Can someone help me with this math problem:
e g g
+ e g g
---------
p a g e
a=____ e=____ g=____ p=_1___
How we started: p = 1, and g is greater than 5.
Thanks!
Can anyone figure this one out?
egg+egg=page using p=1 and g>5. I'm hoping the problem was written incorrectly! Does anyone agree?
a=7
e=8
g=9
p=1
Sure, I can help you with the math problem.
To find the values of a, e, and g, we can use a trial and error method.
Since p = 1, we can rewrite the equation as:
e g g
+ e g g
---------
1 a g e
Since g is greater than 5, it can only be 6, 7, 8, or 9. Let's try each possibility one by one and see which one gives us a valid solution.
1. If g = 6:
e g g
+ e g g
---------
1 a 6 e
Since the sum of two three-digit numbers can only be a maximum value of 1998, it's not possible for the sum to be a four-digit number starting with 1. So, g cannot be 6.
2. If g = 7:
e g g
+ e g g
---------
1 a 7 e
By adding the numbers, we know that e + e = a, which means a must be even. Also, the sum of two three-digit numbers can only be a maximum value of 1998, so the maximum value of 1 + 7 + 7 = 15 for a. Therefore, a must be 2, 4, 6, 8, 10, 12, or 14.
3. If g = 8:
e g g
+ e g g
---------
1 a 8 e
By adding the numbers, we know that e + e + 1 (carry from the previous column) = 8 + e. This means e must be 9 since 8 + 9 = 17 (which carries 1 to the next column).
4. If g = 9:
e g g
+ e g g
---------
1 a 9 e
By adding the numbers, we know that e + e + 1 (carry from the previous column) = 9 + e. This means e must be 4, since 4 + 4 + 1 = 9.
So, the solution is:
9 7 9
+ 9 7 9
---------
1 8 9 4
Therefore, a = 4, e = 9, g = 7, and p = 1.