If S is the infinite sequence S1=9 S2=99 S3=999 SK=10^k-1

+1 E

Question tends to make you just go for 3 & you choose C but as Bunuel pointed out if you use 11 the assumption fails

Hello Bunuel

I have a doubt here.
Since we know after combining the 2 statements that the Prime number will NOT be 3 but any other Prime number that divides at least 1 number in the sequence.
So now we know for SURE that EVERY TERM IS NOT Divisible by a particular Prime number( Which the Questions asks). 11 will also not divide all terms but few only.

SO should the Answer not be "C" ?
Please correct me.

Thankyou

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Why cannot p be 3?

Hi All,
Can any of the moderators amend this post to create a gap between the question and the query posted by morya003?

______________
Done. Thank you.

The only prime that divides all terms in the sequence: 9, 99, 999,.. is 3. So the question is essentially asking: is p=3?
(1) p can be any prime number greater than 2, hence not sufficient.
(2) p can be 3 or 11, hence not sufficient.
Both taken together, again insufficient to identify p as 3.

Answer: E

boomtangboy summarizes aptly that the question tends to prey on the mistake that the test taker will consider 3 as the only possible value from statement (2).

Here we can choose 11 and 3 as test cases and we will get E answer fairly easily

From question stem we know S={9,99,999,9999,99999,999999...................99999999999999....}
We can easily eye ball that this is divisible by either 3 or 11 or a combination of their multiples.

(1) p is greater than 2.
Insufficient :- p can be 3 or 5 or 7 or 11 or 13
In some cases p divides ; in some cases it don't

(2) At least one term in sequence S is divisible by p.
Insufficient again
starting from the second term (99) all terms are divisible by either 3 and 11
But the first team 9 is divisible only by 3 and not by 11

So we cannot for sure whether the prime p that the question stem is referring to is 3 or 11.

BOTH STATEMENT INSUFFICIENT
ANSWER IS E

Question is tricky and seems very tough in the first instance. Though I understood that every term is not divisible by p, but I marked C option due to my understanding that every term is not divisible by p (I did a mistake here... Every term is divisible by 3 and not by 11 as correctly suggested by @bunuel). So, the reading and understanding the language of the question in such problems is very important.

So, finally the solution goes like this.

Given : infinite sequence: S1=9, S2=99, S3=999, ..., SK = 10^K-1
DS: is S1, S2, S3, ......, SK divisible by the prime number p.

Statement 1 : P>2. So, p can be 3,5,7,11,.....
We can clearly see that every term is divisible by 3 but not by 5,7,11.....
NOT SUFFICIENT

Statement 2 : At least one term in sequence S is divisible by p. So, we can see that 9 is divisble by 3 and 99 is divisible by 11.
Here again every term is divisible by 3
But every term is not divisible by 11.
NOT SUFFICIENT

Combined : p>2 and At least one term in sequence S is divisible by p. So, p can be 3 or 11 or...
Its clear that every term is divisible by 3, but every term is not divisible by 11.
NOT SUFFICIENT

Answer E

but s5 is 9999 is clearly divisible by 11 right ?

Question: is every term in S divisible by the prime number p?

Statement 1:
If p = 3: Each term in S is divisible by the prime number 3 (3 is greater than 2)
If p = 11: Each term in S is NOT divisible by the prime number 11 (11 is greater than 2)

Thus, p = 3 or 11 => Hence, NOT Sufficient


Statement 2: At least one term in sequence S is divisible by p

If p = 3: Each term in S is divisible by the prime number 3 => At least one term in S is divisible by the prime number 3
If p = 11: The terms 99, 9999, etc. in S are divisible by the prime number 11 => At least one term in S is divisible by the prime number 11

Thus, p = 3 or 11 => Hence, NOT Sufficient

Combining: Even after combining, we have p = 3 or 11 => Hence, NOT Sufficient

Answer E

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