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If a and b are integers, is b even?

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If a and b are integers, is b even? [#permalink] New post 16 Aug 2010, 05:57
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C
D
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If a and b are integers, is b even?

(1) 3a + 4b is even
(2) 3a + 5b is even
[Reveal] Spoiler: OA
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Re: Number Properties Problem [#permalink] New post 16 Aug 2010, 06:05
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jakolik wrote:
If a and b are integers, is b even?
(1) 3a + 4b is even
(2) 3a + 5b is even


(1) 3a + 4b is even --> if a is even, then it's not necessary for b to be even, may be even or odd. Not sufficient.
(2) 3a + 5b is even --> if a is even, then b is even too, but if a is odd, then b is odd too. Not sufficient.

(1)+(2) Subtract (1) from (2) --> (3a+ 5b)-(3a + 4b)=even_2-even_1 --> b=even_2-even_1=even. Sufficient.

Answer: C.
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Re: If a & b are integers.... [#permalink] New post 28 Oct 2010, 21:19
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monirjewel wrote:
If a and b are integers, is b even?
(1) 3a + 4b is even
(2) 3a + 5b is even


If a sum is even, then both numbers are even or both numbers are odd.

Statement 1: 4b is even, and 3a + 4b is even, which means 3a is even and hence it tells us a is even. This is insufficient.

Statement 2: 3a + 5b is even, this means that either both a and b are even or both a and b are odd, since either way the sum will be even. But that is insufficient too.

Combining both statements, we know that a is even, which means for the second statement to be valid, b also has to be even.

Hence the answer is C.
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Re: Number Properties Problem [#permalink] New post 16 Aug 2010, 06:07
IMO C.
Let me try to explain my approach for this one.
Addition of 2 numbers will be even when both of them are either even or odd.
Now lets check option 1.
3a + 4b is even. 3a can be either even or odd depending value of a. 4b will be even irrespective b being even or odd. Hence, 3a should be even if the sum has to be even. But, we cannot confirm whether b is odd or even with this statement.
Now lets check option 2.
3a + 5b is even. 3a can be either even or odd depending value of a. 5b can be either even or odd depending value of a. Hence, a and b can be either even or odd. This also not sufficient to conclude whether b is even or off.
Combing these 2 statements.
Both 3a + 4b and 3a + 5b are even.
Then as per first statement 3a should be even.
Then 4b and 5b should be even if the sum has to be even in both the cases.
Then b should be even. Hence, C is the answer.
Please let us know the OA and explanation if I am wrong or ambiguous.
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Re: Number Properties Problem [#permalink] New post 17 Aug 2010, 09:13
Statement 1: Not Sufficient
- A must be even
- B can be even or odd

Statement 2: Not sufficient
- Either A & B are both odd OR A & B are both even

Both: Sufficient
- A is even
- When A is even B is also even
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Re: Number Properties Problem [#permalink] New post 02 Jan 2013, 04:22
Bunuel wrote:
jakolik wrote:
If a and b are integers, is b even?
(1) 3a + 4b is even
(2) 3a + 5b is even


(1) 3a + 4b is even --> if a is even, then it's not necessary for b to be even, may be even or odd. Not sufficient.
(2) 3a + 5b is even --> if a is even, then b is even too, but if a is odd, then b is odd too. Not sufficient.

(1)+(2) Subtract (1) from (2) --> (3a+ 5b)-(3a + 4b)=even_2-even_1 --> b=even_2-even_1=even. Sufficient.

Answer: C.



Hi Just a small doubt, can we consider b to be zero.
Is zero treated as an even integer.
Thx
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Re: Number Properties Problem [#permalink] New post 02 Jan 2013, 04:29
Expert's post
tarunjagtap wrote:
Bunuel wrote:
jakolik wrote:
If a and b are integers, is b even?
(1) 3a + 4b is even
(2) 3a + 5b is even


(1) 3a + 4b is even --> if a is even, then it's not necessary for b to be even, may be even or odd. Not sufficient.
(2) 3a + 5b is even --> if a is even, then b is even too, but if a is odd, then b is odd too. Not sufficient.

(1)+(2) Subtract (1) from (2) --> (3a+ 5b)-(3a + 4b)=even_2-even_1 --> b=even_2-even_1=even. Sufficient.

Answer: C.



Hi Just a small doubt, can we consider b to be zero.
Is zero treated as an even integer.
Thx


Zero is an even integer. Zero is nether positive nor negative, but zero is definitely an even number.

An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even (in fact zero is divisible by every integer except zero itself).

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Number Properties Problem   [#permalink] 02 Jan 2013, 04:29
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