If P and Q are integers, is [m][fraction](10^{2P} + Q)/3[/fraction][/m

Question

If P and Q are integers, is  \frac{(10^{2P} + Q)}{3}  an integer?
(1) Q = 5
(2) P*Q is even

GMATbuster's Weekly GMAT Quant Quiz #5 Ques No 2

pk123 · Accepted Answer

If P and Q are integers, is (102P+Q)/3 is an integer

an integer?
(1) Q = 5
(2) P*Q is

Option A : Q= 5, so term is (10^2P +5)/3...now if we have P=0 or P be any integer >0, 
(10^2P +5)/3 will be an integer
Illustration: if P=0, (1+5)/3 = 2
or if P=1, 100+5/3= 105/3= 35

but what if P is negative , if P=-1 it becomes (1/100+ 5)/3, which is not an integer

hence A is not sufficient

Option B: P *Q is even 
which means P is even Q is even
P is odd, Q is even 
P is even and Q is odd

so lets say 10^2+4/3= 104/3 , not an integer 
but 10^2+5= 35 an integer
so B is not sufficient

Lets take both ,
Q=5 and P needs to be even 
if P>0, then expression is definitely an integer as we discussed in Option A explanation
but if P becomes negative , lets take P= -2
so it becomes 1/10000 +5/3 which is not an integer...so even together cant solve the question

Hence E is the answer

lostboy · Answer

A:
We don't need to know the value of P so statement 1 is sufficient. regardless the numerator will be 10...05 and the property for if a number is divisible by 3 is to add the digits. so 1+5 = 6, thus it is an integer

Statement 2 doesn't give us a value for Q so its insufficient

saban · Answer

Stat 1 -

P =0 Q = 5 
Ans 2 
P=-1 Q=5 
Ans 0.01+5 / 3

NS

Stm 2 
P*Q even - either P is even or Q is even - NS

combined 
P is even but if its 2 then int and if -2 then not

so answer is E

raghavshekhar · Answer

A.

We are not worried about the value of P, only Q. The first provides value of Q for which answer can be confirmed Yes as per divisibility rule of 3. 
The second statement mentions the following:
P is even, Q is even
Or
P is even, Q is odd
Or
P is odd, Q is even
There are cases (q=5 or q=6) for which we get both confirmed Yes and No. Thus, not sufficient.

yenbh · Answer

Statement 1: 
Q = 5 --> remainder = 2
When  10^2p  divided by 3 
 + if p= 0,  10^2p  leave remainder =0
 + if p>0, always leaves remainder = 1
 + if p<0, this fraction does not divisible by 3

--> Insufficient

Statement 2:

PQ Even. It is undecided whether  10^(2P)  and Q is even or odd, this leave different remainders when The fraction is divided by 3

--> Insufficient

Combine 2 statements, Q is 5 and PQ is even. This means P must be even.

From analysis in Statement 1 above , P is even and P>0 then we have
-->  \frac{(10^(2P)+Q)}{3}  is divisible by 3

—> Sufficient

Hence, Choice C

redskull1 · Answer

A

To be divisible by 3, the sum of all the numbers should be divisible by 3

A)The sum of all the digits in the first number is 1 whatever be P.Also +5 implies 1+5=6 irrespective of whatever is P.SUFFICIENT.

B)P*Q is even.One case is p is even q is odd.Not always divisible by 3.INSUFFICIENT.

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If P and Q are integers, is [m][fraction](10^{2P} + Q)/3[/fraction][/m

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If P and Q are integers, is [m][fraction](10^{2P} + Q)/3[/fraction][/m

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