What is the value of the positive two-digit integer N?

Question

GMATH  practice exercise (Quant Class 16)

(1) N and the product of the digits of N are 12 units apart.
(2) N is greater than the product of the digits of N.

thyagi · Answer

Let the number N = AB
N can be any number from 10-99.

Statement 1 - 
N -A*B = 12.
N can be 28, 39...
not sufficient..

Statement 2 -
N > A*B
Not sufficient...

Combining both of them still won't help us because N can be either 28 or 39.
Therefore E is the right answer.

Posted from my mobile device

GMATBusters · Answer

Let the number N = 10x+y

St1)

N-xy = 12 or xy - N = 12,

Checking for N-xy = 12 ,

10x+y -xy = 12
or x(10-y)+y = 12
Put y = 0, x is not a digit
also y =1, x is not a digit
.
.
.
For y = 8, x = 2: N = 28
For y = 9, x = 39; N =39

Since we are getting two different values of N,  not sufficient

St2 ) N is greater than product of digits 
N can be any number such as 13, 27 etc
 not sufficient

Combining St1 and St2 , we get
10x+y -xy = 12, 
which gives 2 values of N as explained in St1

Hence,  NOT SUFFICIENT

Answer E

Archit3110 · Answer

given
value of N ; 10a+b
#1
10a+b-ab= 12
no value of ab in sufficient
#2
10a+b>ab
again in sufficient
from 1 & 2
no clear info
in sufficient 
IMO E

fskilnik · Answer

? = N = \left\langle {ab} ightangle \,\,\,\,\,\,\,\left

\left( 1 ight)\,\,\,\left| {\left( {10a + b} ight) - ab} ight| = 12\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,\,\left| {\left( {b - 10} ight)\left( {1 - a} ight)} ight| = 2\,\,\,\,\,\mathop \Leftrightarrow \limits^{\left( {**} ight)} \,\,\,\,\,\left( {10 - b} ight)\left( {a - 1} ight) = 2

\,\,\,\,\, \Rightarrow \,\,\,\,\left( {10 - b\,,\,\,a - 1} ight)\,\,{m{pair}}\,\,\,{m{of}}\,\,{m{positive}}\,\,{m{divisors}}\,\,{m{of}}\,\,2\,\,\,\,\, \Rightarrow \,\,\,\,\left( {10 - b,a - 1} ight) \in \left\{ {\left( {1,2} ight),\left( {2,1} ight)} ight\}

\left. \matrix{
 \left\{ \matrix{
 \,10 - b = 1 \hfill \cr 
 \,a - 1 = 2 \hfill \cr} ight.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {a,b} ight) = \left( {3,9} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 39\,\,{m{viable!}}\,\,\,\, \hfill \cr 
 \left\{ \matrix{
 \,10 - b = 2 \hfill \cr 
 \,a - 1 = 1 \hfill \cr} ight.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left( {a,b} ight) = \left( {2,8} ight)\,\,\,\, \Rightarrow \,\,\,\,\,N = 28\,\,{m{viable!}} \hfill \cr} ight\}\,\,\,\,\, \Rightarrow \,\,\,\,\,N = 28\,\,{m{or}}\,\,39

\left( * ight)\,\,\left( {10a + b} ight) - ab = 12\,\,\,\,\, \Leftrightarrow \,\,\,\,\, - a\left( {b - 10} ight) + b = 12\,\,\,\,\, \Leftrightarrow \,\,\,\,\, - a\left( {b - 10} ight) + b - \underline {10} = 12 - \underline {10} \,\,\,\,\, \Leftrightarrow \,\,\,\,\left( {b - 10} ight)\left( {1 - a} ight) = 2

\left( {**} ight)\,\,\,\left\{ \matrix{
 \,b - 10 < 0\,\,\,\, \Rightarrow \,\,\,\,\left| {b - 10} ight| = 10 - b \hfill \cr 
 \,1 - a \le 0\,\,\,\, \Rightarrow \,\,\,\,\left| {1 - a} ight| = a - 1 \hfill \cr} ight.

\left( 2 ight)\,\,\,\left\{ \matrix{
 \,\left( {{\mathop{m Re}
olimits} } ight){m{Take}}\,\,N = 28\,\,\,\,\,\left \,\,\,\, \Rightarrow \,\,\,? = 28 \hfill \cr 
 \,\left( {{\mathop{m Re}
olimits} } ight){m{Take}}\,\,N = 39\,\,\,\,\,\left \,\,\,\, \Rightarrow \,\,\,? = 39 \hfill \cr} ight.\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left( {m{E}} ight)

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

P.S.: we know 28 and 39 could be found by some "organized manual work"... but the importance of this solution is contained in the techniques presented, not in the numbers themselves!

bumpbot · Answer

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What is the value of the positive two-digit integer N?

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