If x and y are integers, is xy + 1 divisible by 3?

(1) When x is divided by 3, the remainder is 1 --> x=3n+1, if x=1, then 1*y+1 when y=1 not div by 3 and when y=2 div by 3 => insufficient
(2) When y is divided by 9, the remainder is 8 -->y=9m+8, if y=1, then x*1+1 when x=1 not div by 9 and when x=8 div by 9 => insufficient

(1) + (2) :
x=3n+1 and y=9m+8 where m and n are integers=> xy+1=(3n+1)(9m+8)+1 =27mn+9m+24n+8+1 =27mn+9m+24n+9 => this is div by 3 always hence is sufficient => C

Ans. C)
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Given x and y are integers. We have to find out whether xy+1 is divisible by 3 ?

Statement 1: When x is divided by 3, the remainder is 1

Therefore, x can be represented as 3*a+1 ( where a is an integer)
therefore, xy+1 => (3*a+1)*y+1

Since 3*a*y is divisible by 3, we have to find out whether (y+1) is divisible by 3.


Statement 2:
When y is divided by 9, the remainder is 8. Clearly insufficient, since we don't know the value of x.
But, we can say that y => 9*b+8 ( where b is an integer)

Combining 1) and 2), we can say y+1 = 9*b + 9 is divisible by 3. Hence the answer should be C

Question: remainder of (x*y +1)?
(1) Not sufficient. Just tells that x could be equal to 4, 10, 13,.... but nothing about y.
(2) Not sufficient. Just tell that y could be 17, 26, 35,...but no information about x.

(1)+(2) Sufficient.
If x=10, y=26, then (10*26 + 1) = 261 ----->which is divisible by 3, so the remainder is 0
If x=4, y=35, then (4*35 + 1) = 141 ----> divisible by 3, so the remainder equals 0

Answer C

If x and y are integers, is xy + 1 divisible by 3?

(1) When x is divided by 3, the remainder is 1.
(2) When y is divided by 9, the remainder is 8.

Statement 1
x = 3m+1; m is an integer
no information on y, hence insufficient.

Statement 2
y= 9n +8; n is an integer
no information on x, hence insufficient.

Combining statements (1) and (2) and substituting x and y as (3m+1) and (9n+8)
xy +1 = (3m+1)(9n+8)+1 = 27mn+24m+9n+9 = 3*(9mn+8m+3n+3) --> xy+1 is divisible by 3
Hence, both statements together are sufficient

Answer: C

Should be C, since:
xy + 1 = (3ka+1)y = 3ka*y +y = 3kb + 1 + 9kc + 8 = 3kb + 9kc + 9 = multiple of 3.
Hence C.

Statement One: If x is divided by 3, then x is even number, i.e. 4 (since no info is given on Y- INSFFICIENT y can be 1 or 3 for example)

Statement Two: When Y is divided by 9, remainder is 8, I.E. y = 9q+8 , so possible values for Y are 8, 17, 26, 35 etc (INSUFFICIENT)


Combining both we know that x is 4, 7, 10 etc and y is 8, 17, 26, 35 etc. hence xy + 1 divisible by 3


C

My approach:
Stem: If x and y are integers, is xy + 1 divisible by 3?
–> notice that xy+1 is divisible by three if xy is even (but not iff it is even)

(1) When x is divided by 3, the remainder is 1.
x=3q+1
–> not divisible by 3
–> y could be divisible by 3 or it could not be, producing two different answers
–> Hence, NOT SUFFICIENT

(2) When y is divided by 9, the remainder is 8.
–>Hence, the remainder when divided by 3 is also 8
–>However, x could be divisible by 3 or it could not be, producing two different answers
–>Hence, NOT SUFFICIENT

(1) & (2) together
—>From 1, we know that remainder is 1
—>From 2, we know that remainder is 8
—>Iff the remainder product of xy and the sum of the remainder product of xy and 1 add to 3, xy+1 must be divisible

HENCE, (remainder x) * (remainder y) + (1) = 8*1+1= 9, hence, xy+1 is always divisible by 3: C
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If x and y are integers, is x*y + 1 divisible by 3?
--->From the question stem, we need to find out the divisibility of x & y to 3 which should be given in each statement.

(1) When x is divided by 3, the remainder is 1.
3*q + 1 = x
-->3*q + (3 - 2) = x
--->3(q + 1) - 2 = x
However, we have no idea whether or not y is divisible by 3 ---> This statement is NS

(2) When y is divided by 9, the remainder is 8.
9*r + 8 = y
-->9*r + (9 - 1) = y
--->9(r + 1) - 1 = y
However, we have no idea whether or not x is divisible by 3 ---> This statement is NS

(1) + (2): we have enough info to answer that question, and x*y + 1 is not divisible by 3

(1) When x is divided by 3, the remainder is 1: \(x=3q+1=(1,4,7,10…)…y=anything\) insufic.
(2) When y is divided by 9, the remainder is 8: \(y=9p+1=(8,17,26,35…)…x=anything\) insufic.

(1&2) \(\frac{(x=3q+1)(y=9p+1)+1}{3}=\frac{(27pq+9p-3q-1)+1}{3}=\frac{27pq+9p-3q}{3}=\frac{3(9pq+3p-q)}{3}=divisible\) sufic.

Is this last consideration a rule?

(1)\( x=4; y=1; xy+1=4+1=5\) Not Divisible \(3.\)

\(x=4; y=2; xy+1=8+1=9;\) Divisible by \(3\)

\(Insufficient. \)

(2) \(y=17; x=1; xy+1=17+1=18\) Divisible by \(3
\)
\(y=17; x=2, xy+1=34+1=35\) Not Divisible \(3\)

\(Insufficient. \)

Considering Both options.

\(y=17; x=4, xy+1=68+1=69 \ is \ Divisible \ by \ 3 \ Sufficient. \)

The answer is \(C\)
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1) x= 3a+1
Tells us nothing about y, hence insufficient

2) y = 9b+8
....y= 3b+2
Tells us nothing about x hence insuffuicent

1 + 2 -> xy = (3a+1) * (3b+2)
xy = 9ab + 6a + 3b +2
xy + 1 = 9ab + 6a + 3b + 3

You can see that each of the additive terms in the above equation are divisible by 3, hence C is sufficient!