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805+ (Hard)|   Arithmetic|      
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jpu2
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If, in the addition problem above, a, b, c, d, e, f, x, y, a 18 Oct 2025, 22:16
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Don't think you can directly say 6 = z - c, this is assuming that there is no carry over. It just so happens that you can't carry over on this question, but I believe you would have to do a check first to be sure.


sudhir18n
I will try it in a different method... took me 2.5 mints

Problem states : All the 9 digits in the problem are different = no repeats.

(1) SUFFICIENT: Given 3a = f = 6y, the only possible value for y = 1. if y = 2, f> 9. Since y = 1, we know that f = 6 and a = 2.

now the question reads as

2 b c
+ d e 6

= x 1 z

using our knowledge of digits we can rewrite this as

200 + 10b + c + 100d + 10e + 6 = 100x + 10 + z
solving further ;
196 = 100(x - d) - 10(b + e) + 1(z - c)

our idea is to concentrate only on the units digit (z - c);
from the equation
6 = z - c.

lets plug in diffferent values;

z = 9 and c = 3
z = 8 and c = 2 ........ this cant be true as 'a' is already 2
z = 7 and c = 1 ........ this cant be true as 'y' is already 1.

hence Z =9 and X = 3... perfect.


(2) INSUFFICIENT: The statement f - c = 3 .. provides us with different values... not possible.

Hence A.
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navyaashali
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If, in the addition problem above, a, b, c, d, e, f, x, y, a 29 Mar 2026, 06:12
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Hi Sudhir,

given that we are only focused on units digits and statement (1) gives us c + 6 = z; z-c =6 - we don't need to express the equations in hundreds, tens and units right? we can just directly plugin?

sudhir18n
I will try it in a different method... took me 2.5 mints

Problem states : All the 9 digits in the problem are different = no repeats.

(1) SUFFICIENT: Given 3a = f = 6y, the only possible value for y = 1. if y = 2, f> 9. Since y = 1, we know that f = 6 and a = 2.

now the question reads as

2 b c
+ d e 6

= x 1 z

using our knowledge of digits we can rewrite this as

200 + 10b + c + 100d + 10e + 6 = 100x + 10 + z
solving further ;
196 = 100(x - d) - 10(b + e) + 1(z - c)

our idea is to concentrate only on the units digit (z - c);
from the equation
6 = z - c.

lets plug in diffferent values;

z = 9 and c = 3
z = 8 and c = 2 ........ this cant be true as 'a' is already 2
z = 7 and c = 1 ........ this cant be true as 'y' is already 1.

hence Z =9 and X = 3... perfect.


(2) INSUFFICIENT: The statement f - c = 3 .. provides us with different values... not possible.

Hence A.
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If, in the addition problem above, a, b, c, d, e, f, x, y, a 30 Mar 2026, 12:11
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navyaashali
Hi Sudhir,

given that we are only focused on units digits and statement (1) gives us c + 6 = z; z-c =6 - we don't need to express the equations in hundreds, tens and units right? we can just directly plugin?


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Yes, Navya, as long as we get definitive answer of Z, it's not required to know other details.
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