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Math Expert V
Joined: 02 Sep 2009
Posts: 55635
The product of the two-digit numbers above is the three-digi  [#permalink]

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Difficulty:   85% (hard)

Question Stats: 60% (02:21) correct 40% (02:13) wrong based on 769 sessions

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

AB
x BA

The product of the two-digit numbers above is the three-digit number ACA, where A, B and C are three different nonzero digits. If A x B <10, what is the two-digit number AB?

(A) 11
(B) 12
(C) 13
(D) 21
(E) 31

Problem Solving
Question: 174
Category: Arithmetic Operations on rational numbers
Page: 85
Difficulty: 600

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Posts: 55635
Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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1
4
SOLUTION

AB
x BA

The product of the two-digit numbers above is the three-digit number ACA, where A, B and C are three different nonzero digits. If A x B <10, what is the two-digit number AB?

(A) 11
(B) 12
(C) 13
(D) 21
(E) 31

First of all, since the digits must be distinct, then we can eliminate option A (11).

Next, simply plug in the options and see which satisfies AB x BA = ACA:

(B) 12 --> 12*21 --> the units digit is not 1. Discard.
(C) 13 --> 13*31 --> the units digit is not 1. Discard.
(D) 21 --> 21*12 = 252 = AB x BA = ACA. Bingo.

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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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1

21 (AB) x 12 (BA) = 252 (ACA)

All other options do not satisfy

In Option A, it comes up as A = B & AB x BA = ABA; so ignore it

Option B, C & E do not satisfy

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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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Bunuel wrote:
SOLUTION

AB
x BA

The product of the two-digit numbers above is the three-digit number ACA, where A, B and C are three different nonzero digits. If A x B <10, what is the two-digit number AB?

(A) 11
(B) 12
(C) 13
(D) 21
(E) 31

First of all, since the digits must be distinct, then we can eliminate option A (11).

Next, simply plug in the options and see which satisfies AB x BA = ACA:

(B) 12 --> 12*21 --> the units digit is not 1. Discard.
(C) 13 --> 13*31 --> the units digit is not 1. Discard.
(D) 21 --> 21*12 = 252 = AB x BA = ACA. Bingo.

Why is the units digit not 1? Thanks! MBA Section Director V
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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

AB
x BA

The product of the two-digit numbers above is the three-digit number ACA, where A, B and C are three different nonzero digits. If A x B <10, what is the two-digit number AB?

(A) 11
(B) 12
(C) 13
(D) 21
(E) 31

While plugging answer choices is good way to attack this question, it is good to leverage some facts given in the question stem first, so that we need to try out only a few choices.

1) Units digit of AB*BA is A. That means B*A = A, This is possible when B=1. so we have that A1 * 1A = 1C1
2) Since A, B, and C are DIFFERENT non-zero digits and since B=1, we can say that A is not equal to 1

So we have to check only Choice D and E.
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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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I solved it this way:
1) We can eliminate A, C and E
--> A:11: It's said - Three different nonzero digits, so it can't be 11
--> C: Just plug in numbers 13*31= 403 ( It's wrong because 1st and 3rd Digits are the same in the product)
--> E: Same as c 31*13 = 403 ( It's wrong because 1st and 3rd Digits are the same in the product)
So we have just B and D left.

2) If we use the multiplication rules, it is evident that -->
AB
x BA
--------
ACA -------> So, BxA = A so B must be 1; Answer D - 21 is correct
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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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NAL9 wrote:
Bunuel wrote:
SOLUTION

AB
x BA

The product of the two-digit numbers above is the three-digit number ACA, where A, B and C are three different nonzero digits. If A x B <10, what is the two-digit number AB?

(A) 11
(B) 12
(C) 13
(D) 21
(E) 31

First of all, since the digits must be distinct, then we can eliminate option A (11).

Next, simply plug in the options and see which satisfies AB x BA = ACA:

(B) 12 --> 12*21 --> the units digit is not 1. Discard.
(C) 13 --> 13*31 --> the units digit is not 1. Discard.
(D) 21 --> 21*12 = 252 = AB x BA = ACA. Bingo.

Why is the units digit not 1? Thanks! Also am not clear why units digit cannot be 1?
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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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TOB2020 wrote:
NAL9 wrote:
Bunuel wrote:
SOLUTION

AB
x BA

The product of the two-digit numbers above is the three-digit number ACA, where A, B and C are three different nonzero digits. If A x B <10, what is the two-digit number AB?

(A) 11
(B) 12
(C) 13
(D) 21
(E) 31

First of all, since the digits must be distinct, then we can eliminate option A (11).

Next, simply plug in the options and see which satisfies AB x BA = ACA:

(B) 12 --> 12*21 --> the units digit is not 1. Discard.
(C) 13 --> 13*31 --> the units digit is not 1. Discard.
(D) 21 --> 21*12 = 252 = AB x BA = ACA. Bingo.

Why is the units digit not 1? Thanks! Also 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.
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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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[quote="Bunuel"]SOLUTION

AB
x BA

The product of the two-digit numbers above is the three-digit number ACA, where A, B and C are three different nonzero digits. If A x B <10, what is the two-digit number AB?

(A) 11
(B) 12
(C) 13
(D) 21
(E) 31

We are told that all 3 digits are different. So we can reject option (A).

Let us put values and find out the answer.

Option (B) 12
So, 12*21. But here units digit is not A i.e.1. So reject this.
Option (C) 13
So, 13*31. But here units digit is not A i.e.1. So reject this.
Option (D) 21
So, 21*12 = 252 = AB x BA = ACA. This satisfies.

Hence option (D).

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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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AB
*
BA
-----------
ACA

A*B<10 means that B=1 and A can be 2,3,4,5,6,7,8,9. Eliminate A,B,C answers

Test only 21 and 31

21*12=252, it fits

D
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Re: The product of the two-digit numbers above is the three-digi  [#permalink]

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Units digit is A and A*B = A => B=1. Also A,B, and C are different digits. Thus, Discard A, B and C choices.

Upon checking 21 we see that it matches the criteria and is, therefore, the answer choice.
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The product of the two-digit numbers above is the three-digi  [#permalink]

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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

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