Alphametics help please!

Mr_Toad

ValHaller wrote: »

Suppose THREE = 98722 and FOUR = 6054

98722 + 98722 +6054 = 203498

E could easily be 2

No it couldn't.

If E = 2 then the answer ELEVEN would be 2L2V2N and your answer is 203498 which means you have E = 2 and E = 3 and E = 9 which clearly is impossible!

Kaz2904

I'm just going to copy my other post over here too so that you can see my ramblings. If you're interested.


Right, I have got one solution.
I started in the right column.
We know that E=1.
I then did the same as a pp and started working my way through R=?.
I got to R=5 which means N=7.
That means that the number in the next column needs to =11.
11-2=9 so U=9.
I knew the next column was 5+5 (+1carried)+?= Not 0.
That means L=0 because T and F can't as they're leading numbers.
So now we have
THREE
??5 1 1
THREE
?? 5 11
FOUR
?? 95

---

ELEVEN
101?17

I'm left with 8,3,4&6 to fit in.
T must equal 10 and the 0 is used so it can't be 5. If it's 8, it's too much, same with 6 so it must be 4 so means I need to have carried 2 over from previous column so H+H+F=21 (with anything carried).

I hope this makes sense. I don't seem to be able to explain it to DD and she doesn't understand her Dad's explanation.
What a question for homework eh?

Mr_Toad

Kaz2904 wrote: »

We've worked it out on the other thread https://forums.moneysavingexpert.com/discussion/4808089
But to answer ValHaller, your answer doesn't work because you have used 4 for 2 different letters, your E's for ELEVEN are made of different numbers and you haven't used 1.
This is bonkers and I doubt I could solve another without step by step written instructions. And many hints.
I got a different answer to DH. I had 46511+46511+8295=101317 and he got 78011+78011+5390=161412

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

No your answers are fine there is often 3 or 4 correct answers, you only have to find one of them and as long as it works it's OK.

I did say that my solution was one of several probables.

What can be really tricky is what the call a double true alphametic where not only do the numbers add up but the words do too.

THREE
THREE
+FOUR

---

ELEVEN

isn't a double true because 3+3+4=10 and not eleven

THREE
THREE
TWO
TWO
ONE

---

ELEVEN

is a double true.

suejb2

My D.D is in year 6 so I best be prepared for this sort of homework,but like somebody asked really what is the point of it

ValHaller

Kaz2904 wrote: »

But to answer ValHaller, your answer doesn't work because you have used 4 for 2 different letters, your E's for ELEVEN are made of different numbers and you haven't used 1.

That was NOT an answer or even meant to be. The very first word of my post was 'Suppose'.

It was to answer Mr Toad's contention that E = 1 and I showed that at the stage he was at, you could not rule out E = 2

ValHaller

cutestkids wrote: »

Can someone tell me what the point of this type of maths is. I really do not get it and don't understand how it will ever be used in real life,
So is there any actual point to it or is it just one of those things they think up to drive ordinary people round the bend.

There is no direct practical use for 'Alphamatics'. Personally, I detest this class of problem intellectually. But there is a lot of value in the necessary skills of extracting information from the problem by application of understanding a system of rules - arithmetic - and constructing logical arguments leading to a conclusion.

Mr_Toad

suejb2 wrote: »

My D.D is in year 6 so I best be prepared for this sort of homework,but like somebody asked really what is the point of it

Almost everything we do and use is governed by mathematics but for the most part we are shielded from it in everyday life.

All the maths we learn at school is as far as goes for many people but if you're and engineer, architect or computer programmer then you use the stuff we learn at school and a lot more besides, every day.

One of my closest friends is a surveyor, he works with angles and lengths all day and uses trigonometry to make sense of it.

I was a computer programmer and used complicated encryption algorithms on a daily basis. At one point I was writing code that did complex navigational calculations. My notebooks are covered in algebraic formula, calculus and complex calculations.

Some children can play Mozart by the time they are 5, some are gifted footballers, singers or athletes. Others can make the written word dance on the page and some can make numbers do magical things. The only way to find these people is at school while they are young and encourage them to take it to the next level or as high as they want or can go.

Mr_Toad

ValHaller wrote: »

That was NOT an answer or even meant to be. The very first word of my post was 'Suppose'.

It was to answer Mr Toad's contention that E = 1 and I showed that at the stage he was at, you could not rule out E = 2

Actually you can rule it out and the logic is so simple I didn't include it.

THREE
THREE
FOUR

If E = 2 then E + E = R and R is 4 with no carry if R is 4 then U can't be 4 therefore E is NOT 2 and there aren't enough rows to ever carry 3 so E = 1

ValHaller

Mr_Toad wrote: »

Actually you can rule it out and the logic is so simple I didn't include it.

THREE
THREE
FOUR

If E = 2 then E + E = R and R is 4 with no carry if R is 4 then U can't be 4 therefore E is NOT 2 and there aren't enough rows to ever carry 3 so E = 1

Where do you get E + E = R from? There is no immediate result giving a sum of E + E, so you need to provide a bit more explanation

Mr_Toad

ValHaller wrote: »

Where do you get E + E = R from? There is no immediate result giving a sum of E + E, so you need to provide a bit more explanation

Sorry missed a row off, I got distracted.

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

If E=2 the T+T+ the carry would have to be 20 so 9+9=18 so the carry needs to be 2 which means H+H+F must = 22 we've used up the 9(T) and the 8(U) so the most H can be is 7 so 7+7+F = 22 so F = 8 which we can't have as U=8 and we can't have F=7 with a carry of 1 because H=7