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05 Dec 2015, 08:52
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Math Revolution and GMAT Club Contest Starts!
QUESTION #6:
For positive integers a and b, when a^4 - b^4 is divided by 3, what is the remainder?
(1) When a + b is divided by 3, the remainder is 0
(2) When a^2 + b^2 is divided by 3, the remainder is 2
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06 Dec 2015, 13:54
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Wow, I found this one super hard and got it wrong at first, but I've managed to prove that the answer is D, either statement is sufficient. I'm sure i've missed the obvious trick here, but this is how i've done it:
For positive integers a and b, when \(a^4 - b^4\) is divided by 3, what is the remainder?
(1) When \(a + b\) is divided by 3, the remainder is 0
(2) When \(a^2 + b^2\) is divided by 3, the remainder is 2
Firstly note that a and b are positive integers therefore anything constructed from a, b and integer powers of a and b without any division will be an integer. Let's start by factorising the original statement to break it down into easier parts.
\(a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) = (a^2 + b^2)(a + b)(a - b)\)
(1) tells us that \(a + b\) is divisible by 3 as the remainder is 0. Therefore any multiple of \((a + b)\) is divisible by 3. Since we have shown above that \(a^4 - b^4\) is a multiple of \((a + b)\), and since we know that the other terms in the above equation are integers, we know that \(a^4 - b^4\) is divisible by 3 and will therefore have a remainder of 0 when divided by 3. Therefore statement (1) is sufficient and the possible answers are A and D.
(2) tells us that \(a^2 + b^2\) is not divisible by 3, and has a remainder of 2 when divided by 3. From the above we can see that \(a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)\), and therefore we need to test whether \(a^2 + b^2\) having a remainder of 2 means that \((a^2 - b^2)\) has a certain remainder when divided by 3.
Let's consider \(a^2\) and \(b^2\) separately. If \(a^2 + b^2\) has a remainder of 2, then three options are possible:
\(a^2\) has a remainder of 0 and \(b^2\) has a remainder of 2
\(a^2\) and \(b^2\) each have a remainder of 1
\(a^2\) has a remainder of 2 and \(b^2\) has a remainder of 0
Integers can only possibly have remainders 0, 1, or 2, when divided by 3, so each integer fits into one of 3 camps: \(x\), \(x+1\), or \(x+2\), where x is divisible by 3. Therefore every square number is either \(x^2\), \((x+1)^2 = x^2 + 2x + 1\), or \((x+2)^2 = x^2 + 4x + 4\). The only terms here not divisible by 3 are 1 and 4, both of which have a remainder 1. Therefore square numbers are either divisible by 3 or have remainder 1 when divided by 3. This let's us rule out the first and last possibility, so we know that if \(a^2 + b^2\) has a remainder of 2 when divided by 3, \(a^2\) and \(b^2\) each have a remainder of 1.
If \(a^2\) and \(b^2\) each have a remainder of 1, then \((a^2 - b^2)\) has a remainder 0 and is always divisible by 3. In that case \(a^4 - b^4\) is divisible by 3 as it is a multiple of \((a^2 - b^2)\). Therefore (2) is also sufficient to calculate the remainder of \(a^4 - b^4\) when divided by 3. The answer to the question is D, either statement is sufficient on it's own.
For positive integers a and b, when \(a^4 - b^4\) is divided by 3, what is the remainder?
(1) When \(a + b\) is divided by 3, the remainder is 0
(2) When \(a^2 + b^2\) is divided by 3, the remainder is 2
Firstly note that a and b are positive integers therefore anything constructed from a, b and integer powers of a and b without any division will be an integer. Let's start by factorising the original statement to break it down into easier parts.
\(a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) = (a^2 + b^2)(a + b)(a - b)\)
(1) tells us that \(a + b\) is divisible by 3 as the remainder is 0. Therefore any multiple of \((a + b)\) is divisible by 3. Since we have shown above that \(a^4 - b^4\) is a multiple of \((a + b)\), and since we know that the other terms in the above equation are integers, we know that \(a^4 - b^4\) is divisible by 3 and will therefore have a remainder of 0 when divided by 3. Therefore statement (1) is sufficient and the possible answers are A and D.
(2) tells us that \(a^2 + b^2\) is not divisible by 3, and has a remainder of 2 when divided by 3. From the above we can see that \(a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)\), and therefore we need to test whether \(a^2 + b^2\) having a remainder of 2 means that \((a^2 - b^2)\) has a certain remainder when divided by 3.
Let's consider \(a^2\) and \(b^2\) separately. If \(a^2 + b^2\) has a remainder of 2, then three options are possible:
\(a^2\) has a remainder of 0 and \(b^2\) has a remainder of 2
\(a^2\) and \(b^2\) each have a remainder of 1
\(a^2\) has a remainder of 2 and \(b^2\) has a remainder of 0
Integers can only possibly have remainders 0, 1, or 2, when divided by 3, so each integer fits into one of 3 camps: \(x\), \(x+1\), or \(x+2\), where x is divisible by 3. Therefore every square number is either \(x^2\), \((x+1)^2 = x^2 + 2x + 1\), or \((x+2)^2 = x^2 + 4x + 4\). The only terms here not divisible by 3 are 1 and 4, both of which have a remainder 1. Therefore square numbers are either divisible by 3 or have remainder 1 when divided by 3. This let's us rule out the first and last possibility, so we know that if \(a^2 + b^2\) has a remainder of 2 when divided by 3, \(a^2\) and \(b^2\) each have a remainder of 1.
If \(a^2\) and \(b^2\) each have a remainder of 1, then \((a^2 - b^2)\) has a remainder 0 and is always divisible by 3. In that case \(a^4 - b^4\) is divisible by 3 as it is a multiple of \((a^2 - b^2)\). Therefore (2) is also sufficient to calculate the remainder of \(a^4 - b^4\) when divided by 3. The answer to the question is D, either statement is sufficient on it's own.
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05 Dec 2015, 11:58
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I've been struggling to find the answer for this one...very tough and tricky, but if you try algebraic way, then one works and other one not always, and while picking numbers - different results...anyways...below my explanation.
we can rewrite the initial statement:
a^4-b^4 = (a^2+b^2)(a^2-b^2)
(a^2-b^2) can be rewritten as (a+b)(a-b)
or the whole:
a^4-b^4 = (a^2+b^2)(a+b)(a-b)
first statement:
a+b is divisible by 3.
well, in this case, the reminder will always yield 0, since a^4-b^4 is a multiple of 3, because a+b is a multiple of 3.
statement 1 is sufficient.
statement 2:
a^2+b^2, if divided by 3, will yield a remainder of 2.
well, this doesn't tell much. What if a+b is a multiple of 3?
what if a-b is a multiple of 3?
clearly 2 should't be sufficient, right?
well, not exactly!
if you try picking numbers, you can see that all the times, the remainder is 0. and this is what confuses me the most. Maybe I miss something?
a=2 b=1
a^2 + b^2 = 5. 5/3 = 1 and remainder 2.
a^4 - b^4 = 16 - 1 = 15. 15/3 = 5 and remainder 0.
a=4 and b=1
a^2 + b^2 = 16+1 = 17. 17/3 = 5 and remainder 2.
a^4 - b^4 = 256-1 = 255. remainder again 0.
and the pattern is the same for all numbers that are picked and that satisfy the condition of the statement 2.
I really doubt this is a 600-700 level question. Or maybe I am not in the mood for solving math today? :\
we can rewrite the initial statement:
a^4-b^4 = (a^2+b^2)(a^2-b^2)
(a^2-b^2) can be rewritten as (a+b)(a-b)
or the whole:
a^4-b^4 = (a^2+b^2)(a+b)(a-b)
first statement:
a+b is divisible by 3.
well, in this case, the reminder will always yield 0, since a^4-b^4 is a multiple of 3, because a+b is a multiple of 3.
statement 1 is sufficient.
statement 2:
a^2+b^2, if divided by 3, will yield a remainder of 2.
well, this doesn't tell much. What if a+b is a multiple of 3?
what if a-b is a multiple of 3?
clearly 2 should't be sufficient, right?
well, not exactly!
if you try picking numbers, you can see that all the times, the remainder is 0. and this is what confuses me the most. Maybe I miss something?
a=2 b=1
a^2 + b^2 = 5. 5/3 = 1 and remainder 2.
a^4 - b^4 = 16 - 1 = 15. 15/3 = 5 and remainder 0.
a=4 and b=1
a^2 + b^2 = 16+1 = 17. 17/3 = 5 and remainder 2.
a^4 - b^4 = 256-1 = 255. remainder again 0.
and the pattern is the same for all numbers that are picked and that satisfy the condition of the statement 2.
I really doubt this is a 600-700 level question. Or maybe I am not in the mood for solving math today? :\
