mn2010 wrote:
a b c
+
d e f
-----
x y z
-----
If, in the addition problem above, a, b, c, d, e, f, x, y, and z each represent different positive single digits, what is the value of z ?
(1) 3a = f = 6y
(2) f – c = 3
How to solve this in under 2 mins ??
Took me >5 mins to prove
"A's" sufficiency, yet I fear that I may be missing something. Hope this is not from GPrep!!!
a b c
+
d e f
-----
x y z
-----
(1) 3a = f = 6y
y=1; f=6; a=2
2 b c
+
d e 6
-----
x 1 z
-----
Now, y=1 must be the unit's digit of 11 with 1 being carried over to the hundred's place.
So, 2+d+1=x; here "1" is the carried over 1 from y.
3+d=x
x can't be 4 as d can't be 1.
x can't be 5 as d can't be 2.
x can't be 6.
x can be 7 if d=4
x can be 8 if d=5
x can't be 9 as d can't be 6.
Let's try with x=8, d=5
2 b c
+
5 e 6
-----
8 1 z
-----
1,2,5,6,8---used
z can be 3,4,7,9
z=3, it got to be 3 of 13; So, c=7; "b+e" becomes 9+4=13 and 1 carried over from z=13; so 13+1=14; but y=1 NOT 4. Rule out.
z=4, it got to be 4 of 14; c=8(Not possible). Rule out.
z=7, c can't be 1. Rule out.
z=9, c=3; b+e=11; perfect y's "1" properly fits and 1 carries over. ------- This is first case so far. Let's try the other scenario:
Let's try with x=7, d=4
2 b c
+
4 e 6
-----
7 1 z
-----
1,2,4,6,7 used up
3,5,8,9 left.
z=3; it must be 3 of 13; c=7(Not possible). Rule out
z=5; it must be 3 of 15; c=9. b+e+1(carried over)=8+3+1=12; but y=1. Rule out.
z=8; c=2(Not possible). Rule out.
z=9; c=3. b+e=13; but y=1. Rule out.
Thus, only one scenario fits that gave us z=9.
Sufficient.
(2) f – c = 3
a b c
+
d e f
-----
x y z
-----
f=4, c=1, z=5
2 9 1
+
3 8 4
-----
6 7 5
-----
And we already know from 1 that z=9
Not Sufficient.
Ans: "A"