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11/2/2009 11:41:33 AM

ddub
ddub
Posts: 1
I do not know how to begin a Word Arithmetic puzzle. Any suggestions from those of you who are familiar with this type of long division, lettered puzzle?

2/21/2010 3:13:04 PM

jeffloeb
jeffloeb
Posts: 77
Also logic helps, Like e - A = E means A=0 (9 if there is a carrt)

A * BCDE = BCDE generally means that a = 1.

Jeff

2/21/2010 3:14:06 PM

jeffloeb
jeffloeb
Posts: 77
oops. I need 2 watch my typing, I meant, CARRY.

8/13/2010 4:03:36 PM

ThrillerFan
ThrillerFan
Posts: 6
I find Word Arithmetic the most boring of the puzzles in LLM&LP because of their simplicity.

Here are a number of rules to follow:

- For each location where subtraction takes place during your long division problem, any instance where the last digit is something like A - T - A or A - A = T, T must be 0.

- For each instance of A - A = T or A - T = A in any other location, T must be 0 or 9. If T is the first digit of ANY NUMBER in the entire problem, it's 9, not 0.

- Any instance of the divisor being repeated in the problem, the number up top that corresponds to that is a 1

- When doing subtraction, if the 2nd number that you are subtracting from the first number is a digit shorter, the first digit of the first number MUST be 1.

- When doing subtraction, when looking at the last digit, let's say it's B - E = L, if E is an odd number, then the number up top that corresponds AND the divisor MUST BOTH BE ODD as well (since that number was derived from multiplication, and multiplying 2 even numbers or an even and an odd always result in an even number.

- Make a list of ALL letters that are the first digit of any number within the entire problem. NONE of these letters can be 0.

- When doing subtraction, let's say you have CRA - BRE = FSC, from before, you know that S is 9 or 0. If you already know that C and E add up to more than 10, then you know that you had to borrow from the R, and that the A itself wasn't enough when subtracting. Because you had to borrow, you now know that S is 9. Conversely, if E was say, 1, then you know that C is not 9 because if it was, then A would be 0 and you'd have to borrow. Because of that, S would have to be 9, but it can't be because C was already 9. Therefore, C can't be 9, and A can't be 0, and so since the R is NOT borrowed from, then R - R = S and S is 0.

If the quotient is 3 digits or more in length, and the 3 numbers you achieve via multiplication (that are later subtracted) all end in a different letter, then you know the last digit of the divisor is NOT 5 or 0.

In general, the easiest numbers to determine first (based on the problem, varies from problem to problem) are 0, 1, 5, or 9.

Hope this helps.

3/5/2011 11:26:31 PM

twhit
twhit
Posts: 2
dang, thrillerfan, you just confused me and i do these all the time!!

make a list of all the letters (you should have 10). find the easy things first:
abcde
fghik a =1

sic/abcdef
baf
rge
rhe
is s=0

then look at how your letters may be placed based on the basic math:

ann/lines
ann
yyge
unei
auus U comes immediately before Y within the word

also you can use simple anagram logic with your clues you already have and test it in the puzzle to see if the math also works.

once you find what works for you, you will be able to solve these with no problems.

hope this helps.

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