Alphametics help please!

Kaz2904

Hi there, I'm begging here .
DD is in year 7. Her maths homework included a question which I have discovered is an alphametic puzzle. I've never heard of these.
The question is this:
Wilma has changed Wilbur's numbers into letters. Can you replace each letter with a digit to make his addition correct?

THREE
THREE
+FOUR

---

ELEVEN

My friend gave me this link http://mathforum.org/library/drmath/view/56789.html which explains how to solve it but I don't understand it. I am fairly competent at maths and manage to explain most things to her albeit with having to look it up online first sometimes. This has me stumped.
I have spent 2 hours on it so far and have not got a flipping clue.

Is there anyone out there who is able to explain it to me in very simple terms? I understand that there are rules and I understand that the numbers on the left cannot be 0. That's the limit for me. I *think* that T is greater than or equal to 5 but I can't get my head around this.
How do I eliminate letter number combinations?
This is supposed to be homework so I can't see how it is meant to take hours and hours to solve .
I hate school! (I'm kidding but this is making me feel like a dimwit ).

Fag_ash_lil

I wouldn't have a clue either !

If you don't get any help on this thread .... you might have more luck in The Money Savers Arms at the bottom of the board. I know there has been threads like this on there before where people have been able to help out .

Hope you manage to get some help .

Kaz2904

Oh thanks very much. I'm going to go there now and link to this thread to see if anyone can help me/us.
DH is now sitting scratching his head too!

hazyjo

Not able to help, but have you tried switching the existing letters for others? I'm so far working with a-j (each of which will be 0-9). So:

a b c d d
a b c d d
e f g c
d h d i d j

Then making a list like this:

0 1 2 3 4 5 6 7 8 9

and writing each possible letter it could be underneath, ie:




0
1
2
3
4
5
6
7
8
9





a
a
a
a
a
b
b
b
b
b
b
b
b
b
b
c
c
c
c
c
c
c
c
c
c

d
d
d
d
d
d
d
d
d

e
e
e
e
e
e
e
e
e
f
f
f
f
f
f
f
f
f
f
g
g
g
g
g
g
g
g
g
g
h
h
h
h
h
h
h
h
h
h
i
i
i
i
i
i
i
i
i
I
j
j
j
j
j
j
j
j
j
j



Really not sure if that helps you move any further on (not me!) - maybe you can eliminate some numbers? We know that a+a= double figured number as there are 6 numbers in the TOTAL row. Therefore, 'a' can't be under 5.

Totally stumped as to the rest of the eliminations, but wasn't sure if this might make it easier!

My brain won't handle the rest, but I'm guessing that because d+d+c=j and d+d+g=d must mean that there was a number added from the last column, so think 'd' must be above 4, but don't take that as gospel lol!

Good luck!

Let us know when you find out!

Jx

hazyjo

Boll....s - it lost the table I copied. Hope you can work out what I meant!

Jx
PS as you said, we know that 0 can't be a, b, d or e.

Mr_Toad

Kaz2904 wrote: »

Hi there, I'm begging here .
DD is in year 7. Her maths homework included a question which I have discovered is an alphametic puzzle. I've never heard of these.
The question is this:
Wilma has changed Wilbur's numbers into letters. Can you replace each letter with a digit to make his addition correct?

THREE
THREE
+FOUR

---

ELEVEN

My friend gave me this link http://mathforum.org/library/drmath/view/56789.html which explains how to solve it but I don't understand it. I am fairly competent at maths and manage to explain most things to her albeit with having to look it up online first sometimes. This has me stumped.
I have spent 2 hours on it so far and have not got a flipping clue.

Is there anyone out there who is able to explain it to me in very simple terms? I understand that there are rules and I understand that the numbers on the left cannot be 0. That's the limit for me. I *think* that T is greater than or equal to 5 but I can't get my head around this.
How do I eliminate letter number combinations?
This is supposed to be homework so I can't see how it is meant to take hours and hours to solve .
I hate school! (I'm kidding but this is making me feel like a dimwit ).

78011
78011
+5390

---

161412

This is one solution there are probably more. 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

So you then have

THR11
THR11
+FOUR

---

1L1V1N

and so on......

ValHaller

Kaz2904 wrote: »

THREE
THREE
+FOUR

---

ELEVEN

To kick this off:

Adding 2 5 digit numbers can only give another 5 digit number or a 6 digit number beginning with 1. Adding a 4 digit number might make a 6 digit number beginning with 1 or 2

So E from 'ELEVEN' is either 1 or 2


Next

EE + EE + UR = ?EN

So there must be a carry, because

• assuming E = 1 then 11 + 11 + UR must yield a 1 in the 10's
• assuming E = 2 then 22 + 22 + UR must yield a 2 in the 10's

In either case, the UR must generate a carry so that the sum is in the 110's or the 120's

The UR of 'FOUR' must be greater that 66.

Well, it's a start.

puppypants

Whoosh!!! Over the head! xx Good luck! xx

ValHaller

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

Suppose THREE = 98722 and FOUR = 6054

98722 + 98722 +6054 = 203498

E could easily be 2

cutestkids

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.

Kaz2904

We've worked it out on the other thread http://forums.moneysavingexpert.com/showthread.php?t=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!