p and q are different two-digit prime numbers with the same

I did the AD/BCE strategy

Statement 1 - P + Q= 110

Here, we know that every Prime number is gonna be odd, and because the digits are reversed, EVERY digit must be odd. Second. We have to see which combination of odd digits will equal 10 since the ones digit is zero. they are 1 and 9, 3 and 7, and 5 and 5.

We can already eliminate 1 and 9 as a possible combination of digits because 9 isn't prime. If we use two 5's to create a two digit number, that makes 55 and it's not a prime number since 11x5 is 55. So 3 and 7 is left and since P and Q are two digit numbers, let's try 37 and 73 for P and Q. 37+73 = 110 and P IS bigger than Q. If Q is 73, then P is 37. Either way we have enough info to determine which value is bigger so this works so A is a possible answer. Get rid of B, C, and E.

Next let's see statement two and we see the P-Q is 36. Let's see which odd number combinations could result in a difference where the ones digit is 6. They are 9 and 3, 7 and 1, 3 and 7, and 5 and 9. We can get rid of 9 and 3 and 5 and 9 because 9 isn't prime. That leaves 7 and 1, 5 and 9, and 3 and 7. When you combine the digits of 5 with any number for a two digit number in this equation, at some point, 5 is in the ones digit and it ain't prime so get rid of 5 and 9. Next there is 7 and 1 and 3 and 7. combine 7 and 1 first and you see that you get 71 and 17 and see if we can get a result with the ones digit at 6. It's impossible, so 3 and 7 is left so we're back to 37 and 73. We find out that 73-37 is 36 and 73=P and 37=Q and we can tell which value is bigger so that proves sufficiency, and Statement 2 is sufficient.

Since Statements 1 and 2 are sufficient, D is the answer.

In your mind, this shouldn't take longer than 2 minutes, though this response took me like five minutes to type.

I have memorized the first 26 prime number as advised by the gmatclub math-book.
So instantly i could see only 73 (and 37) goes with both the statements. D

Following is how I would approach this question:

"p and q are different two-digit prime numbers with the same digits, but in reversed order."

Since we are talking about prime numbers, the only digits to consider here are 1, 3, 7 and 9.

(1) p + q = 110
I need a sum of something more than 100. I get that if I add something from 10s with something from 90s. So a possible pair is 19 and 91 but 91 is not prime.
I also get something more than 100 is I add something from 30s with something from 70s. So a possible pair is 37 and 73.
There is no other combination that can give me something close to 100.

(You can also think of it as finding numbers equidistant from 55. So something from 30s and something from 70s can be equidistant from something from 50s)

(2) p – q = 36
The difference of the numbers is around 40. Something from 30s and something from 70s can have a difference of around 40. So 37 and 73 is the required pair. There is no other combination possible where the difference could be around 40.

Answer (D)

D it is. Used the same logic as blink005 but it took me more than 2 mins to solve this problem.
Considering that this problem has been tagged as 700-level, is the time taken (a little over 2 mins) excess?

Just notice to two-digit prime numbers

Please correct my approach if it is wrong

Let p be the greater number and of the form 10x+y
q be of the form 10y+x

Statement 1:
-----------------
p+q = 110
11(x+y) = 110
x+y = 10

Only two single digit prime numbers supporting this are 7 and 3. Hence sufficient

Statement 2:
----------------
p - q = 36
9(x-y) = 36
x-y = 4

Only two single digit prime numbers supporting this are 7 and 3. Hence sufficient

Option D

I don't think 1 is a prime number, so I wonder why any of you shortlisted numbers with digits 1 in the run up to your solutions

Yes, 1 is NOT prime. The smallest prime is 2.

Also, we are told that p and q are primes. Now, the digits of p (or q) could be primes and p (or q) could be non-primes. For example, 25, 27, 32, ...

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?

Given: \(p=prime=10x+y\) and \(q=prime=10y+x\), for some non-zero digits \(x\) and \(y\) (any two digit number can be expressed as \(10x+y\) but as both \(p\) and \(q\) are two-digit then \(x\) and \(y\) must both be non-zero digits).

(1) p + q = 110 --> \((10x+y)+(10y+x)=110\) --> \(x+y=10\). Now, if we were not told that \(p\) and \(q\) are primes than they could take many values: 91 and 19, 82 and 28, 73 and 37, 64 and 46, ... Thus the larger number could be: 91, 82, 73, or 64. But as we are told that both \(p\) and \(q\) are primes then they can only be 73 and 37, thus the larger number equals to 73. Sufficient.

(2) p – q = 36 --> \((10x+y)-(10y+x)=110\) --> \(x-y=4\). Again, if we were not told that \(p\) and \(q\) are primes than they could take many values: 95 and 59, 84 and 48, 73 and 37, 62 and 26, 51 and 15. Thus the larger number could be: 95, 84, 73, 62, or 51. But as we are told that both \(p\) and \(q\) are primes then they can only be 73 and 37, thus the larger number equals to 73. Sufficient.

Answer: D.

OPEN DISCUSSION OF THIS QUESTION IS HERE: p-and-q-are-different-two-digit-prime-numbers-with-the-same-98912.html