If a and b are both single-digit positive integers, is a + b a multipl

Question asks if a+b is divisible by 3

1. pick up the numbers 12, 15,18, 21 24, 30 all are multiple of 3 and a+b ( 1+2,1+5,1+9....) is a multiple of 3 - hence sufficient

2. if a is a multiple of 3 and b is a multiple of 3 then a-b is also a multiple of 3
pick up numbers for a-2b, remember this should be a multiple of 3., thus a is multiple of 3 and 2b is a multiple of 3
a = 9 b = 3 then a-2b is 3 ..multiple of 3
a= 24 b= 6 a-2b = 12 multiple of 3..........sufficient

hence D

1) is also the check for 'division by 3'.

I.e. a number is a multiple of 3 if its digits added together are divisible by 3. Therefore, if ab is divisible by 3, then a+b must be divisible by 3.

(1)

Clearly Sufficient as "ab" is a multiple of 3 when a+b is so

You should consider AB as 93, instead of A * B = 27

(2)

a - 2b = 3k, where k is an integer

=> a = 3k + 2b

=> ab is in a format "3k+2bb"

So a + b = 3k + 2b + b = 3(k+b), a multiple of 3

=> ab is a multiple of 3

Sufficient

Answer - D

Tough and Tricky questions: Multiples.

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.

Answer is D, both are sufficient.

1) for any number to be a multiple of 3, the sum of its digits needs to be a multiple of 3.
So therefore, if given ab is a multiple of 3, then a + b is also a multiple of 3.

2) a -2b is a multiple of 3. If we add to this a multiple of 3, the resultant sum should also be a multiple of 3.
a-2b + 3b(multiple of 3) = a+b which by virtue of being a sum of 2 multiple of 3, is also a multiple.

Thus each statement alone is sufficient

I know this is an older post, but can someone clarify how it is we go from => a = 3k + 2b to a + b = 3k + 2b + b = 3(k+b) for statement 2?

thanks

a = 3k + 2b;

Add b to both sides: a + b = (3k + 2b) + b;

a + b = 3k + 3b;

a + b = 3(k + 3).

makes perfect sense now, thank you!!

Before solving the question, you need some knowledge. “The remainder of a certain positive integer divided by 3 and 9 is same as the sum of every digit place of n divided by 3 and 9.
There are 2 variables in the original condition. In order to match the number of variables to the number of equations, we need 2 equations. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that C is the correct answer.
Using the condition 1) and the condition 2) at the same time:
Condition 1) – a multiple of ab=3 is a multiple of a+b=3. Hence, the answer is yes and the condition is sufficient.
Condition 2) – If a+b is a multiple of a-2b=3, then, it becomes a multiple of a-2b+3b=3 plus 3b and a multiple of a+b=3 plus 3b=3(multiple+b)=3. Hence, the answer is yes and the condition is sufficient. Thus, the correct answer is D.
This is 50-51 level question. Please remember the common mistake type 4(B). If the answer is too easily A or B, please consider D.

- For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

1) two digit integer ab is a multiple of 3 <=> (a+b) = 3k => clearly sufficient
2) a-2b =3k => a-2b+3b is also a multiple of 3 => a+b = 3k => sufficient
=> D

Easiest way to solve this by understanding rules of divisibility for 3.
The rule states the sum of the digits should be a multiple of 3.

To find: if (a+b)/3 = integer.

(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.
Then only options for b are 3,6 and 9 not 0 since it states positive integer.
if ab/3 then it needs to be that (a+b)/3
Hence sufficient

(2) a – 2b is a multiple of 3.
if (a-2b)/3 then a/3 -2b/3 is an integer
which means a is divisible by 3 and b is divisible by 3
Hence sufficient

Hence D

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