The product of the two-digit numbers above is the three-digit number

I am not clear why units digit cannot be 1?­

What you need is AB * BA = ACA. You need to find AB. In the product ACA, the unit's digit (A) should be the same as the tens digit of AB (which is also A) i.e. the tens digit of the number you want to find.
You know that 12 * 21 = 252
The units digit of 252 is 2. It should be same as the tens digit of the first number i.e. 12 but the tens digit of 12 is 1.

So the product should instead be written as 21 * 12 = 252. Now, tens digit of AB (i.e. 21) is 2 and units digit of ACA (252) is also 2. They match. So AB = 21 and not 12.

This is what is meant by "units digit is not 1". When you have 12*21, the product's units digit should be the same as the tens digit of 12 so it should be 1. But the product's units digit is not 1; it is 2. Hence, you discard this option.­

Hi,

If this was to be done logically, this is how it would be like

Lets See how two numbers can be multiplied
XY
*YX
--------
\(\hspace{3cm}X^2 \hspace{.3cm} XY\)
\(\hspace{1.3cm} XY \hspace{1cm} Y^2\hspace{.5cm}\) 0
---------------------------------------
\(\hspace{1.3cm} X \hspace{1cm} Z \hspace{.7cm}\) X

Now here we are given certain Conditions , that X and Y are distinct integers and product of XY < 10

The Units Place of product is obtained when we do the following : X*Y +0 = X what can we infer . Using the identity that 1*A = A we can say that Y is definitely 1. And 1*X =X

The Hundreds Place of product is obtained when we do the following : XY + any carry forward from ten's place But . Its told to us that its X . Since we have seen that Y is 1 XY is =X if there is no carry forward from Ten's Place.

The Tens Place of product is obtained when we do the following : \(X^{2}\) +\(Y^{2}\) = Z

But here we Know that Y =1 So \(X^{2}\) +1 gives us the value of Z. Also Since tens Place is a number <10, because any double digit number would lead to carry forward to hundreds place the only integer whose square is less than 10 is 2,3,
But \(2^{2}\)= 4 , \(3^{3}\)=9, If it were 3 then \(3^{2}\) would be 9 and \(Y^{2}\) =1 which would be 10, this would provide carry to hundreds digit. So X is not 3 , hence X is 2.

So we have X as 2 and Y as 1
which is 21

Probus

­Hi Bunuel
Can you please share links to similar type of Questions?

­

­

1. Arithmetic

• Theory
  Mathematical Grouping Symbols
  GMAT Question Patterns: Arithmetic
  The Many Meanings of Minus
  How to Solve Simple Math Equations on the GMAT
  
  • Questions
    GMAT Practice Questions with Fractions and Decimals
    Addition/Subtraction/Multiplication Tables
    ALPHAMETIC QUESTIONS ON THE GMAT
    Symbols Representing Arithmetic Operations Problems
    Conversion Problems
    DS Arithmetic Questions
    PS Arithmetic Questions

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.­