I dont know whether my apporach is right or not..

sqrt{ABC}= 504

here, LHS is sqrt and RHS is integer so LHS must be a n integer value when sqrt is removed.

statement 1 : C=168, nothinh about A and B so insufficient.
statement 2 : A is a perfect square

now A is pefect square, we have value of C=168.

so to make it of even number of power, suppose, B=C=168.

so A,B and C all are of even power, so both the statement are together suffice.B=168 is divisible by 2.
(question stem doesnot mention that all the three are distinct numbers. so we can assume B=C)

Am I right in this?

let us take factors of 504
=2*2*2*7*3*3

LHS has a square root sign. so number of factors double on LHS.
so we require 6 2's, 2 7's and 4 3's to equate LHS to RHS.

statement 1 is clearly insufficient.
statement 2 also insufficient.

1+2 combined,
factors of 168 are 2*2*2*7*3
so balance we require 3 2's, 1 7's and 3 3's to make LHS and RHS equal.
a= 2^2*3^2 and b=2*7*3 satisfy above condition and b is a divisible by 2
a=2^4*3^2 and b=(7*3)/2=21/2 also satisfy above condition but b is not divisible by 2

Hence statements are insufficient.
Ans=E

That is right. But for this problem, ans has to be E. It is not given that numbers are integers.

Hi vishnu440,

abc is always a*b*c, but can be easily mistaken as a 3-digit number..
A 3-digit number will always be mentioned..
say abc is a 3-digit number where a, b are ......etc
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504 can be factorised as follows:
\(\frac{504}{4} = 126\\1+2+6 = 9 \implies 126 = 9 \times 14\)

\(504 = 2^3 \times 3^2 \times 7\)

\(ABC = 2^6 \times 3^4 \times 7^2\)

\(\textbf{(1) } C = 168\)

We know no information about \(B\) or \(A\), so it's only possible to prove that \(B\) is odd.
\(B\) would only be odd if \(C\) were divisible by \(2^6 = 64\)
168 is not divisible by 64.

EDIT: stricken through above, we only know one number, even if C were divisible by 64, then \(B=2\) and \(A = \frac{1}{2}\) would allow for an even

Insufficient.

----

(2) A is a perfect square

It's trivial to see that there are multiple combinations of squares, If \(A = 2^{3^2}\), then there would be no twos to distribute.
Otherwise a two may or may not be distributed to B to make it even or odd.

Insufficient

(1+2)

From (1): \(168 = 2^3 \times 3 \times 7 \implies AB = 2^3 \times 3^3 \times 7\)

Prove by counterexample

\(A = 2^4, B=2^{-1} \times 3^3 \times 7 = \text{ odd }\)

Anything without a fractional power of 2 in \(B\) would be even.

Insufficient


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OA has changed to C now the topics have been merged.

For the OA of E, the numbers do not have to be integers.
Let \(A = 144, B = \frac{21}{2}\). B is not even.
Let \(A = 4, B = 378\). B is even.
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Dude,

504 can be written as \(7 * 3^2 * 2^3\)

so, since \(sqrt(ABC) = 504\), we can say ABC = \(7^2 * 3^4 * 2^6\)

Now, Statement 1 has made it clear that C is an integer.

Now, we also know that A*B = \(7 * 3^3 * 2^3\)

We are not sure if B is even. hence, insufficient.

Now Statement B states that A is a perfect Square. No info about B and C, hence insufficient.

Combining,

We know C and product of A and B.

What do you wanna take the value of A. 1/2, if yes, then take it. I know one thing that in order to make the multiplication of A and B an integer, I MUST have 2 in B.

if you wanna take A = 1/7, again take it ( ) , Since I have the product of A and B even, and A is ODD, I can say B MUST be even.

Also, If I go with the normal integral values, then since I have A a perfect square and 3 2's in the product of A and B, I must infer that B must have atleast one 2 in it.

Hence, sufficient.

Do let me know if you have any doubts.
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For the option C
Let A= 3^2*7^2*2^4
C = 3^2*3^7
In this case B= 3/(7*2)
Which is not divisible by 2

I think it is E not C
Correct me if I am wrong.