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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.

If we take a scenario for part 1

A = 9 B = 3 A x B = 27 A + B = 9 (Divisible by 3)

A = 4 B = 3 A x B = 12 A + B = 7 (Not Divisible by 3)

So it should be insufficient. Please help.

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

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.
_________________

Re: If a and b are both single-digit positive integers, is a + b a multipl [#permalink]

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24 Oct 2014, 15:44

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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.

Re: If a and b are both single-digit positive integers, is a + b a multipl [#permalink]

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25 Oct 2014, 04:14

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Bunuel wrote:

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.

D.

1) ab is a multiple of 3. By basic rule of division, a number is div by 3 if the sum of its digits is div by 3. Reason: 10a+b mod 3 = 0 => (a+b) mod 3 = 0 [since 10 mod 3 = 1] so sufficient.

2) a-2b = 3k a = 3k+2b => a+b = 3k+2b+b = 3(k+b) which is div by 3 so sufficient.
_________________

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.

Re: If a and b are both single-digit positive integers, is a + b a multipl [#permalink]

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20 Nov 2017, 11:13

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