If p, q are different prime numbers greater than 2, which of the follo

Question

If p, q are different prime numbers greater than 2, which of the following can have at most 3 different factors?

A)  2p+q 
B)  p+q 
C)  pq 
D)  p^2q 
E)  p^q

gracie · Answer

p^2 can only have p, p^2, and 1 as factors
e.g., 3, 9, 1; 7, 49, 1; 11, 121, 1
D

philipssonicare · Answer

gracie

Apologies, there was a typo. Option D is  p^2q , not  p^2

ilepton · Answer

Can you please explain the solution to this problem ?

philipssonicare · Answer

ilepton

I have attached the answer from math revolution. I feel it is wrong/I misunderstand though.
2P+Q, if p=5 and q=11
2·5+11=21, 3^1·7^1=(1+1)(1+1)=4. Can’t be A?

martinoco · Answer

As p and q are primes greater than 2, both are odd.

odd1 + odd2 = even
p+q is an even greater than 2, so at least it will have 4 factors: 1, 2, (...), (p+q)
Try p=3 and q=5
p+q=8, Factors: 1, 2, 4, 8

At least 4 Factors: 1, p, q, pq

5 factors: 1, p, p^2, pq, q, (p^2)*q

Remember that p and q are prime numbers greater than 2, so p^q > p^2
At least 5 factor (similar to D): : 1, p, p^2, p^q

So the only possible answer is A.

jfranciscocuencag · Answer

Hello!

Can we always assume that the statements will work for any different prime numbers?

Kind regards!

KanishkM · Answer

At most 3 means they can be 3 or less than 3 as well

p = 3, q = 5, in all the cases

A)  2p+q 
11 => 11^1 => 2 factors ( while calculating powers we add 1 to the power)

B)  p+q 
8 => 2^3 => 4 factors

C)  pq 
15 => 3 * 5 => 4 factors

D)  p^2q 
45 => 3 * 5 * 3 => 6 factors

E)  p^q 
3^5 will again be more than 3 factors

A

KanishkM · Answer

Would like to share my 2 cents on this,

Not necessary,we will have to search for that particular pair which will satisfy the question.

For option A

If you take another pair such as 7,11 => 25 => 3 factors, this will hold good

But when you take 7,13 => 27 => 4 factors, this will fail

Nevertheless the approach mentioned might not be the easiest one to solve this question, Brute force does take time, to get to the answer.

philipssonicare · Answer

chetan2u ,  Gladiator59 ,  VeritasKarishma ,  Bunuel ,  generis 
Please explain how the answer is A when the below is a possibility.
2P+Q, if p=5 and q=11
2·5+11=21, 3^1·7^1=(1+1)(1+1)=4.

generis · Answer

philipssonicare , this question is too clever by half. This sentence uses can in the sense of could —not can in the sense of can only. ( Can only implies must only . You just found out why the question does not mean must only . +1

) Can as a modal verb means to be able to . I am not a fan of questions like this one. The prompt says : If p, q are different prime numbers greater than 2, which of the following can have at most 3 different factors? The prompt means If p, q are different prime numbers greater than 2, in which of the following is it possible for the solution to have at most 3 different factors? Or: which of the following could have at most 3 different factors under the given conditions? If we can come up with even one solution for an option that satisfies the conditions, then that option "can" have at most 3 different factors. Only option A "can" (is able to) have at least one solution that has at most 3 factors. Option A does not have to have only 3 factors every time. The other options will never render a solution with at most 3 factors. Hope that helps.

KarishmaB · Answer

p and q are different prime numbers greater than 2. So they must be both odd prime numbers. 
Let's consider options (C), (D) and (E) first since they are easier.

C)  pq

Factors must be 1, p, q and pq
(Definitely has more than 3 factors)

D)  p^2q

Factors must be 1, p, q, p^2...
(Definitely has more than 3 factors)

E)  p^q

Factors must be 1, p, p^2, p^3 ...
(Definitely has more than 3 factors because q is more than 2)

Now let's look at (B).

B)  p+q

Odd + Odd = Even
So the two odd prime numbers will add up to give an even number. This means that 2 must be a factor of the sum. We also know that factors appear in pairs. If there is a factor x, there is another factor y (distinct or same) which will multiply with x to give the number. e.g. if 2 is a factor of 6, there is a factor 3 which multiplies with 2 to give 6. 
Hence. here also, there must be a complementary factor of 2, say A. 
Factors must be at least 1, 2, A, p+q 
(Definitely has more than 3 factors)

By elimination, answer must be (A).

A)  2p+q 
Say p = 3, q = 5
2p+q = 11
Factors are 1 and 11 (only 2)
Hence 2p + q CAN have at most 3 factors (for some values of p and q).
Note that it does not imply that 2p+q MUST have at most 3 factors for all values of p and q. There could be values of p and q for which it has more than 3 factors such as p = 7, q = 13
2p+q= 27
Factors are 1, 3, 9, 27 
But the questions asks us for the option which CAN have values of p and q such that there are 3 or fewer factors. None of the other options can have any values of p and q such that there are 3 or fewer factors.

Answer (A)

nick1816 · Answer

We can write any prime number greater than 3 in the form of 6k+1 or 6k-1
Now there are 2 cases when 2p+q can be a prime number.
1. p=6a-1 and q=6b+1
2p+q= 6(2a+b)-1=6X-1 which can be prime

2. p=6a+1 and q=6b-1
2p+q=6(2a+b)+1=6Y+1, which can also be a prime

bumpbot · Answer

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If p, q are different prime numbers greater than 2, which of the follo

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If p, q are different prime numbers greater than 2, which of the follo

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