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If \(a\), \(b\) and \(c\) are integers, is \(a*b*c\) divisible by 32?
(1) \(a\), \(b\), and \(c\) are consecutive even integers.
(2) \(ac \lt 0\).
(1) \(a\), \(b\), and \(c\) are consecutive even integers.
(2) \(ac \lt 0\).
16 Sep 2014, 00:56
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Official Solution:
If \(a\), \(b\) and \(c\) are integers, is \(a*b*c\) divisible by 32?
First of all, notice that \(32=2^5\).
(1) \(a\), \(b\), and \(c\) are consecutive even integers.
In this case, \(a*b*c=(2k-2)*(2k)*(2k+2)=2^3*(k-1)*k*(k+1)\), for some integer \(k\). Now, \((k-1)k(k+1)\) is the product of three consecutive integers, so it must be divisible by 6 (the product of \(n\) consecutive integers is ALWAYS divisible by \(n!\); therefore, the product of three consecutive integers must be divisible by \(3!=6\)). Thus, \(a*b*c\) MUST be divisible by \(2^4\) (\(a*b*c=2^3*(k-1)*k*(k+1)=2^3*6*something=2^4*3*something\)) but not necessarily by \(2^5\). Hence, this statement is not sufficient.
For example:
If \(a = 2\), \(b = 4\), and \(c = 6\), then \(abc = 2^4*3\) and in this case, \(a*b*c\) is NOT divisible by 32.
If \(a = 4\), \(b = 6\), and \(c = 8\), then \(abc = 2^6*3\) and in this case, \(a*b*c\) IS divisible by 32.
(2) \(ac \lt 0\).
This merely tells us that \(a\) and \(c\) have different signs, which is clearly insufficient to answer the question.
(1)+(2) From (2), we deduced that either \(a\) or \(c\) is negative. And since (1) states that \(a\), \(b\), and \(c\) are consecutive even integers, then these three integers must be -2, 0, and 2 (this is the only set of THREE consecutive EVEN integers that includes both positive and negative numbers). Thus, \(abc = 0\), which is divisible by 32. Sufficient.
Answer: C
If \(a\), \(b\) and \(c\) are integers, is \(a*b*c\) divisible by 32?
First of all, notice that \(32=2^5\).
(1) \(a\), \(b\), and \(c\) are consecutive even integers.
In this case, \(a*b*c=(2k-2)*(2k)*(2k+2)=2^3*(k-1)*k*(k+1)\), for some integer \(k\). Now, \((k-1)k(k+1)\) is the product of three consecutive integers, so it must be divisible by 6 (the product of \(n\) consecutive integers is ALWAYS divisible by \(n!\); therefore, the product of three consecutive integers must be divisible by \(3!=6\)). Thus, \(a*b*c\) MUST be divisible by \(2^4\) (\(a*b*c=2^3*(k-1)*k*(k+1)=2^3*6*something=2^4*3*something\)) but not necessarily by \(2^5\). Hence, this statement is not sufficient.
For example:
If \(a = 2\), \(b = 4\), and \(c = 6\), then \(abc = 2^4*3\) and in this case, \(a*b*c\) is NOT divisible by 32.
If \(a = 4\), \(b = 6\), and \(c = 8\), then \(abc = 2^6*3\) and in this case, \(a*b*c\) IS divisible by 32.
(2) \(ac \lt 0\).
This merely tells us that \(a\) and \(c\) have different signs, which is clearly insufficient to answer the question.
(1)+(2) From (2), we deduced that either \(a\) or \(c\) is negative. And since (1) states that \(a\), \(b\), and \(c\) are consecutive even integers, then these three integers must be -2, 0, and 2 (this is the only set of THREE consecutive EVEN integers that includes both positive and negative numbers). Thus, \(abc = 0\), which is divisible by 32. Sufficient.
Answer: C
