p and q are different two-digit prime numbers with the same digits, but in reversed order. What is the value of the larger of p and q?
(1) p + q = 110
(2) p - q = 36
OA later
Reverse primes
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Using st 1 Randomly i guessed 91,19 as prime numebrsselango wrote:p and q are different two-digit prime numbers with the same digits, but in reversed order. What is the value of the larger of p and q?
(1) p + q = 110
(2) p - q = 36
OA later
st 2 gives 73,37 as the values..
But value is not unique..IS this a GMATtype?
sorry sorry..91 is not prime...
Now got tat..Pick D
Last edited by gmatmachoman on Wed Jul 07, 2010 5:28 am, edited 2 times in total.
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(1) The only possible combinations is: (37, 73). So, (1) is SUFFICIENT.
(2) The only possible combinations is: (37, 73). So, (2) is SUFFICIENT.
The correct answer is (D).
(2) The only possible combinations is: (37, 73). So, (2) is SUFFICIENT.
The correct answer is (D).
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This one really challenges you to think about number properties.
(1) says that p + q = 110.
We know that p and q are both two-digit primes, which means they are both odd, and that in turn means that they each have an odd number as a units digit. Those units digits also have to be different, since we are dealing with two different numbers.
In addition, the units digits must add up to 10, since 110 ends in a zero.
With those things in mind, only (19, 91) and (37, 73) work. But if you know your multiples of 7, then you know 91 is not prime. You could easily see this by realizing that 91 is 14 greater than 77, and both 77 and 14 are multiples of 7.
Therefore, only (37, 73) works, and (1) is sufficient.
(2) says that p - q = 36
Again, the units digits of p and q have to be odd.
They could be 9 and 3, respectively (since that produces a units digit of 6 for the difference). But 93 - 39 is not 36, and neither number is prime anyway.
They could be 7 and 1, respectively, but 71 - 17 is not 36.
They could be 5 and 9 , respectively, but no two-digit number that ends in 5 is prime.
They could be 3 and 7, respectively, and that works, because 73 - 37 is 36, and both numbers are prime.
They could be 1 and 5, respectively, but again, no two-digit number that ends in 5 is prime.
So only (37, 73) works, and (2) is sufficient.
(1) says that p + q = 110.
We know that p and q are both two-digit primes, which means they are both odd, and that in turn means that they each have an odd number as a units digit. Those units digits also have to be different, since we are dealing with two different numbers.
In addition, the units digits must add up to 10, since 110 ends in a zero.
With those things in mind, only (19, 91) and (37, 73) work. But if you know your multiples of 7, then you know 91 is not prime. You could easily see this by realizing that 91 is 14 greater than 77, and both 77 and 14 are multiples of 7.
Therefore, only (37, 73) works, and (1) is sufficient.
(2) says that p - q = 36
Again, the units digits of p and q have to be odd.
They could be 9 and 3, respectively (since that produces a units digit of 6 for the difference). But 93 - 39 is not 36, and neither number is prime anyway.
They could be 7 and 1, respectively, but 71 - 17 is not 36.
They could be 5 and 9 , respectively, but no two-digit number that ends in 5 is prime.
They could be 3 and 7, respectively, and that works, because 73 - 37 is 36, and both numbers are prime.
They could be 1 and 5, respectively, but again, no two-digit number that ends in 5 is prime.
So only (37, 73) works, and (2) is sufficient.
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@Raz
How do u rate this problem..Mid 600's??
How do u rate this problem..Mid 600's??
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Let the numbers be ab and ba, then
stmt 1 - 10a+b + 10b+a = 110
a + b =10
Only values of a and b that satisfy this equation are 3 and 7. Hence the bigger prime is 73
Stmt 2 a - b = 4 Only numbers that satisfy again are 7 and 3. Again 73
Thus choose D
stmt 1 - 10a+b + 10b+a = 110
a + b =10
Only values of a and b that satisfy this equation are 3 and 7. Hence the bigger prime is 73
Stmt 2 a - b = 4 Only numbers that satisfy again are 7 and 3. Again 73
Thus choose D
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So, essentially:
We know the p and q are prime numbers.
Are they even or odd numbers?
They can not be even because there are no two digit even numbers.
They have to be odd.
So our possibilities of digits are : 1, 3, 5, 7, 9
However, the digits can NOT be 5 because that would result in one of the numbers being divisible by 5. For example, 15 or 51 would result in 15 being divisible by 3 and 5.
Now, we are left with 1, 3, 7, 9
The possibilities are:
13 / 31
17 / 71
19 / 91 - X because 91 is NOT prime
37 / 73
Now the statements:
1) P + Q = 110, only combo that works is 37 + 73. Suff.
2) P - Q = 36. Only combo that works is 73 - 36. Suff.
D
We know the p and q are prime numbers.
Are they even or odd numbers?
They can not be even because there are no two digit even numbers.
They have to be odd.
So our possibilities of digits are : 1, 3, 5, 7, 9
However, the digits can NOT be 5 because that would result in one of the numbers being divisible by 5. For example, 15 or 51 would result in 15 being divisible by 3 and 5.
Now, we are left with 1, 3, 7, 9
The possibilities are:
13 / 31
17 / 71
19 / 91 - X because 91 is NOT prime
37 / 73
Now the statements:
1) P + Q = 110, only combo that works is 37 + 73. Suff.
2) P - Q = 36. Only combo that works is 73 - 36. Suff.
D
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Hi Experts,
Shouldn't we directly select the numbers in the form of 10X+Y and 10Y+X and start solving? An algebraic approach would be a better choice for this question I guess?
Amit
Shouldn't we directly select the numbers in the form of 10X+Y and 10Y+X and start solving? An algebraic approach would be a better choice for this question I guess?
Amit
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Hi gmat_for_life,
You could absolutely approach this question algebraically, but the 'restrictions' are so specific that 'brute force' would work really fast here too.
Since the two-digit numbers (with the same digits reversed) are BOTH prime, we know that NEITHER of the digits can be 0, 2, 4, 6, or 8 (since that would make at least one of the numbers even (and divisible by 2, so not prime). Also, NEITHER of the digits can be 5 (since that would make at least one of the numbers divisible by 5, so not prime).
This leaves you with just 1, 3, 7 and 9 as possible digits. Given that each number is different, and the additional limitations provided by each of the two Facts, a bit of 'playing around' with the digits is all that you really need to do to get to the correct answer.
GMAT assassins aren't born, they're made,
Rich
You could absolutely approach this question algebraically, but the 'restrictions' are so specific that 'brute force' would work really fast here too.
Since the two-digit numbers (with the same digits reversed) are BOTH prime, we know that NEITHER of the digits can be 0, 2, 4, 6, or 8 (since that would make at least one of the numbers even (and divisible by 2, so not prime). Also, NEITHER of the digits can be 5 (since that would make at least one of the numbers divisible by 5, so not prime).
This leaves you with just 1, 3, 7 and 9 as possible digits. Given that each number is different, and the additional limitations provided by each of the two Facts, a bit of 'playing around' with the digits is all that you really need to do to get to the correct answer.
GMAT assassins aren't born, they're made,
Rich