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Each of the letters arranged as below represents a unique integer from $1$ to $9.$ The letters are positioned in the figure such that $(A \times B \times C), (B \times G \times E)$ and $(D \times E \times F)$ are equal. Which integer among the following choices cannot be represented by the letters $A, B, C, D, E, F  \ or \ G?$

A   D
B G E
C   F
  1. $4$
  2. $5$
  3. $6$
  4. $9$
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Migrated from GO Civil 4 years ago by Arjun

6 Answers

Best answer
7 votes
7 votes
As per the given conditions,

$A\times B\times C=B\times G\times E$

$A\times C = G\times E$

If $A=5,$ either $G$ or $E$ has to be $5$ as $5$ is a prime number.

But $5$ can be considered only once as all the numbers are unique.

$\therefore$ It can be concluded that none of the numbers $A,B,C,D,E,F \ \ or \ \ G$  is $5.$

Correct answer is $(B).$
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5 votes
5 votes
for the solution,

we have been given 7 variables.. {A,B,C,D,E,F,G}

And given domain ={1,2,3,4,5,6,7,8,9}

now we cannot choose values ={5 and 7} as the condition A*B*C=B*G*F=D*E*F cannot be satisfied... because atleast one row or column might not fulfill the condition...  = 5answer

.

now for values of given variables...

so required integer values = {1,2,3,4,6,8,9}

now we have = A*B*C*D*E*F*G = 1*2*3*4*6*8*9 = 2^7*3^4

                      = (A*B*C)^2 * G = 2^7 * 3^4  (as ABC=DEF)

 NOW FROM THIS FOUR VARIABLES ONE MUST BE 2.

And if we choose any one from A,B,C  to be 2 than G will be a multiple of 2^5 which will not be in the given domain.

therefore, G=2.

now (A*B*C)^2 = (2^6 * 3^4)

    A*B*C = 2^3 * 3^2  = 72

               = 9*8*1 or  3*4*6

 ALSO, B*G*F = 72 (AS G=2)

             B*F = 36 (9*4) 

SO , the given solution would look like 

1   6
9 2 4
8   3

or , you can inter-change the values between 1,8 and 3,6 or 9,8(such that ABC=BGF=DEF)=72 

and this can be done in 8ways...

1 votes
1 votes

Given: Each letter has unique integer value. 

(A × B × C) = (B × G × E) = (D × E × F) 

Now AxBxC= BxGxE // cut B from both side

So, AxC = GxE

Similarly, BxG=DxF

Now, AxC value will be like that 2 more values should be belongs to set which give same result as AxC

(e.g. , A=3 or 2, B=2 or 3 then G=6 or 1, E=1 or 6)

Similarly, for BxG = DxF

Now, as @parabol mentioned above, 5 and 7 are prime numbers and we have to eliminate both 5 and 7.

Now, you can see, (1, 2, 4, 8) 1 set for AC and GE.

now if you make combination then you can find easily that with value 72 it will satisfy given condition.

 (A × B × C) = (B × G × E) = (D × E × F) 

(8x9x1)=(9x2x4)=(6x4x3) =72

0 votes
0 votes

It's $5$ Because there is no other number between $1$ to $9$ other than $5$ has $5$ as its factor.  Since $ABC = BGE = DEF$ if one keeps some value in place of alphabet, other triplets must have that value as one of its factor to make triplets equal.  This can't be achieved with $5$ & $7$.

 

More Explanation

Let's say $A*B*C=144= 2^{4}*3^2$, Now some ways to make product $144$ are $2^{3}*(2*3)*4= 8*6*3$ and $2^{2}*2*3^{2}=4*2*9$. So, you could be able to make product equal with different integers just because you selected integers which could be derived from factorization of other integers to make product equal. But this could not be achieved with prime numbers. 

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