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16 Sep 2014, 00:16
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If \(x\) and \(y\) are positive integers, is \(x^{16} - y^8 + 345y^2\) divisible by 15?
(1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20
(2) \(y = x^2\)
(1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20
(2) \(y = x^2\)
16 Sep 2014, 00:16
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Bookmarks
Official Solution:
If \(x\) and \(y\) are positive integers, is \(x^{16} - y^8 + 345y^2\) divisible by 15?
Firstly, note that since \(345y^2\) is divisible by 15, we can ignore it, as it won't affect the remainder.
The question then becomes: is \(x^{16} - y^8\) divisible by 15?
(1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20.
In this case, both \(x\) and \(y\) could be multiples of 15 as well (e.g., \(x = 25*15\) and \(y = 20*15\)) and in this case, \(x^{16} - y^8 = 15 \cdot (...)\). This would be divisible by 15. However, one could be a multiple of 15 and the other not (e.g., \(x = 25* 15\) and \(y = 20\)), and in this case, \(x^{16} - y^8\) would not be divisible by 15, as we cannot factor out 15 from \(x^{16} - y^8\). Not sufficient.
(2) \(y = x^2\).
Substitute \(y\) with \(x^2\): \(x^{16} - y^8 = x^{16} - (x^2)^8 = x^{16} - x^{16} = 0\). Zero is divisible by 15 (zero is divisible by every integer except zero itself). Sufficient.
Notes for statement (1):
If integers \(a\) and \(b\) are both multiples of some integer \(k > 1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
• Example: \(a = 6\) and \(b = 9\), both divisible by 3: \(a + b = 15\) and \(a - b = -3\), again both divisible by 3.
If out of integers \(a\) and \(b\), one is a multiple of some integer \(k > 1\) and the other is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
• Example: \(a = 6\), divisible by 3 and \(b = 5\), not divisible by 3: \(a + b = 11\) and \(a - b = 1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k > 1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
• Example: \(a = 5\) and \(b = 4\), neither is divisible by 3: \(a + b = 9\), is divisible by 3 and \(a - b = 1\), is not divisible by 3
• OR: \(a = 6\) and \(b = 3\), neither is divisible by 5: \(a + b = 9\) and \(a - b = 3\), neither is divisible by 5
• OR: \(a = 2\) and \(b = 2\), neither is divisible by 4: \(a + b = 4\) and \(a - b = 0\), both are divisible by 4.
Thus, according to the above, the fact that \(x\) is a multiple of 25 and \(y\) is a multiple of 20 is insufficient to determine whether \(x^{16} - y^8\) is divisible by 15.
Answer: B
If \(x\) and \(y\) are positive integers, is \(x^{16} - y^8 + 345y^2\) divisible by 15?
Firstly, note that since \(345y^2\) is divisible by 15, we can ignore it, as it won't affect the remainder.
The question then becomes: is \(x^{16} - y^8\) divisible by 15?
(1) \(x\) is a multiple of 25, and \(y\) is a multiple of 20.
In this case, both \(x\) and \(y\) could be multiples of 15 as well (e.g., \(x = 25*15\) and \(y = 20*15\)) and in this case, \(x^{16} - y^8 = 15 \cdot (...)\). This would be divisible by 15. However, one could be a multiple of 15 and the other not (e.g., \(x = 25* 15\) and \(y = 20\)), and in this case, \(x^{16} - y^8\) would not be divisible by 15, as we cannot factor out 15 from \(x^{16} - y^8\). Not sufficient.
(2) \(y = x^2\).
Substitute \(y\) with \(x^2\): \(x^{16} - y^8 = x^{16} - (x^2)^8 = x^{16} - x^{16} = 0\). Zero is divisible by 15 (zero is divisible by every integer except zero itself). Sufficient.
Notes for statement (1):
If integers \(a\) and \(b\) are both multiples of some integer \(k > 1\) (divisible by \(k\)), then their sum and difference will also be a multiple of \(k\) (divisible by \(k\)):
• Example: \(a = 6\) and \(b = 9\), both divisible by 3: \(a + b = 15\) and \(a - b = -3\), again both divisible by 3.
If out of integers \(a\) and \(b\), one is a multiple of some integer \(k > 1\) and the other is not, then their sum and difference will NOT be a multiple of \(k\) (divisible by \(k\)):
• Example: \(a = 6\), divisible by 3 and \(b = 5\), not divisible by 3: \(a + b = 11\) and \(a - b = 1\), neither is divisible by 3.
If integers \(a\) and \(b\) both are NOT multiples of some integer \(k > 1\) (divisible by \(k\)), then their sum and difference may or may not be a multiple of \(k\) (divisible by \(k\)):
• Example: \(a = 5\) and \(b = 4\), neither is divisible by 3: \(a + b = 9\), is divisible by 3 and \(a - b = 1\), is not divisible by 3
• OR: \(a = 6\) and \(b = 3\), neither is divisible by 5: \(a + b = 9\) and \(a - b = 3\), neither is divisible by 5
• OR: \(a = 2\) and \(b = 2\), neither is divisible by 4: \(a + b = 4\) and \(a - b = 0\), both are divisible by 4.
Thus, according to the above, the fact that \(x\) is a multiple of 25 and \(y\) is a multiple of 20 is insufficient to determine whether \(x^{16} - y^8\) is divisible by 15.
Answer: B
General Discussion
