If S is a sequence of numbers of the form X, XX, XXX, XXXX, XXXXX, …,

B.
Trial and Error method.
P will be 2,3,5,7
So 2nd statement is true in any case.
1st statement is disproved if one takes X as (say) 4 and P as 3.

My Take: B
I always solve these kind of questions by taking an example of numbers.
1) possible values of p =3/5/7 (lets take p=3)
to make x a multiple of p, X should be 6, but x can be 5 as well. Hence, NOT SUFFICIENT.

2) obviously its very much sufficient as Every number in series will be multiple of P if, X is multiple of P. This is telling what exactly question is asking.

If S is a series of numbers of the form X, XX, XXX, XXXX, XXXXX, …, where X is a non-zero digit, is every number in this series a multiple of the same prime number P?

(1) P is an odd number such that P < X
Since X is a one digit number it can take any value from 1-9
P is prime and P is odd and since P is less than one digit number X, we know that our prime is an odd prime less than the highest one digit number 9
therefore P=3,5,7
So many possible cases
P can be 3 but the elements in the set can be {4,44,444,444} NOT a multiple of P(3)
P can be 5 and the element in the set can be {55,555...5555,,,} YES it is a Multiple of P(5)
Insufficient

(2) X is a multiple of P
P is a prime, so our our range of primes is single digit prime=> 2,3,5,7 (because the lowest element X in the set is single digit)
The set can look like
When P= 2 then S={2,22,222,2222...}
When P= 3 then S= {3,33,333,3333,33333,..}
When P= 5 then S={5,55,555,5555,...} or
When P= 7 then S={7,77,777.....}

NOTICE HOW THE CONDITION STATED IN THE FIRST STATEMENT THAT P<X BECOMES REDUNTANT ; BECAUSE EVEN IF WE REMOVE THE FIRST ELEMENT (SINCE IN CASE OF FIRST ELEMENT P=X) , STILL THE REST OF THE ELEMENT OF THE SET WILL BE MULTIPLES OF P

Sufficient
All element {X,XX,XXX...} in the set {S} are multiple of the prime {P}

ANSWER IS B

Note that P < X.
So P = 5 and X = 5 is not valid. But P = 3 and X = 6 is a valid example for YES.

(2) X is a multiple of P
P is a prime, so our our range of primes is single digit prime=> 2,3,5,7 (because the lowest element X in the set is single digit)
The set can look like
When P= 2 then S={2,22,222,2222...}
When P= 3 then S= {3,33,333,3333,33333,..}
When P= 5 then S={5,55,555,5555,...} or
When P= 7 then S={7,77,777.....}

NOTICE HOW THE CONDITION STATED IN THE FIRST STATEMENT THAT P<X BECOMES REDUNTANT ; BECAUSE EVEN IF WE REMOVE THE FIRST ELEMENT (SINCE IN CASE OF FIRST ELEMENT P=X) , STILL THE REST OF THE ELEMENT OF THE SET WILL BE MULTIPLES OF P

Sufficient
All element {X,XX,XXX...} in the set {S} are multiple of the prime {P}

ANSWER IS B[/quote]

Note that P < X (the digit) in statement 1. It means that if P = 2, X = 4 or 6 or 8. It means X cannot be 2. The comparison is not between P and the first number. It is between P and the digit X.

According to statement 2:
If P = 2, we have {2, 22, 222, 2222 ...} or {4, 44, 444, 4444...} or {6, 66, 666...}
etc
If P = 3, we have {3, 33, 333...} or {6, 66, 666...} etc

the sequence looks something like x( 1,11,111,1111,........) 111 divisible by 3 , 1111 divisible by 11... thus for all the terms to be a multiple of a prime P , x has to = P or x is a multiple of P

from 1

p not = x ... x could be a multiple of p ..... insuff

from 2

p = xm ... suff


B

according to st1, X can be or cannot be divisible by P.
by st2, just test each scenario => P = 2, or = 3, = 5, =7.

for st 2
p= 2, x=6,66,666,6666,66666....
p=3, x =6,66,666,6666,66666....

here we have two different value of 'p' in case of 6.
how can st2 be the answer.