In the correctly worked addition problem above, A, B, C, D

Hi Bunuel ,

Just curious to know why A=1,B=3,C=5 and sum = 9 and the resulting number 96 cannot be the case. It says any digits so why EFG =096 cant be a possibility??

From the stem we can assume that E is not 0.

To add to what Bunuel said, the question stem tells us that when we add three 2 digit numbers, we get a 3 digit number. Had you obtained a 2 digit number as the sum, you would have wirtten the addition as

AD
BD
CD
---
FG

and not EFG.

Hi, Why cant it be AD =56; BD =76; CD =96 and EFG =218

In this case the sum is E+F+G = 11 only.

56 + 76 + 96 = 228, not 218 and in this case E and F are not distinct.

Hi Bunuel
In first statement ,Case 1,why have we assumed that E can only be =1

If A, B, and C are 1, 3, and 5, then the sum is less than 200.

Hello Bunuel

I had a scenario like this

3 4
5 4
+ 7 4
------------
1 6 2

and the other can be

5 8
7 8
+ 9 8
-----------
2 3 4

Then how can A be sufficient?

The question asks "What is the sum of E, F, and G ?". In both valid cases (ii and iii) the sum of E, F, and G is 9. Check the solution it clearly explains this:

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 ?

AD
BD
CD
---
EFG

Notice that E can be only 1 or 2 (no sum of 3 two-digit numbers can give number more than 297).

(1) A, B, and C are consecutive odd integers. 3 cases are possible:

(i) A, B, and C are 1, 3, and 5 (it doesn't matter which is which) --> 1+3+5=9 then E (hundreds digit) can only be 1, which is not possible since we are told that the digits are distinct and we already have 1 (A, B, or C);

(ii) A, B, and C are 3, 5, and 7 (it doesn't matter which is which) --> 3+5+7=15 then E (hundreds digit) can only be 1. So, D can be 0, 2, 4, 6, 8, or 9. After trial and error we can get that only D=4 will give all distinct digits:
34
54
74
---
162

E+F+G=9.

(iii) A, B, and C are 5, 7, and 9 (it doesn't matter which is which) --> 5+7+9=21 then E (hundreds digit) can only be 2. So, D can be 1, 3, 4, 6, or 8. After trial and error we can get that only D=8 will give all distinct digits:
58
78
98
---
234

E+F+G=9.

So, as you can see in both valid cases (ii and iii) the sum of E, F, and G is 9. Sufficient.

(2) E = 2. After some trial and error you can find that several numbers can be found which will give different values for the sum of E, F, and G, for example: 58+78+98=234 and 38+78+98=214. Not sufficient.

Answer: A.

Sorry, I misread the question.

Correct option : A

we have :
1. AD + BD + CD = EFG (D unit is common for all Integer AD, BD, CD)
2. Correctly worked addition problem above, A, B, C, D, E, F, and G are distinct digits.
(Statement :1) A, B, and C are consecutive odd integers Eg. {1,3,5,7,9}
(Statement :2) E = 2

To Find : What is the sum of E, F, and G ?

Statement 1 :
Sum of two digit interger of 3 numbers, gives us 3 digit integer Number
D has to be same for all 3 integer in Unit place.
A, B, C has to be odd integer, keeping unit digit D constant
{11,31,51,71,91} are the set, and unit must give us 3 digit integer
easy : 91+71+51 = 213 = (E + F + G) = (2 + 1+ 3) = 6
Statement 1 sufficient , eliminates option D, and E

Statement 2 : E = 2, this cannot give us Solution but can be used as a check indicator,
Statement 2 is No Sufficieint - Eliminates B, and C

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 = 9

Thus, 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

While concluding for statement 1 in the case where A,B,C were 1,3,5 why can E not be Zero? E can be zero if D is 2.

Yes, it might have been clearer if the question had explicitly stated that E cannot be 0. However, even without that clarification, writing a two-digit number as "0FG" wouldn't make sense, since we don’t start a two-digit number with 0.

I'm wondering, in exam conditions, how realistically are we expected to solve this with a high degree of accuracy given the 2-minute constraint? It seems impossible. It might be better to guess and move along, maybe?

If A, B and C are 5, 7 and 9, and D = 8, then E = 2, F = 3, and G = 4, so there are 2 cases that give different digits, so wouldn't that make the right answer Both Statements together are not sufficient?

The question asks: What is the sum of E, F, and G ?

For both valid cases, the sum of E, F, and G from (1) is 9.