Let p and q each represent a digit from 0 through 9 and let pq represe

Question

Let p and q each represent a digit from 0 through 9 and let pq represent a 2-digit positive prime number such that \(20 < pq < 99\). For example, if \(p = 2\) and \(q = 3\), then the 2-digit prime number \(pq\) is 23. If the 2-digit integer \(qp\) is 27 greater than \(pq\), what is the sum of the digits p and q?

A. \(5\)

B. \(7\)

C. \(9\)

D. \(11\)

E. \(13\)

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Attachment:

Screenshot 2023-12-10 004311.png

MartyMurray · Accepted Answer

Let p and q each represent a digit from 0 through 9 and let pq represent a 2-digit positive prime number such that \(20 < pq < 99\). For example, if \(p = 2\) and \(q = 3\), then the 2-digit prime number \(pq\) is 23. If the 2-digit integer \(qp\) is 27 greater than \(pq\), what is the sum of the digits p and q?

A. \(5\)

B. \(7\)

C. \(9\)

D. \(11\)

E. \(13\)

Show Spoiler

Attachment:

Screenshot 2023-12-10 004311.png

The value of  2 -digit number  qp  is  10q + p .

The value of  2 -digit number  pq  is  10p + q .

So, if  qp - pq = 27 , then  (10q + p) - (10p + q) = 9q - 9p = 27 . Thus,  q - p = 3 .

Since  q  must be  3  greater than  p , the only possibilities for  pq > 20  are  25 ,  36 ,  47 ,  58 , and  69 .

25 ,  36 ,  58 , and  69  are not prime. So,  47  is the only possible  pq . We can confirm that  74 - 47 = 27 .

Thus the sum of  p  and  q  is  4 + 7 = 11 .

Correct Answer   D

RahulJain293 · Answer

I like this question! Got it wrong in my mock but found interesting way to solve it...

Since we know q = p +3, we can narrow down possible options of p & q based on the scope and question stem

p can be 2,3,4,5,6 (NOT 7,8,9 since it has to smaller than Q by 3)
q out of 5,6,7,8,9 CAN ONLY be 7 & 9 since it is UNIT's digit and prime..

Hence we can quickly make all possible combos and eliminate to find that ONLY 47 is where the difference is 3 between the digits and we can also double check by adding 27 to 47 to get 74 (swapped digits)

Hence 11

Archit3110 · Answer

given
10q+p- 10p-q = 27
q-p = 3
possible number such that pq is prime ; 47
sum is 11
option D

shivanks · Answer

can any one explain to me why A is wrong ?
41 is a prime no and when we interchange the digits then we got 14 and according to question 
(41-14=27) then the some of p+q is 5 how it is wrong?

Archit3110 · Answer

Check question for range where pq can fall in..
 shivanks

20<pq<99

Posted from my mobile device

Jake7Wimmer · Answer

I actually think it's just easy to list out the primes starting with 23 and flipping them to see your options. Doesn't take too long because they're aren't actually that many.

After listing out a few you're kinda just left with 41 -> 14 and 47 -> 74.

If you read the question, the lather number needs to be larger so 47 -> 74 is the only answer and adding the sums = 11.

Really only takes like 30 seconds this question.

firuzaadilzadeh · Answer

why is 63-36=27 is wrong? it satisfies the equation. some questions just make me sick!

Bunuel · Answer

We are told that the smaller number, pq, is prime, whereas 36 in your example is not. You should read questions more carefully.

KarishmaB · Answer

We are given that qp - pq = 27

(10q + p) - (10p + q) = 27 
 9(q-p) = 27 
 (q-p) = 3

Hence various values of p and q are possible. Neither p nor q can be 0 because both pq and qp are two digit numbers. 
p = 1, q = 4, Then pq = 14 (Not prime)
p = 2, q = 5, Then pq = 25 (Not prime)
p = 3, q = 6, Then pq = 36 (Not prime)
p = 4, q = 7, Then pq = 47 (Prime!)

We needed pq to be prime (but not qp necessarily) and when p = 4, q = 7, pq is prime. Here p + q = 11.
Other values of p and q are also possible but we will not evaluate them because this is a PS question and here there will be only one correct answer.

Hence  answer (D)

hemanthPA · Answer

Given pq = 23, 29, 31, 39, 41, 43, 47, 53, 57, 59, 61, 67....97 
 
10p + q = 27 + 10q + p
P + q=?

10p + q + 27 = 10q + p
 27 = 9q – 9p
27/9 = q-p = 3

Therefore q can be 5,7,9 (Q cannot be even, and q must be greater than 3)
And p can be 2, 4, 6

So possible combinations are, 25, 47, 69
25 and 69 aren’t prime, so 47 is the only option

Lets check qp = 27 + pq, which is 27 + 47 = 74

Therefore, q+p = 7 + 4 =11

Danou · Answer

Why did you add the coefficient "10" to the equation?

Krunaal · Answer

2-digit integer  qp  can be written as  10q + p  because  q  is 10's digit and  p  is units digit. Example, if  qp  = 74, here  q =7 and  p =4
10 q  +  p  = 10*(7) + 4 = 70 + 4 = 74.

KarishmaB · Answer

Danou

Look there are 2 ways of talking about a two digit number. I can say a two digit number is x. I can say when we reverse its digits, the number becomes y. If the difference between them is 9, I can write it as x - y = 9. But how do I solve it now? I have lost some important information - that digits in y are reverse of digits in x.

So in this case I can say that say the first two digit number is pq (where p is tens digit and q is units digit). Then the second number is qp (where q is tens digit and p is unit digit). Now this captures the info of reversing of the digits. But when I want to now calculate the difference between them, I must write it as
10p + q because the value of pq = 10p+q. 
Essentially, when p = 3 and q = 2, pq = 32 = 3*10 + 2 
I must take the place value of the digit into account too.

A three digit number per = 100p + 10q + r where p, q and r are the digits of the number (say 8, 5 and 1). 
851 = 100*8 + 10*5 + 1

Kinshook · Answer

Let p and q each represent a digit from 0 through 9 and let pq represent a 2-digit positive prime number such that  20 < pq < 99 . For example, if  p = 2  and  q = 3 , then the 2-digit prime number  pq  is 23.

If the 2-digit integer  qp  is 27 greater than  pq , what is the sum of the digits p and q?

20 < 10p + q < 99

(10q + p) - (10p + q) = 27
9(q-p) = 27
q - p = 3
q = p + 3 < 10
p < 7

20 < 10p + p + 3 < 99
20 < 11p + 3 < 99
17 < 11p < 96

p = {2,3,4,5,6}

p = 2 ; q = p+3 = 5; 25 is NOT a prime number
p = 3 ; q = p+3 = 6; 36 is NOT a prime number
p = 4 ; q = p+3 = 7; 47 is a prime number
p = 5 ; q = p+3 = 8; 58 is NOT a prime number
p = 6 ; q = p+3 = 9; 69 is NOT a prime number

pq = 47
Sum of digits of pq = 4+ 7 = 11

IMO D

DanTheGMATMan · Answer

Classic digit swap scenario, set up as usual:
 https://www.youtube.com/watch?v=E5lH7E-oGjo

Em1le · Answer

they specified it is a prime number, I did the same mistake.

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Let p and q each represent a digit from 0 through 9 and let pq represe

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GMAT Problem Solving (PS) Questions

Let p and q each represent a digit from 0 through 9 and let pq represe

Prep Toolkit

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