Basically, the symbol is the multiplication operation.

If you couple two number from the bottom, for instance 6 and 5 do not work 6*5=30^2=900

Going backward, 3*2=6^2=30 but the answer are only A and B 2*1=2^2=4

Because 3*4=12^2=144 so 3 is excluded because is in both operations.

However, I do not infer from the question why 3 cannot be used in 36.

A little bit ambiguous.


Regards

For all integers \(a\) and \(b\), where \(b>a>0\), if \(a \odot b =(ab)^2\) and \(a \odot b <10\), then which of the following could be the value of \(a\) ?

A. 1

B. 2

C. 3

D. 4

E. 5

F. 6


Kudos for the right answer and explanation

Hi,

I do get it how we get the value of a=1, however, I am not able to understand how we get 2 as the value of a. Because if we take 2=a the b has to be atleast 3 as they are integers. If so the (2 x3)^2=36.


Could you please help?

Many thanks!

Regards

Sorry I have posted the question already.

Is difficult to find out sometimes

let me know about the explanation above. If not, let me know

\(a \odot b = (ab)^2\)
\(a \odot b <100\)

Hence \((ab)^2 < 100\)
-10 < ab < 10

Now it is given that b>a>0 So both are positive

so 0<ab<10

Now if we take a = 1 ; b can be 2,3,...9

a=2 ; b can be 3 & 4

Now a can not be 3 as b>a so b will be 4 and ab = 12, but we are given ab<10

So a can be 1 & 2

Answer A B

Almost had it! Just didn't re-read the question carefully at the end. I got A-E, but did not realize that it cannot be the larger of the two integers. Shucks!