hi GHong14
next time try to hide the answer in spoiler. as to this problem, one of the ways
(1) 10a+b=3k, where k is an integer
b=3k-10a, then a+b=a+(3k-10a)=3k-9a=3*(k-3a) divisible by 3
(2) the same story
a-2b=3m, a=3m+2b,
a+b=3m+2b+b=3m+3b=3(m+b) divisible by 3
D seems right answer
Multiples Number Property
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Adam@Knewton
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I really don't want to get into the concepts behind this because it has to do with very odd, interesting things that happen when you apply the principles of modular arithmetic to the number 3 (or 9) which has special properties in base 10. Suffice to say this:
Divisibility problems are often better thought of as remainder problems, conceptually. To be divisible by 3 is to have a remainder of 0 when divided by 3. What's interesting about three is that, with certain numbers, multiplying does not change the remainder. For instance, 10. For example, 7 divided by 3 has a remainder of 1; similarly, 70 divided by 3 has a remainder of 1. This is not a coincidence. So, if "a" is a multiple of 3, then "10a" is a multiple of 3.
For a + b to be divisible by 3, then (the remainder when a is divided by 3) + (the remainder when b is divided by 3) must = 3. Since 10a doesn't change the remainder situation for a, Statement (1) is Sufficient.
Let's all agree not to think this way on the test! Ever! Clock's explanation is elegant and straightforward. Here's how to do it with plugging in numbers, knowing only that if we see a "coincidence," it's probably not a coincidence:
Let a=1. Now b=2 (for Statement 1 to be valid). Now a+b=3; answer Yes.
Let a=1, but let's have b=8 (Statement 1 must remain valid). Now a+b=0; answer again Yes.
Let a=5 (something totally different), so we need to have b=4 (for example). Now a+b=0; answer AGAIN Yes.
Enough. I'm convinced. Sufficient.
The same kind of approach will work on Statement (2).
For a Statement that would be Insufficient (be careful, GHong -- the Statements can never ever ever be untrue, just Insufficient, as by rule the statements are always true!), we could use a number that doesn't have that cool remainder property, such as 5:
Statement (3): 5a+b is divisible by 3.
So, if a=1 and b=1, a+b=2 and the answer is "no."
However, if a=3 and b=3 (this will work for any such situation, obviously, since everything is divisible by 3), then a+b=6 and the answer is "yes." Insufficient.
Divisibility problems are often better thought of as remainder problems, conceptually. To be divisible by 3 is to have a remainder of 0 when divided by 3. What's interesting about three is that, with certain numbers, multiplying does not change the remainder. For instance, 10. For example, 7 divided by 3 has a remainder of 1; similarly, 70 divided by 3 has a remainder of 1. This is not a coincidence. So, if "a" is a multiple of 3, then "10a" is a multiple of 3.
For a + b to be divisible by 3, then (the remainder when a is divided by 3) + (the remainder when b is divided by 3) must = 3. Since 10a doesn't change the remainder situation for a, Statement (1) is Sufficient.
Let's all agree not to think this way on the test! Ever! Clock's explanation is elegant and straightforward. Here's how to do it with plugging in numbers, knowing only that if we see a "coincidence," it's probably not a coincidence:
Let a=1. Now b=2 (for Statement 1 to be valid). Now a+b=3; answer Yes.
Let a=1, but let's have b=8 (Statement 1 must remain valid). Now a+b=0; answer again Yes.
Let a=5 (something totally different), so we need to have b=4 (for example). Now a+b=0; answer AGAIN Yes.
Enough. I'm convinced. Sufficient.
The same kind of approach will work on Statement (2).
For a Statement that would be Insufficient (be careful, GHong -- the Statements can never ever ever be untrue, just Insufficient, as by rule the statements are always true!), we could use a number that doesn't have that cool remainder property, such as 5:
Statement (3): 5a+b is divisible by 3.
So, if a=1 and b=1, a+b=2 and the answer is "no."
However, if a=3 and b=3 (this will work for any such situation, obviously, since everything is divisible by 3), then a+b=6 and the answer is "yes." Insufficient.
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Yes! Great catch, Vinnaray; I totally missed it. This question MUST say that a and b are integers (so, for the above explanations, let's pretend it does). Otherwise, Vinnaray's example would make Statement (1) Insufficient (and unnecessarily confuse the number properties this question is designed to test).vinnaray wrote:Hi,
Shouldn't the question also say that a,b are integers?
w/o that ,statement 1 could have a= (1/5) b= 1 , making a+b not a multiple of 3.
Would this be right?
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(a+b)/3 -? true only when (a+b)/3GHong14 wrote:Is a+b a multiple of 3?
1. 10a+b is a multiple of 3
2. a -2b is a multiple of 3
D
st(1) 10a+b= 9a + a+b OR (10a+b)/3 = 9a/3 +(a+b)/3 Sufficient
st(2) a-2b= a+b - 3b OR (a-2b)/3 =(a+b)/3 -3b/3 Sufficient
answer is D
