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When the digits of twodigit, positive integer M are reversed, the res
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30 Sep 2014, 06:56
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When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. If M > N, what is the value of M? (1) The integer (M  N) has 12 unique factors. (2) The integer (M  N) is a multiple of 9. Official Explanation: A. A good way to approach this problem is to look at the possible range of (M  N). The largest difference between digits would be 91  19 = 72, and the smallest would be something like 2112 = 9. So you're working with a relatively limited range for the value of (M  N).
That should get you thinking: 12 is a lot of factors. A number has to be relatively large just to have a large number of factors, so with a maximum value of 72 it's unlikely that more than a few number will have that. So start with 72. Its factors include:
1 and 72; 2 and 36; 3 and 24; 4 and 18; 6 and 12; 8 and 9. That's 12 total factors. And if you break it down into primes, it's 2 * 2 * 2 * 3 * 3, a combination of the two lowest prime factors available. For a smaller number to have as many factors, it doesn't have many options other than to turn those 3s into 2s. But try it using 2 * 2 * 2 * 2 * 3. That's 48, and 48 doesn't have 12 factors:
1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8. That's only 10 factors.
Because you can't find another difference that has 12 factors, M must be 91. And note that you can use the Unique Factors Trick to more quickly do the above:
1) Express the number as the product of prime numbers. (72 = 2 * 2 * 2 * 3 * 3)
2) Express that product using exponents for each prime base. (72=23∗32)
3) Forget about the bases and concentrate on the exponents (in this case 3 and 2)
4) Add one to each exponent (making them 4 and 3 in this case)
5) Multiply the exponents and you'll have your number of total factors.
If you use that in reverse here, you'll see that to get 12 as the total number of factors, you can have exponents of either 4 and 2 or 5 and 1. And since the smallest use of 5 and 1 would be 25∗31=96, which is out of the possible range, statement 1 must be sufficient.
Statement 2, on the other hand, is not sufficient. Note that you can express (A  B) algebraically using tens and units digits. For the twodigit number xy, for example, in which x and y are each digits (and not numbers to be multiplied), you'd algebraically say that the value is 10x + y, as x is the tens digit and y the units. That makes (A  B) equal to:
10x + y  (10y + x)
That simplifies to:
10x + y  10y  x
Which is 9x  9y, and can be expressed as 9(x  y). Statement 2 then doesn't tell us anything we don't already know; (A  B) MUST BE a multiple of 9, so we don't get any new information.
The correct answer is A.
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When the digits of twodigit, positive integer M are reversed, the res
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06 Jan 2018, 19:49
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 \(M  N\): \(M  N = [10T + U]  [10U + T] = 10T + U  10U  T = 9T  9U = 9(TU)\) This tells us that M  N is equal to a multiple of 9. Since the problem tells us that M  N is a "twodigit" number, this drastically reduces our options. There are only so many twodigit multiples of 9: 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99. Since switching the digits of this number must result in a twodigit number as well, we can eliminate 90, since 9 is a singledigit number. We can also eliminate any number that would result in \(N\) being bigger than or equal to \(M\). (The problem explicitly states "\(M>N\)".) This drastically reduces our options for \(MN\). The only possible numbers left are: 54, 63, 72, and 81. 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 MN has 12 unique factors. We just need to look at these 4 numbers to see if they have 12 unique factors. With the exception of perfect squares, the factors of a number always come in complementary pairs. Since 81 is a perfect square, it will have an odd number of factors. In terms of complementary pairs, the factors of 81 are: \(1*81\) \(3*27\) \(9*9\) (Notice that, if we are counting "unique factors" we can't count 9 twice. 81 has 5 total unique factors.)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 twodigit number, this means only 19). So, in the end, a quick check of the divisibility of 19 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 54 and 63. 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. Statement #1 is sufficient. It should be very easy to see that Statement #2 is a trap. All of our possible numbers (54, 63, 72, and 81) 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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When the digits of twodigit, positive integer M are reversed, the res
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23 Oct 2014, 16:11
I solved the question like this : As per the give info , M = 10a + b ; N =10b + a Also, M>N
Lets analyse statements now: Stmt 1 : The integer (M  N) has 12 unique factors.
M  N = (10a + b)  (10b + a) = 9a  9b so, M  N = 9 (ab) = 3^2 * (ab)
Stmt1 says M  N has 12 factors , this implies that (ab) should be a number with power as 3.
Remember the rule , if prime factorization of the integer ,N = X^p * Y^q * Z^r , then the number of factors of N = (p+1)*(q+1)*(r+1)
So, ab should be a number with power as 3. This implies it should be 8 , which is 2^3.
Hence, (a  b) = 8 , this implies a= 9 and b =1
So, M= 91 and N = 19 , coz given is M>N
Stmt1: Sufficient
Stmt 2 : The integer (M  N) is a multiple of 9.
M  N = (10a + b)  (10b + a) = 9a  9b so, M  N = 9 (ab) . This is already a multiple of 9.
Therefore, (ab) can be any integer. Hence we cannot narrow down the values of a & b to find M.
Stmt2 : Insufficient




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Re: When the digits of twodigit, positive integer M are reversed, the res
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30 Sep 2014, 06:56
Did not quite get the hang of the provided OA.
Can a math expert please provide an alternative approach?



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Re: When the digits of twodigit, positive integer M are reversed, the res
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24 Oct 2014, 06:39
pratikshr wrote: When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. If M > N, what is the value of M? (1) The integer (M  N) has 12 unique factors. (2) The integer (M  N) is a multiple of 9. Official Explanation: A. A good way to approach this problem is to look at the possible range of (M  N). The largest difference between digits would be 91  19 = 72, and the smallest would be something like 2112 = 9. So you're working with a relatively limited range for the value of (M  N).
That should get you thinking: 12 is a lot of factors. A number has to be relatively large just to have a large number of factors, so with a maximum value of 72 it's unlikely that more than a few number will have that. So start with 72. Its factors include:
1 and 72; 2 and 36; 3 and 24; 4 and 18; 6 and 12; 8 and 9. That's 12 total factors. And if you break it down into primes, it's 2 * 2 * 2 * 3 * 3, a combination of the two lowest prime factors available. For a smaller number to have as many factors, it doesn't have many options other than to turn those 3s into 2s. But try it using 2 * 2 * 2 * 2 * 3. That's 48, and 48 doesn't have 12 factors:
1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8. That's only 10 factors.
Because you can't find another difference that has 12 factors, M must be 91. And note that you can use the Unique Factors Trick to more quickly do the above:
1) Express the number as the product of prime numbers. (72 = 2 * 2 * 2 * 3 * 3)
2) Express that product using exponents for each prime base. (72=23∗32)
3) Forget about the bases and concentrate on the exponents (in this case 3 and 2)
4) Add one to each exponent (making them 4 and 3 in this case)
5) Multiply the exponents and you'll have your number of total factors.
If you use that in reverse here, you'll see that to get 12 as the total number of factors, you can have exponents of either 4 and 2 or 5 and 1. And since the smallest use of 5 and 1 would be 25∗31=96, which is out of the possible range, statement 1 must be sufficient.
Statement 2, on the other hand, is not sufficient. Note that you can express (A  B) algebraically using tens and units digits. For the twodigit number xy, for example, in which x and y are each digits (and not numbers to be multiplied), you'd algebraically say that the value is 10x + y, as x is the tens digit and y the units. That makes (A  B) equal to:
10x + y  (10y + x)
That simplifies to:
10x + y  10y  x
Which is 9x  9y, and can be expressed as 9(x  y). Statement 2 then doesn't tell us anything we don't already know; (A  B) MUST BE a multiple of 9, so we don't get any new information.
The correct answer is A. A. Let M = 10a+b so N = 10b+a (1) The integer (M  N) has 12 unique factors MN = 9(ab) = 3^2(ab) 3^2 already has 3 factors. so (ab) has 4 factors. so (ab) is of the form x*y or k^3 (where x, y, and k are prime numbers apart from 3) (i) if (ab) is of the form x*y then x,y can be 2,5,7,... max difference between a and b is 8, which doesn't even satisfy the smallest product of 2*5. so this case is invalid. (ii) if (ab) is of the form k^3 note that (ab) > 1 other wise MN would only have 3 factors. 2^3 = 8 , 3^3 = 27 (not possible) so (ab) = 8 which is possible only for a=9,b=1. hence, A alone is sufficient. (2) The integer (M  N) is a multiple of 9 This is already known. Not useful...hence insufficient.
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Re: When the digits of twodigit, positive integer M are reversed, the res
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24 Oct 2014, 08:59
From the premise, we already know that MN is always a multiple of 9, so that immediately rules out statement 2 as insufficient. This is because when a positive 2 digit integer's digits are reverse, the difference between the two numbers is a multiple of 9.
I eventually got statement 1 to be sufficient. However, could someone clarify the rules with prime factorizationI'm still a bit confused.



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Re: When the digits of twodigit, positive integer M are reversed, the res
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03 Aug 2015, 13:27
pratikshr wrote: When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. If M > N, what is the value of M? (1) The integer (M  N) has 12 unique factors. (2) The integer (M  N) is a multiple of 9. Official Explanation: A. A good way to approach this problem is to look at the possible range of (M  N). The largest difference between digits would be 91  19 = 72, and the smallest would be something like 2112 = 9. So you're working with a relatively limited range for the value of (M  N).
That should get you thinking: 12 is a lot of factors. A number has to be relatively large just to have a large number of factors, so with a maximum value of 72 it's unlikely that more than a few number will have that. So start with 72. Its factors include:
1 and 72; 2 and 36; 3 and 24; 4 and 18; 6 and 12; 8 and 9. That's 12 total factors. And if you break it down into primes, it's 2 * 2 * 2 * 3 * 3, a combination of the two lowest prime factors available. For a smaller number to have as many factors, it doesn't have many options other than to turn those 3s into 2s. But try it using 2 * 2 * 2 * 2 * 3. That's 48, and 48 doesn't have 12 factors:
1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8. That's only 10 factors.
Because you can't find another difference that has 12 factors, M must be 91. And note that you can use the Unique Factors Trick to more quickly do the above:
1) Express the number as the product of prime numbers. (72 = 2 * 2 * 2 * 3 * 3)
2) Express that product using exponents for each prime base. (72=23∗32)
3) Forget about the bases and concentrate on the exponents (in this case 3 and 2)
4) Add one to each exponent (making them 4 and 3 in this case)
5) Multiply the exponents and you'll have your number of total factors.
If you use that in reverse here, you'll see that to get 12 as the total number of factors, you can have exponents of either 4 and 2 or 5 and 1. And since the smallest use of 5 and 1 would be 25∗31=96, which is out of the possible range, statement 1 must be sufficient.
Statement 2, on the other hand, is not sufficient. Note that you can express (A  B) algebraically using tens and units digits. For the twodigit number xy, for example, in which x and y are each digits (and not numbers to be multiplied), you'd algebraically say that the value is 10x + y, as x is the tens digit and y the units. That makes (A  B) equal to:
10x + y  (10y + x)
That simplifies to:
10x + y  10y  x
Which is 9x  9y, and can be expressed as 9(x  y). Statement 2 then doesn't tell us anything we don't already know; (A  B) MUST BE a multiple of 9, so we don't get any new information.
The correct answer is A. Bunuel, Could you explain how is the first point sufficient using a simpler solution. For the second point, 'The integer (M  N) is a multiple of 9' I followed the below logic. two digit multiples of 9 are 18,27,36,45,54,63,72,81,90,99. Out of these 18,27,36,45 and 99 cannot be taken into consideration because it is given that M>N. only possible considerations are 54,63,72,81,90. The only number satisfying the condition where MN should be a multiple of 9 is 54. M=54 and flipping the digits you get N=45. MN= 9 which is the only multiple possible. Am i missing anything here Bunuel. Need your help!!!!



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Re: When the digits of twodigit, positive integer M are reversed, the res
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03 Aug 2015, 13:53
arshu27 wrote: pratikshr wrote: When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. If M > N, what is the value of M? (1) The integer (M  N) has 12 unique factors. (2) The integer (M  N) is a multiple of 9. Official Explanation: A. A good way to approach this problem is to look at the possible range of (M  N). The largest difference between digits would be 91  19 = 72, and the smallest would be something like 2112 = 9. So you're working with a relatively limited range for the value of (M  N).
That should get you thinking: 12 is a lot of factors. A number has to be relatively large just to have a large number of factors, so with a maximum value of 72 it's unlikely that more than a few number will have that. So start with 72. Its factors include:
1 and 72; 2 and 36; 3 and 24; 4 and 18; 6 and 12; 8 and 9. That's 12 total factors. And if you break it down into primes, it's 2 * 2 * 2 * 3 * 3, a combination of the two lowest prime factors available. For a smaller number to have as many factors, it doesn't have many options other than to turn those 3s into 2s. But try it using 2 * 2 * 2 * 2 * 3. That's 48, and 48 doesn't have 12 factors:
1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8. That's only 10 factors.
Because you can't find another difference that has 12 factors, M must be 91. And note that you can use the Unique Factors Trick to more quickly do the above:
1) Express the number as the product of prime numbers. (72 = 2 * 2 * 2 * 3 * 3)
2) Express that product using exponents for each prime base. (72=23∗32)
3) Forget about the bases and concentrate on the exponents (in this case 3 and 2)
4) Add one to each exponent (making them 4 and 3 in this case)
5) Multiply the exponents and you'll have your number of total factors.
If you use that in reverse here, you'll see that to get 12 as the total number of factors, you can have exponents of either 4 and 2 or 5 and 1. And since the smallest use of 5 and 1 would be 25∗31=96, which is out of the possible range, statement 1 must be sufficient.
Statement 2, on the other hand, is not sufficient. Note that you can express (A  B) algebraically using tens and units digits. For the twodigit number xy, for example, in which x and y are each digits (and not numbers to be multiplied), you'd algebraically say that the value is 10x + y, as x is the tens digit and y the units. That makes (A  B) equal to:
10x + y  (10y + x)
That simplifies to:
10x + y  10y  x
Which is 9x  9y, and can be expressed as 9(x  y). Statement 2 then doesn't tell us anything we don't already know; (A  B) MUST BE a multiple of 9, so we don't get any new information.
The correct answer is A. Bunuel, Could you explain how is the first point sufficient using a simpler solution. For the second point, 'The integer (M  N) is a multiple of 9' I followed the below logic. two digit multiples of 9 are 18,27,36,45,54,63,72,81,90,99. Out of these 18,27,36,45 and 99 cannot be taken into consideration because it is given that M>N. only possible considerations are 54,63,72,81,90. The only number satisfying the condition where MN should be a multiple of 9 is 54. M=54 and flipping the digits you get N=45. MN= 9 which is the only multiple possible. Am i missing anything here Bunuel. Need your help!!!! Even 63 (&36) follows that MN = multiple of 9. 6336 = 27 = 9*3. Thus you also have 63 as one of the possible answers.



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Re: When the digits of twodigit, positive integer M are reversed, the res
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03 Aug 2015, 13:58
Engr2012 wrote: arshu27 wrote: pratikshr wrote: When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. If M > N, what is the value of M? (1) The integer (M  N) has 12 unique factors. (2) The integer (M  N) is a multiple of 9. Official Explanation: A. A good way to approach this problem is to look at the possible range of (M  N). The largest difference between digits would be 91  19 = 72, and the smallest would be something like 2112 = 9. So you're working with a relatively limited range for the value of (M  N).
That should get you thinking: 12 is a lot of factors. A number has to be relatively large just to have a large number of factors, so with a maximum value of 72 it's unlikely that more than a few number will have that. So start with 72. Its factors include:
1 and 72; 2 and 36; 3 and 24; 4 and 18; 6 and 12; 8 and 9. That's 12 total factors. And if you break it down into primes, it's 2 * 2 * 2 * 3 * 3, a combination of the two lowest prime factors available. For a smaller number to have as many factors, it doesn't have many options other than to turn those 3s into 2s. But try it using 2 * 2 * 2 * 2 * 3. That's 48, and 48 doesn't have 12 factors:
1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8. That's only 10 factors.
Because you can't find another difference that has 12 factors, M must be 91. And note that you can use the Unique Factors Trick to more quickly do the above:
1) Express the number as the product of prime numbers. (72 = 2 * 2 * 2 * 3 * 3)
2) Express that product using exponents for each prime base. (72=23∗32)
3) Forget about the bases and concentrate on the exponents (in this case 3 and 2)
4) Add one to each exponent (making them 4 and 3 in this case)
5) Multiply the exponents and you'll have your number of total factors.
If you use that in reverse here, you'll see that to get 12 as the total number of factors, you can have exponents of either 4 and 2 or 5 and 1. And since the smallest use of 5 and 1 would be 25∗31=96, which is out of the possible range, statement 1 must be sufficient.
Statement 2, on the other hand, is not sufficient. Note that you can express (A  B) algebraically using tens and units digits. For the twodigit number xy, for example, in which x and y are each digits (and not numbers to be multiplied), you'd algebraically say that the value is 10x + y, as x is the tens digit and y the units. That makes (A  B) equal to:
10x + y  (10y + x)
That simplifies to:
10x + y  10y  x
Which is 9x  9y, and can be expressed as 9(x  y). Statement 2 then doesn't tell us anything we don't already know; (A  B) MUST BE a multiple of 9, so we don't get any new information.
The correct answer is A. Bunuel, Could you explain how is the first point sufficient using a simpler solution. For the second point, 'The integer (M  N) is a multiple of 9' I followed the below logic. two digit multiples of 9 are 18,27,36,45,54,63,72,81,90,99. Out of these 18,27,36,45 and 99 cannot be taken into consideration because it is given that M>N. only possible considerations are 54,63,72,81,90. The only number satisfying the condition where MN should be a multiple of 9 is 54. M=54 and flipping the digits you get N=45. MN= 9 which is the only multiple possible. Am i missing anything here Bunuel. Need your help!!!! Even 63 (&36) follows that MN = multiple of 9. 6336 = 27 = 9*3. Thus you also have 63 as one of the possible answers. Thanks Bunuel.. totally ignored that point. could you explain how the first point is sufficient. I did not understand the earlier solutions at all.



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Re: When the digits of twodigit, positive integer M are reversed, the res
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23 Aug 2015, 19:52
A good way to approach this problem is to look at the possible range of (M  N). The largest difference between digits would be 91  19 = 72, and the smallest would be something like 2112 = 9. So you're working with a relatively limited range for the value of (M  N) Express the number 72 as 2x2x2x3x3 = 2^3x3^2 Total number of factors can be found by ignoring base and taking only the exponents i.e 3 and 2. Add 1 to them and multiply. i.e (3+1)x(2+1)= 4x3 = 12 factors available. If you use that in reverse here, you'll see that to get 12 as the total number of factors, you can have exponents of either 4 and 2 or 5 and 1. And since the smallest use of 5 and 1 would be 2^5∗3^1=96, which is out of the possible range, statement 1 must be sufficient. Statement 2,is not sufficient. Note that you can express (A  B) algebraically using tens and units digits. For the twodigit number xy, for example, in which x and y are each digits (and not numbers to be multiplied), you'd algebraically say that the value is 10x + y, as x is the tens digit and y the units. That makes (A  B) equal to: 10x + y  (10y + x) That simplifies to: 10x + y  10y  x Which is 9x  9y, and can be expressed as 9(x  y). Statement 2 then doesn't tell us anything we don't already know; (A  B) MUST BE a multiple of 9, so we don't get any new information.



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Re: When the digits of twodigit, positive integer M are reversed, the res
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21 May 2016, 14:59
Statement 1:
I would show M as 10t + u and N as 10u + t. Where t is the tens digit of M and u is the units digit.
So 10t + u  10u  t = 9t  9u. We can factor out a 9 here 9(tu) has 12 unique factors.
9 itself has a prime factorization of 3^2 so to find the number of unique factors of 9 we add 1 to the 2 (taken from 3^2) telling us 9 has 3 unique factors.
In order to get 12 unique factors we need a number to the power of 3.
The only number to the power of 3 when two single digits are subtracted from each other is 8, 2^3.
There (MN) is equal to 3^2.2^3. Which as twelve unique factors 3.4 = 12.
Sufficient.
Statement 2:
Doesn't give us any new info.



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Re: When the digits of twodigit, positive integer M are reversed, the res
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30 Dec 2016, 08:29
Hi all,
I totally understand the logic, but what about the combination of the integers 8 and 0: 8008=72 which has 12 factors.
It says nothing about nonzero digits...
Thanks in advance!



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Re: When the digits of twodigit, positive integer M are reversed, the res
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30 Dec 2016, 08:34
nachobs wrote: Hi all,
I totally understand the logic, but what about the combination of the integers 8 and 0: 8008=72 which has 12 factors.
It says nothing about nonzero digits...
Thanks in advance! When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. 08 is just 8, so it's not a twodigit integer, it's a sing;edigit integer.
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Re: When the digits of twodigit, positive integer M are reversed, the res
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02 Jan 2017, 23:36
[quote="pratikshr"]When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. If M > N, what is the value of M?
(1) The integer (M  N) has 12 unique factors.
(2) The integer (M  N) is a multiple of 9.
let m = 10a+b and n = 10b+a , a>b , 2<=a<=9
from 1
9a9b = mn = 9(ab) ,let (ab) = k thus n = 3^2 *k and since mn has 12 unique factors and since 9 has 3 unique factors (1,3,9) then k has 10 factors including 1.
ab = can only yield ( 1 , 2,3,4,5,6,7,8)
number of factors of (3^2)*x^p is 3*(p+1) thus only ab = 8 works since number of factors of 3^2 * 2^3 = 3*4 = 12
and since 2<=a<=9 thus a = 9 and b = 1 and thus m = 91 and N = 19..........suff
from 2
telling us what we already know ... insuff
A



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Re: When the digits of twodigit, positive integer M are reversed, the res
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15 Apr 2017, 16:35
thefibonacci wrote: pratikshr wrote: When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. If M > N, what is the value of M? (1) The integer (M  N) has 12 unique factors. (2) The integer (M  N) is a multiple of 9. Official Explanation: A. A good way to approach this problem is to look at the possible range of (M  N). The largest difference between digits would be 91  19 = 72, and the smallest would be something like 2112 = 9. So you're working with a relatively limited range for the value of (M  N).
That should get you thinking: 12 is a lot of factors. A number has to be relatively large just to have a large number of factors, so with a maximum value of 72 it's unlikely that more than a few number will have that. So start with 72. Its factors include:
1 and 72; 2 and 36; 3 and 24; 4 and 18; 6 and 12; 8 and 9. That's 12 total factors. And if you break it down into primes, it's 2 * 2 * 2 * 3 * 3, a combination of the two lowest prime factors available. For a smaller number to have as many factors, it doesn't have many options other than to turn those 3s into 2s. But try it using 2 * 2 * 2 * 2 * 3. That's 48, and 48 doesn't have 12 factors:
1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8. That's only 10 factors.
Because you can't find another difference that has 12 factors, M must be 91. And note that you can use the Unique Factors Trick to more quickly do the above:
1) Express the number as the product of prime numbers. (72 = 2 * 2 * 2 * 3 * 3)
2) Express that product using exponents for each prime base. (72=23∗32)
3) Forget about the bases and concentrate on the exponents (in this case 3 and 2)
4) Add one to each exponent (making them 4 and 3 in this case)
5) Multiply the exponents and you'll have your number of total factors.
If you use that in reverse here, you'll see that to get 12 as the total number of factors, you can have exponents of either 4 and 2 or 5 and 1. And since the smallest use of 5 and 1 would be 25∗31=96, which is out of the possible range, statement 1 must be sufficient.
Statement 2, on the other hand, is not sufficient. Note that you can express (A  B) algebraically using tens and units digits. For the twodigit number xy, for example, in which x and y are each digits (and not numbers to be multiplied), you'd algebraically say that the value is 10x + y, as x is the tens digit and y the units. That makes (A  B) equal to:
10x + y  (10y + x)
That simplifies to:
10x + y  10y  x
Which is 9x  9y, and can be expressed as 9(x  y). Statement 2 then doesn't tell us anything we don't already know; (A  B) MUST BE a multiple of 9, so we don't get any new information.
The correct answer is A. A. Let M = 10a+b so N = 10b+a (1) The integer (M  N) has 12 unique factors MN = 9(ab) = 3^2(ab) 3^2 already has 3 factors. so (ab) has 4 factors. so (ab) is of the form x*y or k^3 (where x, y, and k are prime numbers apart from 3) (i) if (ab) is of the form x*y then x,y can be 2,5,7,... max difference between a and b is 8, which doesn't even satisfy the smallest product of 2*5. so this case is invalid. (ii) if (ab) is of the form k^3 note that (ab) > 1 other wise MN would only have 3 factors. 2^3 = 8 , 3^3 = 27 (not possible) so (ab) = 8 which is possible only for a=9,b=1. hence, A alone is sufficient. (2) The integer (M  N) is a multiple of 9 This is already known. Not useful...hence insufficient. "max difference between a and b is 8, which doesn't even satisfy the smallest product of 2*5. so this case is invalid. " thank you



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Re: When the digits of twodigit, positive integer M are reversed, the res
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15 Apr 2017, 16:52
Bunuel wrote: nachobs wrote: Hi all,
I totally understand the logic, but what about the combination of the integers 8 and 0: 8008=72 which has 12 factors.
It says nothing about nonzero digits...
Thanks in advance! When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. 08 is just 8, so it's not a twodigit integer, it's a sing;edigit integer. @bunuel VeritasPrep gives the following solution; however, I think there is an error where state the possibilities for exponents are 5 and 1 and 4 and 2 wouldn't 4 and 2 be (4+1)(2+1)= 15 which is clearly not 12? A. A good way to approach this problem is to look at the possible range of (M  N). The largest difference between digits would be 91  19 = 72, and the smallest would be something like 2112 = 9. So you're working with a relatively limited range for the value of (M  N). That should get you thinking: 12 is a lot of factors. A number has to be relatively large just to have a large number of factors, so with a maximum value of 72 it's unlikely that more than a few number will have that. So start with 72. Its factors include: 1 and 72; 2 and 36; 3 and 24; 4 and 18; 6 and 12; 8 and 9. That's 12 total factors. And if you break it down into primes, it's 2 * 2 * 2 * 3 * 3, a combination of the two lowest prime factors available. For a smaller number to have as many factors, it doesn't have many options other than to turn those 3s into 2s. But try it using 2 * 2 * 2 * 2 * 3. That's 48, and 48 doesn't have 12 factors: 1 and 48; 2 and 24; 3 and 16; 4 and 12; 6 and 8. That's only 10 factors. Because you can't find another difference that has 12 factors, M must be 91. And note that you can use the Unique Factors Trick to more quickly do the above: 1) Express the number as the product of prime numbers. (72 = 2 * 2 * 2 * 3 * 3) 2) Express that product using exponents for each prime base. ( 72=23∗32 ) 3) Forget about the bases and concentrate on the exponents (in this case 3 and 2) 4) Add one to each exponent (making them 4 and 3 in this case) 5) Multiply the exponents and you'll have your number of total factors. If you use that in reverse here, you'll see that to get 12 as the total number of factors, you can have exponents of either 4 and 2 or 5 and 1. And since the smallest use of 5 and 1 would be 25∗31=96 , which is out of the possible range, statement 1 must be sufficient. Statement 2, on the other hand, is not sufficient. Note that you can express (A  B) algebraically using tens and units digits. For the twodigit number xy, for example, in which x and y are each digits (and not numbers to be multiplied), you'd algebraically say that the value is 10x + y, as x is the tens digit and y the units. That makes (A  B) equal to: 10x + y  (10y + x) That simplifies to: 10x + y  10y  x Which is 9x  9y, and can be expressed as 9(x  y). Statement 2 then doesn't tell us anything we don't already know; (A  B) MUST BE a multiple of 9, so we don't get any new information. The correct answer is A.



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Re: When the digits of twodigit, positive integer M are reversed, the res
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15 Apr 2017, 17:03
yezz wrote: pratikshr wrote: When the digits of twodigit, positive integer M are reversed, the result is the twodigit, positive integer N. If M > N, what is the value of M?
(1) The integer (M  N) has 12 unique factors.
(2) The integer (M  N) is a multiple of 9.
let m = 10a+b and n = 10b+a , a>b , 2<=a<=9
from 1
9a9b = mn = 9(ab) ,let (ab) = k thus n = 3^2 *k and since mn has 12 unique factors and since 9 has 3 unique factors (1,3,9) then k has 10 factors including 1.
ab = can only yield ( 1 , 2,3,4,5,6,7,8)
number of factors of (3^2)*x^p is 3*(p+1) thus only ab = 8 works since number of factors of 3^2 * 2^3 = 3*4 = 12
and since 2<=a<=9 thus a = 9 and b = 1 and thus m = 91 and N = 19..........suff
from 2
telling us what we already know ... insuff
A But how can "K" have ten factors? If the answer is 3^2 x (ab) then shouldn't k have 4 factors?



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Re: When the digits of twodigit, positive integer M are reversed, the res
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07 Sep 2017, 05:05
AutoBot wrote: I solved the question like this : As per the give info , M = 10a + b ; N =10b + a Also, M>N
Lets analyse statements now: Stmt 1 : The integer (M  N) has 12 unique factors.
M  N = (10a + b)  (10b + a) = 9a  9b so, M  N = 9 (ab) = 3^2 * (ab)
Stmt1 says M  N has 12 factors , this implies that (ab) should be a number with power as 3.
Remember the rule , if prime factorization of the integer ,N = X^p * Y^q * Z^r , then the number of factors of N = (p+1)*(q+1)*(r+1)
So, ab should be a number with power as 3. This implies it should be 8 , which is 2^3.
Hence, (a  b) = 8 , this implies a= 9 and b =1
So, M= 91 and N = 19 , coz given is M>N
Stmt1: Sufficient
Stmt 2 : The integer (M  N) is a multiple of 9.
M  N = (10a + b)  (10b + a) = 9a  9b so, M  N = 9 (ab) . This is already a multiple of 9.
Therefore, (ab) can be any integer. Hence we cannot narrow down the values of a & b to find M.
Stmt2 : Insufficient Hello, thanks for the explanation. But regarding statement 1, can't ab be equal to 2 which would eventually make N>M. (therefore IS) Can someone please explain? Thanks!



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Re: When the digits of twodigit, positive integer M are reversed, the res
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27 Nov 2017, 12:08
Let \(M = 10a + b\), \(N = 10b +\)a
Statement 1: (MN) has 12 unique factors. \(M  N = 10a + b  10b  a\) \(= 9a  9b = 9(ab) = 3^2 * (ab)\) since \(MN\) has 12 unique factors, \(3^2 *(ab)\) must be of the form \(3^2 * prime^3\), so that total unique factors \((2+1) * (3+1) = 12\) now '8' is only digit that fits \(prime^3\) which is \(2^3\). so \((ab) = 8\) => \(a = 9, b = 1\) OR\(a = 8, b = 0\) but if a were 8 and b were 0, then reverse 08 won't be two digit positive integer (as prompted by the question) so a must be 9 and b = 1, => M = 91 and N = 19 > Sufficient
Statement 2: Doesn't tell us anything new, we already know from statement 1, \(M  N = 9(ab)\) is multiple of 9 InSufficient.
Answer (A)



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Re: When the digits of twodigit, positive integer M are reversed, the res
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25 Feb 2018, 00:18
AutoBot wrote: I solved the question like this : As per the give info , M = 10a + b ; N =10b + a Also, M>N
Lets analyse statements now: Stmt 1 : The integer (M  N) has 12 unique factors.
M  N = (10a + b)  (10b + a) = 9a  9b so, M  N = 9 (ab) = 3^2 * (ab)
Stmt1 says M  N has 12 factors , this implies that (ab) should be a number with power as 3.
Remember the rule , if prime factorization of the integer ,N = X^p * Y^q * Z^r , then the number of factors of N = (p+1)*(q+1)*(r+1)
So, ab should be a number with power as 3. This implies it should be 8 , which is 2^3.
Hence, (a  b) = 8 , this implies a= 9 and b =1
So, M= 91 and N = 19 , coz given is M>N
Stmt1: Sufficient
Stmt 2 : The integer (M  N) is a multiple of 9.
M  N = (10a + b)  (10b + a) = 9a  9b so, M  N = 9 (ab) . This is already a multiple of 9.
Therefore, (ab) can be any integer. Hence we cannot narrow down the values of a & b to find M.
Stmt2 : Insufficient Can someone brief me the above highlighted area. Please.




Re: When the digits of twodigit, positive integer M are reversed, the res &nbs
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