Hard Math Problem!?

This is from a math contest.


-A+B+C+D+E = FG


-FG is a 2 digit number


-FG is equal to 10F+G


-A,B,C,D,E,F,G are all diff. digits.





If FG is at its greatest, what is hte value of G?

Hard Math Problem!?
32 is the greatest possibility for FG, so G = 2. You can figure this out by a process of elimination.





FG must be less than 35 since 9+8+7+6+5 = 35, and in this case, G = E, so this is not a possibility.


9+8+7+6+4 = 34, but E=G still.


9+8+7+5+4 = 33, but now F=G. (9+8+7+6+3 has the same problem.)





9+8+6+5+4 = 32. This is the greatest possibility for FG. Hence G = 2.





I hope this helps!
Reply:9+8+7+6+5=35


1+2+3+4+5=15


Checking for numbers from 15 to 35 which can be divided into two factors such that 10 times one factor plus the other factor equals the number...


15-19: false because 2*10 is higher and neither of the factors can be 1 because the addition would be 10 higher than the number itself.


20: false because 10F+G can't be between 0 and 30.


21: false because 10F+G can't be between 0 and 37.


22: false because 10F+G can't be between 0 and 31.


23: false because it is prime.


24: false because 10F+G can't be between 0 and 32.


25: false because 10F+G can't be between 0 and 55.


26: false because 10F+G can't be between 0 and 33.


27: false because 10F+G can't be between 0 and 39.


28: false because 10F+G can't be between 0 and 34.


29: false because it is prime.


30: false because 10F+G can't be between 0 and 35.


31: false because it is prime.


32: false because 10F+G can't be between 0 and 36.


33: false because 10F+G can't be between 0 and 41.


34: false because 10F+G can't be between 0 and 37.


35: false because 10F+G can't be between 0 and 57.


Therefore I conclude that the problem (if I interpreted it correctly) doesn't have any answer.
Reply:my guess is 8

gardenia