Find the number AB

[Da_Gr8_Mperor]

AB and BA represent two 2-digit integers, where A is not equal to B. CDE represents a 3-digit integer. If:

 

A B

*B A

C D E

 

1). A=E

2). A=C

[800Bob]

What is the question?

[Da_Gr8_Mperor]

Whts the number AB

[manish8109]

Ans is B

 

Number is 21 * 12 =252

[GMAT-HELP]

Stmt 1:

A = E

=> B = 1

Insufficient as 21*12 or 31*13 or 41*14 and so on.. can satisfy the equation

 

Stmt 2:

A = C

=> A=C=2

as A=1 => 12*21 = 252 where C = 2;

A= 2=>21*12 = 252 where C=2 (satisfies)

A =3=>31*13 = 403 where C > A

all other combinations will yield C > A

 

Sufficient (21*12 = 252)

Ans B.

[iisan]

A bit convoluted approach:

 

(10a+b)(10b+a)=100c+10d+e

100ab+10(a^2+b^2)+ab = 100c+10d+e -- (1)

 

Since c

That is, ab = e --- (2)

Therefore,

a^2 + b^2 = d or a^2 + b^2 = d + 10x (where 0

ab = c or ab + x = c --- (4)

 

i) a = e

Since ab = e, b = 1.

But, 1 + e^2 = d or 1 + e^2 = d + 10x

And, c = e or c = e + x

(INSUFF)

 

ii) a = c

From (4), bc + x = c. Only possible if b = 1 and x = 0.

From (3), 1 + c^2 = d

Since, c 0 or 1 and d

Therefore, a = c = 2 and d = 5 with e = ab = 2

Number is 21 and the multiple is 252

(SUFF)

 

Ans B.

[Cedars]

Stmt 1:

A = E

=> B = 1

Insufficient as 21*12 or 31*13 or 41*14 and so on.. can satisfy the equation

 

Stmt 2:

A = C

=> A=C=2

as A=1 => 12*21 = 252 where C = 2;

A= 2=>21*12 = 252 where C=2 (satisfies)

A =3=>31*13 = 403 where C > A

all other combinations will yield C > A

 

Sufficient (21*12 = 252)

Ans B.

 

GMAT-HELP,

 

Did you just assume the values in Stmt 1: B=1 and in Stmt 2: A=C=2?

 

Thanks:hmm:

[GMAT-HELP]

GMAT-HELP,

 

Did you just assume the values in Stmt 1: B=1 and in Stmt 2: A=C=2?

 

Thanks:hmm:

Cedars,

For Stmt 1:

A = E

=> Multiplying the units digit from the qn:

B*A = E

=> B = E/A

=> B = 1 as A = E from the qn stem stmt 1.

 

For Stmt 2:

A = C (see the value substitution below for A =1, 2 & 3 to figure out the value of C)

A=1 => 12*21 = 252 where C = 2;

A= 2=>21*12 = 252 where C=2 (satisfies)

A =3=>31*13 = 403 where C > A

all other combinations will yield C > A

Only A = C= 2 satisfies the qn stem.

[gjgjgj]

 

For Stmt 2:

A = C (see the value substitution below for A =1, 2 & 3 to figure out the value of C)

A=1 => 12*21 = 252 where C = 2;

A= 2=>21*12 = 252 where C=2 (satisfies)

A =3=>31*13 = 403 where C > A

all other combinations will yield C > A

 

Only A = C= 2 satisfies the qn stem.

 

In qn stem statement 2 , is there a way to guess that B will be 1 or is it just hit and trial ?

 

Because all the examples that you have mentioned above have the value of B as 1.