Two and Two | | DJS

Two and Two

I’m well known for insisting that two and two might not be four. See the essay entitled much the same as this page. No, you may not print that one.

As an end-of-term bit of fun, following on from hexaflexagons, I asked a class to find any of the many solutions to

                  TWO  + TWO  = FOUR   and       T W O  + T W O  =  F I V E

where answers include     734 + 734 = 1468    and     5  3  2  +   5  3  2   =  1 0 6 4.

Assume that no two letters can stand for the same digit. Assume all answers are single digits, so T can’t be twelve. Leading digits (T and F in this case) cannot be zero. Looking for answers, T has to be big enough for the F to be 1 and F can’t be anything else. Immediately, nothing else can be 1. T has to be {5,6,7,8,9}.

For two+two=five, ‘two’ could be entirely odds and ‘ive’ entirely even or vice versa. For two+two=four, the same applies and you want the ‘O’ to re-occur (but not recur). I count just 7 solutions. Common failures claimed as successes have a repeated digit.

The bottom set class I gave this to produced the amused reaction that they’d shown lots of ways of getting Five but none of finding Four. Considering I’ve spent a lot of time telling them that 2+2 is (can be) five, they were greatly amused, but not interested in taking the problem further (I discovered later that they’d asked some older students who said their class found several ‘four’ solutions).

For those wanting to use algebra TWO is 100T+10W + O

This problem and others of the same general type opens a whole field of potential excitement for a class, especially a class that likes pushing buttons on calculators. Success can come from random work, but the most rewards come to those genuinely trying to apply some system or to do some mathematical thinking. Put that another way: a low-satndard class will be excited and occupied and a few of them will find merit in greater organisation, even sharing out some work in a team sense. The rrecording of results is itsdelf an admninistrative problem that some find difficult (“I had a solution but I can’t find it after I pressed equals”).

Let us have some terminology: an integer problem works for the words and as integers; a real problem allows each integer word to have a range of unity  (n-0.5 ≤ N < n+0.5, for integer n and spelling N). My classes much preferred ‘valid' problems, saying that reals committed them to far too much effort and required them to agree with my (per)versions of spelled number. An invalid but fun problem has the possible real values wrong (Two²=Seven, below). One who decided that valid problems were misleading enough to call ‘dishonest' called an invalid problem 'totally dishonest'. It’s all about defining your terminology sufficiently.
For those who find that difficult, 6≤2+5≤8 when you allow a half-unit error on each of the four, but we would not attempt, say “two + five = nine” which possibly has a solution in letters, but is not what we were calling valid.  Those who were ‘really into’ this stayed with entirely honest calculations, so two+five=seven was ruled as test-able. [It fails: The S can only be 1, the F only 9 and so the E must be 0, but then the E of five forces O=N, so no solution. That on its own is pretty good maths at KS3 and 4.]

Staying with integer addition, ONE + FOUR = FIVE is honest and has many solutions suitable for motivating a class into the exploration. I usually see (saw, have seen) several students in each class in which this is attempted begin to grasp that there are sets of possibilities they can reject and conversations spring up around the room using vocabulary one actually likes to hear used in a maths room.

FOUR + FOUR = EIGHT obviously has E=1 and F>4 and solutions include 5239, 5382, 5436 (I didn’t hunt any further, from E=5 and I (the letter) rising.

ONE + ONE = TWO has, my classes tell me, has solutions only where O is 2 or 4. I have eight different values of ONE.

Year Nine found answers to FOUR + FOUR = EIGHT; Soorya says there are (only) two solutions; Zach says there are (only) two solutions to the problem FOUR + FIVE = NINE. I put their answers at the bottom in colour.

I found a solution to THREE + FOUR = SEVEN. At ten unknowns it proved too open for analysis and I tried no further than E=1. The other constraint is from the R. Year Nine found two more.

I think I showed that the similar THREE + FIVE is EIGHT cannot be done (far below). [nor does each of these in words: 5+7=12, 3+8=11, 3+9=12] It is quite instructive for students to show that some such problems have no solution, another such is FOUR + EIGHT = TWELVE (too few digits).

It is also instructive when someone finds a second solution to any problem (“So now we know there’s more than one solution, how many are there? Can you count them / write them out / find them all?” “How do you know there are no more?”).

Not yet tackled (future classes) are the the integer problems including 1+7=8, summing to ten, nor the counting of solutions to the others here. All other additions fail up to 9+10, but I have not yet tried the teens for addition. I’m thinking surely  ..teen+ ..teen= t… ty… won’t often work – but I need to check this.

Moving on to valid / real problems and introducing multiplication, the best problems have few solutions, such as this multiplication:

TWO x TWO = THREE    (which it can, if 1.5 < “TWO” < 1.75).

This has only one solution. It can be seen early that TWO < 317, else TWO² >10⁵  and since √2x104 = 141 we conclude 100 ≤  TWO < 141.  So T=1, and because we can’t have repeated digits, 10²≤TWO≤139 already. Squares ending in 0, 1, 5 & 6 end in that same digit (making O=E), so O can only be in the set {2,3,4,7,8,9}. W≠O  reduces the search (already) to only 16 possibilities. So you can easily be exhaustive, which constitutes a proof.

Multiplication proves to be a rich source of such problems / puzzles. I have been far from exhaustive, but offer some results of a weak week at school (end of term issues across the campus, erratic numbers in class, no possible planning, all planning invalidated; you know the routine).

ONE x ONE = THREE proved not only manageable but helpful. E must be zero and 14<ON<32 or there are too many digits, which leaves only four cases to be found.

Incidentally, for those itching to complain that this is an invalid problem, I agree that ‘1’ 2 ≠ 3, since 2.5≤‘3’<3.5 therefore 1.58<√(’3’)<1.87, both of which round to ‘Two’ to one sig.fig. These issues were discussed on another page, possibly Precision 1 Q6&7.

ONE x TWO = THREE also has solutions. I continue to have useful argument with classes over whether this is a ‘valid’ problem; 0.5 ≤ ONE < 1.5 ≤ TWO ≤ 2.5, so the product lies between 0.75 and 3.75, so arguably between One and Four, once we’ve agreed what we mean by our terminology (opportunity for a “Simon says” - type game, if Simon will play). For this problem the repeated E continues to help (and hinder); I stopped at finding a second solution, having decided that O=1 was required.

I passed out the challenge to do this in other languages, but have no response as yet. Numbers in Pinyin stay short and so it doesn’t work. I seem to remember showing that there were no solutions in addition in French, but offer that as a challenge in the hope it prompts some correspondence.

Looking for valid or ‘honest’ statements, we found some were impossible, such as

FOUR x FIVE = TWENTY (max length of answer is five digits, not the six of twenty).

I found one solution to TWO x SIX=TWELVE but cannot show it is unique except by being exhaustive - I failed to find a second solution but didn’t try terribly hard; it took me long enough to find just the one solution. The TW repeat is an effective constraint.

I eventually explored the ‘teen’s pretty thoroughly using Excel to test .ee. endings. I found a single cube for nineteen, exactly one possible fifteen [TEN³=FIFTEEN]. This is an dishonest problem, since ³√15 = 2.47  and that is not in the ‘honesty range' of 9.5<’10’to one sig fig<15.

I found no possible solutions for the other teens.

I have spent ages chasing three x six = eighteen with as yet no success nor proof it can’t be done. (12633 x 987 = 1246771 is my closest yet)

I’d love to show a solution to these:

ELEVEN x TWO = TWENTYTWO  (the two is possibly 120 or mayhap 160)

FIVE x SIX = THIRTY                     (all ten digits required).


Multiplication needs checking for possibility, probably up to 10x10.


Reverting to standard algebraic understanding, we have had problems with understanding factorising recently (the exams, you know) and I wanted some fun; these problems revert to treating the letters of a word as being multiplied as in ordinary algebra so we tried nominating long words, one of which resulted in:


We also had           (idiot +nit) / twit = (ido+n)/tw          (“I don’t win”, apparently)

With any response, this algebra would become a separate, third, page for end-of-term ideas.

DJS 20130617-23

I offer this challenge (homework): for you to make / create an algebraic expression of the form (a+b) / c = d or =d/e where each of (a b c d e} is a word or decherable as having meaning.  I’d love it to have the result being something like I2M / U  (I’m over you). The best I’ve managed in two minutes is 


A minute later I have 


(or the denominator might be HALT MANY ARROUSD); 


HALT YO MAN ARROUSYD        YOU           is achieved.


[2005-2007 I was waiting for divorce to occur, not wanting it at all; I suspect this was written in that period]

revisit, one minute in,        aol is diy)/cool holidays                             (  = is over school?)

rate of tapping foot / past point of 0 rages = ??                      (ft/s,  a speed of tapping?)

How about         (bridges + beers ) / (sweat + teaser)  ??      (An oldie !!)

Soorya says FOUR is 8352 or 7293

Zach says FOUR + FIVE is 1960+1475 or 2970+2586

TWO&TWO is FOUR:  734, 765, 836, 846, 857, 867, 928, 938,

FOUR + FOUR = EIGHT:  FOUR = 5239, 5382 and several more...

THREE + FOUR has ten digits and solutions include these THREEs: 36255, 26455, 17544

THREE + FOUR has ten digits and includes 36255+8902=45157, 26455+8704=35159, 17544+6805=24349

Invalid problems:

WON2 = SEVEN                     (no it isn’t, minimum value is 1.87, answer 241)

TWO2 = SEVEN                     (no it isn’t, minimum value is 2.55, TWO=192, 237, 307)

SIX2  = SEVEN                      (no solution, shown, though SEx = 103)

WOT3 = NINETEEN               (minimum value of 191/3 is 2.65; only one possible NINETEEN)

TEN3 = FIFTEEN               (minimum value of 151/3 is 2.47; only one possible FIFTEEN)

√SEVEN has ten possibilities where the E fits  and S≠V≠N

10609, 21316, 24649, 26569, 36864, 42025, 56169, 58081, 90601, 94249,

Nineteen has only one possible value for cubes, 29218112 = xyT3. I chose an anagaram of TWO 

Less (evem less) well spelled, and found while hunting for ...TEEN solutions, we have




SON3 = THETEEN  one solution only.

After some hours, three x six = eighteen still has no solutions for E=1

DJS 20130623

TWO x TWO = THREE      1382 = 19044

ONE + ONE = TWO    ONE = 231, 276, 281, 427, 432, 457, 467, 482.

THREE + FOUR = SEVEN  32611 + 8906 = 41517

ONExONE=THREE: ONE = 170, 180, 240, 290 (not 230 ‘cos of the unwanted 2)

ONExTWO=THREE includes 108x461 and 104x561

SIX x TWO = TWELVE    986x345=340170

WOT = 308,      TEN3=FIFTEEN 1763=5451776

SITTTEEN=70444997, AENYTEEN = 35287552, XISSTEEN = 95443993, WON =241,

THREE + FIVE = EIGHT has (a) E+E=T or (b) 2E=T+10 AND (c)E=T+1. So a&c imply that 2E = T=2T+2 => E=-1, T=-2  (rejected), while b&c imply that E=9, T=8. Great: but now V=H is forced upon us, making the problem insoluble (unsolvable).

BR(IDG+E) / AT(W+ER) = bridge over troubled water

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