Alphametics help please!

ValHaller

Mr_Toad wrote: »

E+E+U = E for that to be true U would have to = 8

I don't agree with this

E + E + R could carry 1 out.

So only 1 of these is true

** E + E + U = E
** E + E + U + 1 = E

Your logic of your contention

Mr_Toad wrote: »

Actually you can rule it (that E = 2) out and the logic is so simple I didn't include it.

is far from simple.

FBaby

Fancy giving something this complex to kids who've been at secondary school for 6 weeks or so!

Unless there are clear rules to follow to get to the answer, and these rules were learnt at school, I would have thought this homework was nothing more than a case of 'give it a try guys, but don't waste your energy on it, you are not expected to find the answer'. Don't pester your parents about it either or you will find them so absorbed, you might lose out on your evening meals for the rest of the week'!

Mr_Toad

ValHaller wrote: »

I don't agree with this

E + E + R could carry 1 out.

So only 1 of these is true

** E + E + U = E
** E + E + U + 1 = E

Your logic of your contention

is far from simple.

You're thinking conventionally and starting from right to left as in normal maths. If you carry on you'll eventually get to the point where it all falls apart and E can't = 2 and this is a perfectly valid method of solving this. All I've done is skip this avenue of exploration.

You are right that E+E+R could carry one but for that R would need to be at least 6 or more. It's at this point you need to start looking at the other end of the puzzle and think about the carry numbers and what you need. And what you need are the larger numbers.

So with this in mind I'm starting at the other end with the EL if we make the assumption that E=2 then ELE = 2?2 and to achieve this the left most T+T+the carry must = 20 at the very least to get the 2 in the left hand column. Again, the trick is to look at what you need to carry from each column.

T=9 would work. T=8 would need a carry of 4 to make 20 and you can't do that with three single digit numbers even if they were all 9!

Also the second column HAS to add up to 22 to give the E in the answer and the 2 to carry.

Using [STRIKE]simple[/STRIKE] logic working from left to right and concentrating on the carry you need the larger numbers and we've already used 9 and if we use your logic then 6 is also used so we have 8, 7 & 5 Can these be used to generate the carry numbers we need AND meet the other criteria of the puzzle?

The answer is no, it works so far then falls apart.

You could just keep questioning my logic saying you don't agree or you could have a go with your logic and see if you prove the E can =2

I've worked out four solutions that work and in all of them E=1

I think part of the problem is that I look at something like this puzzle and do most of what I've tried to write down in my head. I've dismissed E=2 and wasted no time trying to prove it can, apart from these posts.

It's like a crossword, some people can do them in no time, others take ages and have to work very hard to come up with the answers. I can't explain, not very well anyway, how I do these things I just can and to me the logic process is simple, whether that is true or not I can find the solutions.

Spendless

I also think it was homework along the lines of 'see if anyone can do this' rather than 'official' homework that they'd be in bother for not doing. My 2 years older son has had occassional homework along the lines of cracking a code, not that exact example though. Whether he does it depends on how easy he finds it - mixed in with whether he's listening in the first place! Glad The Arms could help out though, just as well you din't ask on Fri evening when they're all drunk. The cheeky replies gave me a smile.:D

ValHaller

Mr_Toad wrote: »

You're thinking conventionally and starting from right to left as in normal maths. If you carry on you'll eventually get to the point where it all falls apart and E can't = 2 and this is a perfectly valid method of solving this. All I've done is skip this avenue of exploration.

You are right that E+E+R could carry one but for that R would need to be at least 6 or more. It's at this point you need to start looking at the other end of the puzzle and think about the carry numbers and what you need. And what you need are the larger numbers.

So with this in mind I'm starting at the other end with the EL if we make the assumption that E=2 then ELE = 2?2 and to achieve this the left most T+T+the carry must = 20 at the very least to get the 2 in the left hand column. Again, the trick is to look at what you need to carry from each column.

T=9 would work. T=8 would need a carry of 4 to make 20 and you can't do that with three single digit numbers even if they were all 9!

Also the second column HAS to add up to 22 to give the E in the answer and the 2 to carry.

Using [STRIKE]simple[/STRIKE] logic working from left to right and concentrating on the carry you need the larger numbers and we've already used 9 and if we use your logic then 6 is also used so we have 8, 7 & 5 Can these be used to generate the carry numbers we need AND meet the other criteria of the puzzle?

The answer is no, it works so far then falls apart.

You could just keep questioning my logic saying you don't agree or you could have a go with your logic and see if you prove the E can =2

I've worked out four solutions that work and in all of them E=1

I think part of the problem is that I look at something like this puzzle and do most of what I've tried to write down in my head. I've dismissed E=2 and wasted no time trying to prove it can, apart from these posts.

It's like a crossword, some people can do them in no time, others take ages and have to work very hard to come up with the answers. I can't explain, not very well anyway, how I do these things I just can and to me the logic process is simple, whether that is true or not I can find the solutions.

No I am not thinking conventionally. Like you I started at the left.

But when you started at the left, you immediately - and wrongly - jumped to a conclusion that E MUST be 1. Now in the event it probably IS 1, but for where you got to and with the logic you have used, you need to consider that it COULD BE 2.

It is legitimate to proceed as though E = 1 and drive this to a solution at which point you can say E = 1. But that is not to say that E could not be 2 - there may be other solutions

All I am saying on E = 2 is that while you might well be right to rule it out at some stage, the very first argument you used to show that E = 1 actually showed that E = 1 or E = 2

suejb2

This thread seems to have gone off on a tangent (see what I did there?) It seems to be a match between Mr Toad And Vallhala.......Next service please?

ValHaller

suejb2 wrote: »

This thread seems to have gone off on a tangent (see what I did there?) It seems to be a match between Mr Toad And Vallhala.......Next service please?

I would not call it a tangent. It is very much focussed on the initial problem.

Mr T has indeed got a solution, but unfortunately his logic along the way is not good, so it is a bit misleading for anyone who is hoping to learn how to crack this class of problem.

Mr_Toad

ValHaller wrote: »

Suppose THREE = 98722 and FOUR = 6054

98722 + 98722 +6054 = 203498

E could easily be 2

OK new direction. So far we've been concentrating on my logic and I've been happy to do so because it helps me visualise my thought process.

The above example is wrong it is not a solution. Valhalla has FOUR=6054 and the answer as 203498 the numbers work but not the letters as there are two letters for 0

You can't have the O=0 in FOUR and the L=0 in ELEVEN or as in this example E=2 and 3 and 9!

If the above is the basis for Valhalla's logic that E could be 2 then that logic is clearly flawed.

I may not have expressed myself and my logic clearly but it works the above doesn't.

ValHaller

Mr_Toad wrote: »

ValHaller wrote: »

Suppose THREE = 98722 and FOUR = 6054

98722 + 98722 +6054 = 203498

E could easily be 2

OK new direction. So far we've been concentrating on my logic and I've been happy to do so because it helps me visualise my thought process.

The above example is wrong it is not a solution. Valhalla has FOUR=6054 and the answer as 203498 the numbers work but not the letters as there are two letters for 0
It was never proposed as a solution.

You can't have the O=0 in FOUR and the L=0 in ELEVEN or as in this example E=2 and 3 and 9!

If the above is the basis for Valhalla's logic that E could be 2 then that logic is clearly flawed.

I may not have expressed myself and my logic clearly but it works the above doesn't. It was never intended to work as a solution, it was intended to show that your statement E = 1 is not supported by the logic you used.

It was never proposed as a solution!!! Will you please accept that it was never intended as anything more than a counterexample to your initial argument below

Mr_Toad wrote: »

Your starting point is that E = 1 since the only carry from adding T + T is 1 given that the largest number T can be is 9 and 9 + 9 = 18

The starting point cannot be simply E = 1 based on your argument. The carry from T + T can be 2. So what you define as the starting point should be E = 1 or 2.

Clearly you are too wrapped up in your solution to understand the point which I am making.

go_cat

Has anyone else read this and not understood a word of it