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In the above multiplication, A, B, C, D, E, F, G and H are the different digits of number such that none of them is divisible by 5. 
If it is known that D is twice that of F, answer the following questions
 
Q. What is the vale of DEFA × FE
  • a)
    AAADCA
  • b)
    GGDCGG
  • c)
    AADCAA
  • d)
    FFHDFF
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
In the above multiplication, A, B, C, D, E, F, G and H are the differe...
Since none of the digits is divisible by 5, Digits can be 1,2,3,4,6,7,8 and 9 only.
From the above we see that C + 0 = 6. Thus C = 6.
Looking at the second digit of the product. Unit place of C +  C = B. Thus unit digit of 6+6(12) is 2. Thus B = 2.
The product simplifies into 
We observe that unit places of D × G = 6 The pair of digits which gives 6 at unit digits on their product are (1,6), (2,3), (2,8), (4,4), (4,9) (6,6) and (7,8). Since we already know that C = 6, B = 2  and  D ≠ G only remaining possible pairs are (4,9) and (7,8). 
Thus (D,G) can be (4,9) or (9,4) or (7,8) or (8,7)
Upon further observation, we see that FGAD × 2 = HFG6 which is a 4-digit number. Thus the maximum possible value of F = 4 (if F≥5 then the product will be at least 5- digit. Thus F = 1 or 3 or 4
We are given that D = 2 × F Thus F = 4, D= 8 which leaves G = 7
The remaining alphabets are A, E and H. The remaining digits to be allocated are 1,3 and 9. 
Since one of the numbers for the product is 27. Thus the final product E27H26 should be divisible by 27. Thus it should be divisible by 9 also. For a number to be divisible by 9, the sum of its digits has to be a multiple of 9
Thus  E+2+7+H+2+6 = 9k or E + H = 9k−17 or E + H = 9k′ + 1 where k =  k′+2
Thus E + H = 1 or 10
E + H = 1 is not possible as both are positive digts. Thus E + H = 10 and it happens when (E,H) = (1,9) or (9,1). Thus it will mean that A = 3 is the only possible scenario. Now we know that the 2 numbers whose product is taken is  4738 and 27. We can input the values and get the following
Thus we obtain the following allocation
A × C = 3 × 6 = 18
C+D+F = 6+8+4 = 18. Option D is correct
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Most Upvoted Answer
In the above multiplication, A, B, C, D, E, F, G and H are the differe...
Since none of the digits is divisible by 5, Digits can be 1,2,3,4,6,7,8 and 9 only.
From the above we see that C + 0 = 6. Thus C = 6.
Looking at the second digit of the product. Unit place of C +  C = B. Thus unit digit of 6+6(12) is 2. Thus B = 2.
The product simplifies into 
We observe that unit places of D × G = 6 The pair of digits which gives 6 at unit digits on their product are (1,6), (2,3), (2,8), (4,4), (4,9) (6,6) and (7,8). Since we already know that C = 6, B = 2  and  D ≠ G only remaining possible pairs are (4,9) and (7,8). 
Thus (D,G) can be (4,9) or (9,4) or (7,8) or (8,7)
Upon further observation, we see that FGAD × 2 = HFG6 which is a 4-digit number. Thus the maximum possible value of F = 4 (if F≥5 then the product will be at least 5- digit. Thus F = 1 or 3 or 4
We are given that D = 2 × F Thus F = 4, D= 8 which leaves G = 7
The remaining alphabets are A, E and H. The remaining digits to be allocated are 1,3 and 9. 
Since one of the numbers for the product is 27. Thus the final product E27H26 should be divisible by 27. Thus it should be divisible by 9 also. For a number to be divisible by 9, the sum of its digits has to be a multiple of 9
Thus  E+2+7+H+2+6 = 9k or E + H = 9k−17 or E + H = 9k′ + 1 where k =  k′+2
Thus E + H = 1 or 10
E + H = 1 is not possible as both are positive digts. Thus E + H = 10 and it happens when (E,H) = (1,9) or (9,1). Thus it will mean that A = 3 is the only possible scenario. Now we know that the 2 numbers whose product is taken is  4738 and 27. We can input the values and get the following
Thus we obtain the following allocation
A × C = 3 × 6 = 18
C+D+F = 6+8+4 = 18. Option D is correct
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Community Answer
In the above multiplication, A, B, C, D, E, F, G and H are the differe...
Since none of the digits is divisible by 5, Digits can be 1,2,3,4,6,7,8 and 9 only.
From the above we see that C + 0 = 6. Thus C = 6.
Looking at the second digit of the product. Unit place of C +  C = B. Thus unit digit of 6+6(12) is 2. Thus B = 2.
The product simplifies into 
We observe that unit places of D × G = 6 The pair of digits which gives 6 at unit digits on their product are (1,6), (2,3), (2,8), (4,4), (4,9) (6,6) and (7,8). Since we already know that C = 6, B = 2  and  D ≠ G only remaining possible pairs are (4,9) and (7,8). 
Thus (D,G) can be (4,9) or (9,4) or (7,8) or (8,7)
Upon further observation, we see that FGAD × 2 = HFG6 which is a 4-digit number. Thus the maximum possible value of F = 4 (if F≥5 then the product will be at least 5- digit. Thus F = 1 or 3 or 4
We are given that D = 2 × F Thus F = 4, D= 8 which leaves G = 7
The remaining alphabets are A, E and H. The remaining digits to be allocated are 1,3 and 9. 
Since one of the numbers for the product is 27. Thus the final product E27H26 should be divisible by 27. Thus it should be divisible by 9 also. For a number to be divisible by 9, the sum of its digits has to be a multiple of 9
Thus  E+2+7+H+2+6 = 9k or E + H = 9k−17 or E + H = 9k′ + 1 where k =  k′+2
Thus E + H = 1 or 10
E + H = 1 is not possible as both are positive digts. Thus E + H = 10 and it happens when (E,H) = (1,9) or (9,1). Thus it will mean that A = 3 is the only possible scenario. Now we know that the 2 numbers whose product is taken is  4738 and 27. We can input the values and get the following
Thus we obtain the following allocation
A × C = 3 × 6 = 18
C+D+F = 6+8+4 = 18. Option D is correct
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If one gives an answer before one hears, it is ones folly and shame. It is arrogant to answer before one hears. Humility does not presume that it knows precisely what a person is asking until the questioner has finished asking the question. How many times have Ijumped to a wrong conclusion by starting to formulate my answer before I heard the whole question! Often it is the last word in the question that turns the whole thing around and makes you realize that the questioner is not asking what you thought he was.Not answering a question before you hear it all, honours and respects the person asking the question. It treats the person as though his words really matter. It is belittling to the other person if you finish his question for him. Carefully listening to a question often reveals that the question has several layers and is really more than one question. Several questions are all mixed into one. When you see this, you can break down the question into parts and answer them one at a time. You will not see such subtleties if you are hasty with your answer and not careful in your listening.Q.Which of the following can be used to describe Several questions are all mixed into one ?

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In the above multiplication, A, B, C, D, E, F, G and H are the different digits of number such that none of them is divisible by 5.If it is known that D is twice that of F, answer the following questionsQ. What is the vale of DEFA × FEa)AAADCAb)GGDCGGc)AADCAAd)FFHDFFCorrect answer is option 'A'. Can you explain this answer?
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