In the addition problem on the right, different letters represent different digits and A is bigger than P.
What three-digit number does EEE, the sum,
represent?
PA + PA + PA = EEE
Just wondering if anyone would have a technique to solve this if not then trial and error will do
Sorry meant to have it as
AP + AP + AP = EEE
To solve this addition problem, you can use a technique called "logic and deduction" rather than relying solely on trial and error. Here's how you can approach it:
1. Start by examining the units place: Since PA + PA + PA equals EEE, we know that A + A + A equals the units digit E. Therefore, E must be a multiple of 3 (since the sum of three As is a multiple of 3).
2. Consider the carry-over to the tens place: Since A is bigger than P, the sum of three As plus the carry-over from the previous column (if any) will be greater than or equal to 30, ensuring that the two-digit sum in the tens place is at least 0-3.
3. Examine the tens place: Since the two-digit sum in the tens place ranges from 0-3, there are four possibilities: 0, 1, 2, and 3.
4. Calculate the possible values for E: Take one of the possible values from the tens place (e.g., 0) and add it to A + A + A (which equals E) from the units place. If the sum is a multiple of 3, then that value is a possible value for E. Repeat this process for each possibility in the tens place.
By systematically following these steps and testing each possibility, you can find the value(s) for E and determine the three-digit number EEE.