9 and R = 2

5. T = 8, W = 6, O = 7, F = 1, U = 3 and R = 4

6. T = 9, W = 2, O = 8, F = 1, U = 5 and R = 6

7. T = 9, W = 3, O = 8, F = 1, U = 7 and R = 6

To find all seven possible solutions we used the following reasoning:

- F must be 1 because if two identical three-digit numbers are added together in order to get a four-digit answer, the thousands digit must be 1. No other letter can be 1.

- O has to be 4, 5, 6, 7 or 8 for it to have the same value in all three places.

- T, W and O cannot be zero.

- Zero can only appear as R when O is 5.

- T can only be 7, 8 or 9 in order to carry a '1' to the thousands column when two Ts are added together (and make sure that O is the same in all three places).

Well done, both of you - you've thought really hard about this!

Some children also had a go at the second part of this activity. Imeth explained their ideas about which cryptarithms would be possible, and why:

As for the second part, you can use similar logic, but for ONE + TWO = THREE, two 3-digit numbers add to a 4-digit at maximum. ONE + THREE = FOUR is also impossible: a 5-digit > 4-digit, and there's no such thing as negative digits.

Youyou from Sha Tin Junior School had some similar ideas, and also came up with some of their own cryptarithms:

Meanwhile, ONE+ONE=TWO is possible.

ONE+TWO=THREE is impossible. It is impossible for a three-digit number to be multiplied by two and equal to a product higher than 9999.

ONE+THREE=FOUR is also impossible. Five digit numbers are not lower than four digit numbers.

FOUR+FIVE=NINE is possible.

Cryptarithm subtractions can be composed and said to be possible if they comply with the rules of arithmetic. Here are some possible cryptarithm subtractions:

TWO-ONE=ONE

FOUR-TWO=TWO

NINE-FIVE=FOUR

NINE-FOUR=FIVE

SEVEN-FOUR=THREE

Shaunak from Ganit Manthan also sent in some cryptarithms that they came up with:

If the number of digits in any of the numbers have more digits than possible, the cryptarithm is impossible. For example, in ONE + TWO = THREE, three has more digits than possible, because the maximum value of the sum of two 3-digit numbers is 1998, which is a 4-digit number, and the value for THREE must be a 5-digit number.

Some cryptarithms are ONE + ONE = TWO, FOUR + FIVE = NINE, and POINT + ZERO = ENERGY. There can be many such cryptarithms.

I made a cryptarithm subtraction – EIGHT – SEVEN = ONE.

Thank you all for sharing these ideas with us! We had some similar solutions sent in by Kimberly from Clifton Hill Primary School in Australia, and from the children at Ganit Kreeda, Vicharvatika, India. You might like to take a look at their full solutions to see what their ideas were:

Thank you as well to the following children who also sent in similar solutions to this activity: Uinifareti from Sahuaro, AZ, USA; Tasman, Georgia, Zoe, Lucas, Zoe, Leon and Jack from Clifton Hill Primary School in Australia; Dhruv from The Glasgow Academy in the UK; Kanaa from Ganit Kreeda, Vicharvatika, India; the children from Indooroopilly State School in Australia; Macy, Isabel, Elliot, Izzy, Sammy and Bobby from Richmond Methodist Primary School; Evan from Mountains Christian College in Australia; Tejas, Ruhi, Valerie, Mithravinda, Kabir, Nithya, Deethya, Uday, Udit, Kashvi, Swathi, Arnav and Samarth from Ganit Kreeda, Vicharvatika, India; and Claire and George from Shatin Junior School in Hong Kong.

### Why do this problem?

This problem offers an opportunity to practise addition in a more interesting and challenging context than is usual. It requires students to work systematically, record their progress efficiently and apply their understanding of place value.

Possible approach

*This printable worksheet may be useful:*

*Two and Two*

Write the following sum on the board and ask students to complete it.

8 | 6 | 1 | ||

+ | ? | ? | ? | |

1 | 4 | 2 | 7 |

Expect justifications for any suggestions.

b | b | |

+ | c | b |

a | c | c |

"How can we approach this?"

**TWO + TWO = FOUR**

To finish off, students could present their approach to the rest of the class, with emphasis on explaining clearly why they are convinced that they have found all the possible solutions.

### Key questions

What must "F" be?

What does that tell you about "T"?

Are you certain you have considered all the possibilities?

### Possible support

Suggest students start with Spell by Numbers

### Possible extension

Suggest students find other word sums that work.

Here are some possibilities that they might consider:

**ONE + ONE = TWO**

**ONE + TWO = THREE**

**ONE + THREE = FOUR**

**FOUR + FIVE = NINE**

Why are some impossible?

Can they make a word subtraction?