2 votes

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 |

- $4$
- $5$
- $6$
- $9$

2 votes

Best answer

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 $(B).$

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

$A\times C = G\times E$

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 $(B).$

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.

0

in the question it is mentioned that all letters represent a unique integer.. then how can all triplets have the same integers?

0

Rather that makes for answer... I'm not talking about repeating integers but using integers having factors. @aditi19

0 votes

Here it is given that,

ABC=BGE=DEF

this means that,

AC=GE and BG=DF

Now 4*2=1*8

Here,i can assign A=4,C=2,G=1,E=8

Now,6*3=2*9

Here,i can assign A=6,C=3,G=2,E=9

In above 2 cases,i can find a assignment of a letter to 4,6,9. Only number left is 5. Hence it is the answer.

Also if A or C is 5, then the product GE must be one of 5,10,15,20,25,30,35 40,45 which requires one of G or E to be 5. This violates,unique assignment property given in question.

ABC=BGE=DEF

this means that,

AC=GE and BG=DF

Now 4*2=1*8

Here,i can assign A=4,C=2,G=1,E=8

Now,6*3=2*9

Here,i can assign A=6,C=3,G=2,E=9

In above 2 cases,i can find a assignment of a letter to 4,6,9. Only number left is 5. Hence it is the answer.

Also if A or C is 5, then the product GE must be one of 5,10,15,20,25,30,35 40,45 which requires one of G or E to be 5. This violates,unique assignment property given in question.