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11 Dec 2015, 11:09
Kudos
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A.
Statement 1 alone is sufficient
a^4-b^4 can be written as (a^2+b^2)*(a^2-b^2) which can further be written as (a^2+b^2)*(a+b)*(a-b)
Since a+b is divisible by 3, the above expression leaves 0 as remainder when divided by 3
Statement2:
a^2+b^2 leaves remainder 2 when divided by 3 , but we have no idea about remainder for a^2-b^2 when divided by 3, therefore it is not sufficient.
Statement 1 alone is sufficient
a^4-b^4 can be written as (a^2+b^2)*(a^2-b^2) which can further be written as (a^2+b^2)*(a+b)*(a-b)
Since a+b is divisible by 3, the above expression leaves 0 as remainder when divided by 3
Statement2:
a^2+b^2 leaves remainder 2 when divided by 3 , but we have no idea about remainder for a^2-b^2 when divided by 3, therefore it is not sufficient.
11 Dec 2015, 11:11
Kudos
Bookmarks
A.
Statement 1 alone is sufficient
a^4-b^4 can be written as (a^2+b^2)*(a^2-b^2) which can further be written as (a^2+b^2)*(a+b)*(a-b)
Since a+b is divisible by 3, the above expression leaves 0 as remainder when divided by 3
Statement2:
a^2+b^2 leaves remainder 2 when divided by 3 , but we have no idea about remainder for a^2-b^2 when divided by 3, therefore it is not sufficient.
Statement 1 alone is sufficient
a^4-b^4 can be written as (a^2+b^2)*(a^2-b^2) which can further be written as (a^2+b^2)*(a+b)*(a-b)
Since a+b is divisible by 3, the above expression leaves 0 as remainder when divided by 3
Statement2:
a^2+b^2 leaves remainder 2 when divided by 3 , but we have no idea about remainder for a^2-b^2 when divided by 3, therefore it is not sufficient.
11 Dec 2015, 23:49
Kudos
Bookmarks
a^4 - b^4 is divided by 3
a^4-b^4= (a+b)(a-b)(a^2 + b^2)
1) When a + b is divided by 3, the remainder is 0
Since a+b is a factor, therefore a^4 - b^4 divided by 3 will also give a remainder of 0 and factor.
Hence choice 1 is sufficient.
(2) When a^2 + b^2 is divided by 3, the remainder is 2
When we square a number that gives a remainder of 1 when divided by 3, it gives remainder of 1 again. For example
4 gives a remainder of 1 when divided by 3, and 16 also gives a remainder of 1 when divided by 3
When we square a number that gives a remainder of 2 when divided by 3, it gives remainder of 1. For example
5 gives a remainder of 1 when divided by 3, and 25 also gives a remainder of 1 when divided by 3.
So point to note is When we square a number that gives a remainder of 2 when divided by 3, it always gives remainder of 1
Therefore point to infer from the statement, a^2 + b^2 is divided by 3, the remainder is 2, is both a and b are not multiples of 3. They must be numbers that either leave a remainder of 1 or 2 for a^2+b^2 to leave a remainder of 2 when divided by 3.
If we select numbers that either leave 1 or 2 as remainder when divided by 3 for the equation a^2 + b^2 , then the individual numbers added up or subtracted, a+b or a-b always turn out to be a factor of 3 and therefore always leave remainder of 0. This can be tried with any numbers with the above conditions and will hold true.
Hence this choice is also sufficient.
As each statement alone is sufficient , therefore answer is choice D.
a^4-b^4= (a+b)(a-b)(a^2 + b^2)
1) When a + b is divided by 3, the remainder is 0
Since a+b is a factor, therefore a^4 - b^4 divided by 3 will also give a remainder of 0 and factor.
Hence choice 1 is sufficient.
(2) When a^2 + b^2 is divided by 3, the remainder is 2
When we square a number that gives a remainder of 1 when divided by 3, it gives remainder of 1 again. For example
4 gives a remainder of 1 when divided by 3, and 16 also gives a remainder of 1 when divided by 3
When we square a number that gives a remainder of 2 when divided by 3, it gives remainder of 1. For example
5 gives a remainder of 1 when divided by 3, and 25 also gives a remainder of 1 when divided by 3.
So point to note is When we square a number that gives a remainder of 2 when divided by 3, it always gives remainder of 1
Therefore point to infer from the statement, a^2 + b^2 is divided by 3, the remainder is 2, is both a and b are not multiples of 3. They must be numbers that either leave a remainder of 1 or 2 for a^2+b^2 to leave a remainder of 2 when divided by 3.
If we select numbers that either leave 1 or 2 as remainder when divided by 3 for the equation a^2 + b^2 , then the individual numbers added up or subtracted, a+b or a-b always turn out to be a factor of 3 and therefore always leave remainder of 0. This can be tried with any numbers with the above conditions and will hold true.
Hence this choice is also sufficient.
As each statement alone is sufficient , therefore answer is choice D.
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