+0

# Common Sense or Math?

+1
2
1
+178

My school algebra teacher gave us this question, and it stumped all of us, even the generational geniuses in our class (not me).

She told us it had something to do with common sense and not variables, but not one person could figure it out. So here I am, consulting to hopefully find an answer! Thanks!

Here is the question.

If $$\frac{EIGHT}{FOUR} = TWO$$, then what is the value of $$THREE$$

Like I've said, I still have absolutely no idea how to come up with an answer, but if you can, great job!

Thanks!

May 1, 2024

#1
+474
0

To solve the equation EIGHT / FOUR = TWO, where each letter stands for a different digit, we can start by considering the possible values of the letters.

Since we are dividing a four-digit number by a four-digit number and obtaining a three-digit number, we know that E and F must be 9 and 1, respectively, as they are the leading digits in the division.

With this information, let's construct the division:


9 1
-------
T W O


Now, let's consider the possible values for T and W.

Since 91 divided by 4 is not a whole number, T must be greater than 2. Let's try T = 3:


9 1
-------
3 W O


For the division to yield a three-digit result, the number formed by the last two digits of the dividend (1W) must be divisible by 4. Therefore, W must be an even number. The only even digit left is 2.


9 1
-------
3 2 O


Now, let's find the value of O. Since 91 divided by 4 is 22 with a remainder of 3, O must be 3.


9 1
-------
3 2 3


Therefore, the solution to the equation EIGHT / FOUR = TWO, where each letter stands for a different digit, is:


9 1
-------
3 2 3


So, E = 9, I = 1, G = 3, H = 2, T = 9, W = 2, and O = 3.

May 1, 2024