Here is the full "GMAT Jujitsu" for this question:
The first thing we need to do is come up with a way to "mathematize" this problem. We need to translate the concept of "tens digit" and "units digit" into something that we can algebraically manipulate. Note that if the tens digit = \(T\) and the units digit = \(U\), the original number is
not calculated by multiplying \(T\) by \(U\). For example, 41 is not equal to \(4*1\), instead, it it equal to \(4*10 + 1\).
Thus, the number \(M\) must be equal to \(10T + U\), where \(T\) is the tens column value and \(U\) is the units column value. Since \(N\) switches the digits, \(N = 10U + T\). This allows us to solve for \(M - N\):
\(M - N = [10T + U] - [10U + T] = 10T + U - 10U - T = 9T - 9U = 9(T-U)\)
This tells us that M - N is equal to a multiple of 9. Since the problem tells us that M - N is a "two-digit" number, this drastically reduces our options. There are only so many two-digit multiples of 9. Of those multiples, 81, 90, and 99 wouldn't work, since this this would mean \(T-U = 9\), \(10\), or \(11\) respectively and the largest difference for \(T-U\) we can get is \(9-0 = 9\). Even this would result in N=9, which is NOT a two-digit number. Our only options are:
18, 27, 36, 45, 54, 63, and 72
With so few options, it is often easiest to just check each number to see if it follows the "rules" of the problem.
Statement #1 tells us that M-N has 12 unique factors. This is A LOT of unique factors. We just need to look at these seven possibilities to see if they have 12 unique factors. With the exception of perfect squares, the factors of a number always come in complementary pairs, a small factor and a larger factor. 36 is a perfect square, so it has an odd number of factors. This one can easily be eliminated. 18 and 27 aren't even close to having 12 factors.
45 has six factors:
\(1*45\)
\(3*15\)
\(5*9\)
54 has eight factors:
\(1*54\)
\(2*27\)
\(3*18\)
\(6*9\)
63 has six factors:
\(1*63\)
\(3*21\)
\(7*9\)
72 has twelve factors:
\(1*72\)
\(2*36\)
\(3*24\)
\(4*18\)
\(6*12\)
\(8*9\)
If you realize that each complementary pair of factors is a small value multiplied by another larger value, the math even becomes simpler. Since we are only counting factors, we don't need to even know what they are. Thus, you really only need to check the divisibility of the number up through the approximate square root of the number (and for a two-digit number, this means only 1 through 9). So, in the end, a quick check of the divisibility of 1 through 9 allows you to count pairs without needing to actually factor the numbers to the bitter end. If I were doing this problem on the GMAT, my scratch paper would look like this for 72:
\(1*72\)
\(2*\)__
\(3*\)__
\(4*\)__
\(6*\)__
\(8*9\)
You could also do the same thing for other numbers. Using this strategy allows you to rapidly check a series of numbers for factors. In the end, we quickly narrowed down our list of possible numbers to one number: 72. Now, if
\(M - N = 72 = 9(T-U)\), then
\(T-U=8\). That is a big range between the tens digit and the ones digit. \(9-1 = 8\). We can't have \(8-0 = 8\) because the transposed number \(N\) would then have a \(0\) in the tens digit, making a single-digit number. We only have one option.
\(M = 91\) and
\(N=19\). Statement #1 is sufficient.
It should be very easy to see that Statement #2 is a trap. All of our possible numbers (18, 27, 36, 45, 54, 63, and 72) are divisible by 9, so Statement #2 tells us nothing new that we didn't already know. It can't be sufficient.
Since Statement #1 sufficient and Statement #2 insufficient,
the answer is A.
Addendum: some people like to memorize the equation for the unique number of factors for any given number. It requires you to prime factor the number, then count the number of times each prime factor appears. If the prime factorization of \(N = A^x * B^y * C^z\) , then the number of factors of \(N\) is \((x+1)*(y+1)*(z+1)\). However, the utility of this equation is very limited. It is often quicker to use divisibility rules and count complementary pairs of factors than it is to fully prime factorize a number.
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Aaron PondVeritas Prep Teacher of the YearVisit me at https://www.veritasprep.com/gmat/aaron-pond/ if you would like to learn even more "GMAT Jujitsu"!