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Question
In the following multiplication, each of the letters denote a different integer. Each letter stands for the same integer throughout where A stands for 2 and E stands for 6. A B A × C A ------------------ D A D E C E X ------------------ E F G DWhat is the sum total of B A D ?
In the following multiplication, each of the letters denote a different integer. Each letter stands for the same integer throughout where A stands for 2 and E stands for 6.
A B A
× C A
------------------
D A D
E C E X
------------------
E F G D
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Solution
The correct option is D 7
Given , A=2 and E=6
2 B 2
x C 2
D 2 D
6 C 6 x
6 F G D
In first multiplication
22=D=4 and 2B=2 (unit digit) so B can be either 1 or 6, if it is 6, then the product is 26=12, where 1 is a carry over but, at the Hundred's place also 22=4. Thus 2B=2 and not 12 where B=1. In second multiplication.
C2=6 so C is either 3 or 8, if it C=8, then the Ten's place multiplication would have become BC+1=1C+1, but it remains 1C=C, thus C=3.
Therefore,
2 1 2
x 3 2
4 2 4
6 3 6 x
6 7 8 4
So, B = 1, A = 2 and D = 4 : B+A+D = 1+2+4 = 7 (option C)
| Given , A=2 and E=6 2 B 2 x C 2 D 2 D 6 C 6 x 6 F G D In first multiplication 22=D=4 and 2B=2 (unit digit) so B can be either 1 or 6, if it is 6, then the product is 26=12, where 1 is a carry over but, at the Hundred's place also 22=4. Thus 2B=2 and not 12 where B=1. In second multiplication. C2=6 so C is either 3 or 8, if it C=8, then the Ten's place multiplication would have become BC+1=1C+1, but it remains 1C=C, thus C=3. Therefore, 2 1 2 x 3 2 4 2 4 6 3 6 x 6 7 8 4 So, B = 1, A = 2 and D = 4 : B+A+D = 1+2+4 = 7 (option C) |
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