Nermal wrote:If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.

From definition if the sum of the digits of an interger is a multiple of 3, the integer is a multiple of 3. a+b is a multiple of 3, so is ab or ba. So 1 is Suff:

For 2 to be a multiple of 3, both a and 2b must be. Since 2 is even, we know b must be a multiple of 3. So a/3 +b/3 is an integer or (a+b)/3, hence a+b must be a multiple of 3

Sufficient.

Choose D.

dtweah wrote:

Nermal wrote:If a and b are both single-digit positive integers, is a + b a multiple of 3?

(1) The two-digit number "ab" (where a is in the tens place and b is in the ones place) is a multiple of 3.

(2) a - 2b is a multiple of 3.

From definition if the sum of the digits of an interger is a multiple of 3, the integer is a multiple of 3. a+b is a multiple of 3, so is ab or ba. So 1 is Suff:

For 2 to be a multiple of 3, both a and 2b must be. Since 2 is even, we know b must be a multiple of 3. So a/3 +b/3 is an integer or (a+b)/3, hence a+b must be a multiple of 3

Sufficient.

Choose D.

IMO B..

Given ab is a multiple of 3 .. A - multiple or b - multiple.. for ab to be a multiple of 3..
or both a and b are multiple of 3..
therefore .. a + b/3 is not necessarily a multiple of 3.. therefore St 1 is insufficient

IMO D

rule of divisibility for 3 is addition of numbers shud be multiple of 3....thus if ab multiple of 3 implies that a+b multiple of 3

statement 2 proves "a" multiple of 3 and "2b" multiple of 3 since 2 not divisible by 3 implies b divisible by 3... If a divisible by 3 and b divisible by 3 then a-b also divisible by 3.

PAB2706 wrote:IMO D

rule of divisibility for 3 is addition of numbers shud be multiple of 3....thus if ab multiple of 3 implies that a+b multiple of 3

statement 2 proves "a" multiple of 3 and "2b" multiple of 3 since 2 not divisible by 3 implies b divisible by 3... If a divisible by 3 and b divisible by 3 then a-b also divisible by 3.

your logic is wrong; consider a = 5, b = 1, a - 2b = 3, which is a multiple of 3 but a or b are by no means a multiple of 3. The answer is D but not according to your reasoning. Picking numbers is the best strategy

Remember do not confuse the theorem; if x divides a and x divides b then x divides any linear combination of a and b. if x divides a linear combination of a and b, then IT DOES NOT MEAN that x divides a and x divides b

IMO D but I have different reasoning for condition 2.

a-2b is same as a+b-3b.

therefore, according to condition B

a+b-3b=3c where c is an integer.
a+b=3c+3b
a+b=3(c+b)
Therefore, a+b is a mulitple of 3.