MGMAT Question: Reverse Primes

[shahan25]

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

Why is is D not the right answer?
The explanation given is as follows:

Suppose that one of the digits is even. In that case, one of p and q will be even"”but there are no two-digit even primes. This indicates that both digits need to be odd, drawn from the set {1, 3, 5, 7, 9}. By a similar reasoning, we can rule out the possibility of 5 being one of the digits, as that would lead to one of the numbers ending in 5 (and therefore being divisible by 5). Lastly, we observe that the two digits cannot be identical, because that would make p and q equal.

In light of the above arguments, the only combinations of numbers that satisfy the conditions are as follows:
(13, 31), (17, 71), (19, 91), (37, 73) and (79, 97). Let us now consider the statements.

I disagree with the 19/91 combination as 91 is clearly not a prime and hence can't be p/q. Can you please explain?

[esledge]

You are right. 91 = 7*13. I checked our database, and this has already been edited to show (D) as the answer, with the correct p and q pairs listed as justification. It's likely you looked at an old version, but nice catch!

[singh.181]

Sorry for bringing up this thread :(

Can expert help me in interpreting the question?
I interpreted it as "which one is larger i.e either p or q"?

Stat 1: gives 37 and 73, so either p or q could be either of the values

Stat 2: gives p> q since p-q = 36.

So, choose B.

Please explain.

[jnelson0612]

Sorry for bringing up this thread :(

Can expert help me in interpreting the question?
I interpreted it as "which one is larger i.e either p or q"?

Here's the problem. You interpreted it differently from what is asked. Under your interpretation, the data is sufficient if we can say "p is larger" or "q is larger".

Look at what the question says: What is the value of the larger of p and q?

So to be sufficient, the data has to allow us to determine whether p or q is larger and the VALUE of that larger variable.

Hope this helps!

[kinshukpharma]

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

Why is is D not the right answer?
The explanation given is as follows:
By a similar reasoning, we can rule out the possibility of 5 being one of the digits, as that would lead to one of the numbers ending in 5 (and therefore being divisible by 5).

I dont understand why 5 has been eliminated. Can someone please explain?

[mithunsam]

If 5 is one of the digits, then atleast one of p or q will have unit digit 5. However, all numbers with unit digit 5 are divisible by 5. So they are not prime and we should exclude them.

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

Why is is D not the right answer?
The explanation given is as follows:
By a similar reasoning, we can rule out the possibility of 5 being one of the digits, as that would lead to one of the numbers ending in 5 (and therefore being divisible by 5).

I dont understand why 5 has been eliminated. Can someone please explain?

[tim]

thanks mithunsam..

[jafsaq]

An alternative..

(2) P-Q = 36. Since both are prime, both P & Q are positive. So, P>Q

(1) P+Q = 110

In order to prove, if P > Q or not, all we have to do is to see if we can find value of P & Q with the given condition of p + Q = 110

Lets assume the unknown digits are X & Y.
So, P = 10X + Y and Q = 10Y + X

Given that P + Q = 110

(10X + Y) + (10Y + X) = 110

11X +11Y = 110

X + Y = 10

Since, P & Q are prime, X & Y must be Odd.

Of all combinations of X & Y that can yield sum of 10, only 3 and 7 are odd. ( Other combinations such as 1 & 9 ; 2 & 8 ; 4 & 6 do not have both odds ).

We can stop the calculation here. Since we can find P & Q with X & Y, we can find which is greater of P & Q.

[jnelson0612]

Nice, I like it! Thanks! :-)

[krishnan.anju1987]

An alternative..

(2) P-Q = 36. Since both are prime, both P & Q are positive. So, P>Q

(1) P+Q = 110

In order to prove, if P > Q or not, all we have to do is to see if we can find value of P & Q with the given condition of p + Q = 110

Lets assume the unknown digits are X & Y.
So, P = 10X + Y and Q = 10Y + X

Given that P + Q = 110

(10X + Y) + (10Y + X) = 110

11X +11Y = 110

X + Y = 10

Since, P & Q are prime, X & Y must be Odd.

Of all combinations of X & Y that can yield sum of 10, only 3 and 7 are odd. ( Other combinations such as 1 & 9 ; 2 & 8 ; 4 & 6 do not have both odds ).

We can stop the calculation here. Since we can find P & Q with X & Y, we can find which is greater of P & Q.

Hi, I used the same approach and the thing that I did not understand is the part where you mentioned that other combinations are not both odd. 1 and 9 are odd and so are 3 and 7, 5 and 5. I think the process of elimination of these numbers is that they one of them is at least not prime. e.g 55 and 91 etc.

[tim]

it sounds like you do have a correct understanding of this one (including of the error the previous poster made); let us know if there is still something you need help with..