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23 Apr 2019, 13:42
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
50% (02:33) correct
50%
(02:11)
wrong
based on 219
sessions
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In the addition shown below A, B, C, and D are distinct digits. How many different values are possible for D ?
(A) 2
(B) 4
(C) 7
(D) 8
(E) 9
Attachment:
f247ec6fe2d36c608337f8763645addb01b26338.png [ 1.85 KiB | Viewed 5010 times ]
Archit3110
27 Apr 2019, 03:20
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Bunuel
total digits = 10
if a,b,c occupy 1 digit position each and all are distinct then value of D integer can be 10-3 ; 7
IMO C
General Discussion
26 Apr 2019, 05:05
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Bunuel
Key points :
1) All are distinct
2) All are single digit integers.
In any addition we always start from the right hand side.
Let us assume D =1 then A+B= 1 or 0 +1=1 or 1+0=1 so if A =1 then B=0 or Vice versa
let us take A=1 and B=0 and proceed, in the second column if D=1 then C has to be 0 this contradicts point 1 above , as all integers are supposed to be distinct. C and B both cannot be 0, hence D cannot be 1
Let us assume D=2 then A=0 and B=2 or vice versa , we can't have both as 1, as all are supposed to be distinct.
Put D=2 in second column then C=0, this again violates point 1 , hence D cannot be 2
Let us assume D=3 then first column A+B=1+2 or 2+1 let us take first case A=1 and B=2 , second column put d=3 then C=0 ,Third Column B + A= D so 2+1=3, Fourth Column B+C= B so 2+0=2, Fifth Column A+B=D so 1+2=3
So D can be 3 .
There is a pattern here in all those cases in which C=0 and is distinct ,we have a valid value for D.
So possible values of D are = 3,4,5,6,7,8 and 9, hence 7 values .
Ans C.
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