EACH LETTER IN EACH ADDITION PROBLEM STANDS FOR A NUMBER BETWEEN 0 AND 9.NO TWO LETTERS STAND FOR THE SAME NUMBER IN EACH QUESTION. FIND THE VALUE FOR EACH LETTER.FOR EACH PROBLEM, USE THE NUMBERS IN THE BOX NEXT TO THE PROBLEM.
T E N (5,6,1,2,7)
+ O N
_________
Something seems to be missing.
THOSE WERE THE INSTRUCTION(THIS IS THIRD GRADE MATH)
If you were not given letters for the answer, then just pick any values you want!
Answer=EGG
clearly, E=T+1 because of the carry. So, T is 1,5,6
G is even (since it is N+N), so it is 2,6
Now just a little trial and error yields
671
+51
------
722
To find the value for each letter in the addition problem, we need to assign a unique number between 0 and 9 to each letter. Let's go step by step to solve this problem.
1. Look at the units place:
Since the units digit in the sum is N, it means that T + O must equal N, so the value for T + O would be a number between 0 and 9.
2. Carrying over:
If T + O is greater than 9, we need to carry over a digit to the tens place. In this case, we don't have any information about the carry-over, so we'll move on to the next step.
3. Tens place:
Since the tens digit in the sum is E, it means that E + N must equal E (since there is no carry-over). In other words, N must be 0.
4. Now we know that T + O = N = 0, and since each letter represents a unique number, we can conclude that T = 1 and O = 9 (since the only remaining numbers are 5, 6, and 7).
So, T = 1, O = 9, N = 0.
The addition problem looks like this now:
1 9 1
1 0
_______