carcass wrote:
In the correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits. What is the sum of E, F, and G ?
(1) A, B, and C are consecutive odd integers
(2) E = 2
Attachment:
challprobadd.jpg
We have:
AD
BD
+CD
------
EFG
We know that A, B, C are consecutive odd: 1,3,5 or 3,5,7 or 5,7,9
We can discard the set 1,3,5.
This is because the carry in this case would be only 1, i.e. E=1, which is not possible since all digits should be distinct.
Case 1: If we use 3,5,7: 3+5+7 = 15
So, there must be a carry from the units place else F=5 will match B=5
To have a carry from the units place, D must be either 4, 6, 8 or 9 (other digits either won't give a carry, or would be equal to one of A, B or C)
# With 4, we have:
34
54
+74
-----
162 => Satisfies all conditions. Thus,
E+F+G = 1+6+2 = 9# 6 cannot be used since the carry would be 1, resulting in F=6, which would be equal to D=6
# 8 cannot be used since the carry would be 2, resulting in F=7, which would be equal to C=7
# 8 cannot be used since we would have G=7, which would be equal to C=7
Case 2: If we use 5,7,9: 5+7+9 = 21
Possible values of D = 6, 7, 8, 9
Note: 1 is not possible since we would have D=F=1
2 is not possible since we would have D=E=2
3 is not possible since we would have G=C=9
4 is not possible since we would have G=E=2
5,7 or 9 is not possible since we would have D equal to one of A, B or C
6 is not possible since we would have F=E=2
8 is the only possible case:
58
78
+98
-----
234 => Satisfies all conditions. Thus,
E+F+G = 2+3+4 = 9Thus, Statement 1 is sufficient since we get a consistent value of E+F+G=9
Statement 2:
We only know that E = 2
There can be many possible scenarios for that:
For example: 97+87+67 = 251
97+87+47 = 231, etc.
Thus, it is not sufficient
Answer A