pq

Question

pq<70, where both P and Q are distinct odd primes. Determine PQ.

(1) PQ is one greater than a power of two

(2) The sum of the digits of PQ is a prime number




Please explain your answer

eschn3am · Answer

1. PQ is one greater than a power of two. So PQ is either

3, 5, 9, 17, 33, 65 (not counting 2^0, because that doesn't fit the rule later anyway)


2. The sum of the digits of PQ is a prime number

3, 5, 9, 8, 6, 11

now it can't be 9 because there aren't any distinct odd primes that multiply together to = 9. but that still leaves us with 

3 (3*1), 5 (5*1), 17 (17*1), 33 (3*11) and 65 (5*13)

and the sum of the digits of PQ

3, 5, 8, 6, 11

so that rules out 17 (1+7=8) and 33 (3+3=6)

but we're still left with 3, 5 and 65 fitting both statements.

E. Insufficient

tarek99 · Answer

the OA is C actually. i'm trying to figure out how. anybody?

asdert · Answer

We all agree that 1, 2 INSUFF.

Taking both into consideration, we are looking for 2 odd primes that when multiplied, the result is a power of 2+1 and that the sum of the digits equal another prime.

3*11 = 33 = (2^5)+1 and 3+1+1 =5
we also have:
5*13 = 65 = (2^6)+1, but 5+1+3=9, which is not a prime.

Answer is 3(11)=33= C

FN · Answer

I get E

lets see
1) says PQ can be 65, 33, 17, 9 5 however since p and q are odd primes..only 65 and 33 count

2) the sum of the numbers is prime number

well 2+3=5 also counts 5+2=7 also counts

1+2 together there is no number that meets both statements..

pq<70, where both P and Q are distinct odd primes.

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