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26 May 2017, 04:42
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If \(a\), \(b\), and \(c\) are integers. Is \(abc + ab + c\) an odd?
(1) \(ac = odd\)
(2) \(ab = odd\)
(1) \(ac = odd\)
(2) \(ab = odd\)
26 May 2017, 04:42
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Official Solution:
If you modify the original condition and the question, from \(abc + ab + c = odd\)?, \(ab(c + 1) + c = odd\)?, you get the same result as \(c = odd\)?. This is because if \(c = odd\), you get \(c + 1 = even\), then \(ab(c + 1) + c = ab(even) + odd = even + odd = odd\), hence yes
In the case of con 1), if \(ac = odd\), the condition is yes and sufficient.
In the case of con 2), by the law of "CMT 4(B: if you get A or B too easily, consider D)", when \(ab = odd\),
(1) If \(c = odd\), you get \(ab(c + 1) + c = odd(odd + 1) + odd = odd(even) + odd = odd\), hence yes.
(2) If \(c = even\), you get \(ab(c + 1) + c=odd(even + 1)=odd(odd) + even = odd\), hence, in the case of (1) and case of (2), it is always yes and sufficient. Therefore, the answer is D.
Answer: D
If you modify the original condition and the question, from \(abc + ab + c = odd\)?, \(ab(c + 1) + c = odd\)?, you get the same result as \(c = odd\)?. This is because if \(c = odd\), you get \(c + 1 = even\), then \(ab(c + 1) + c = ab(even) + odd = even + odd = odd\), hence yes
In the case of con 1), if \(ac = odd\), the condition is yes and sufficient.
In the case of con 2), by the law of "CMT 4(B: if you get A or B too easily, consider D)", when \(ab = odd\),
(1) If \(c = odd\), you get \(ab(c + 1) + c = odd(odd + 1) + odd = odd(even) + odd = odd\), hence yes.
(2) If \(c = even\), you get \(ab(c + 1) + c=odd(even + 1)=odd(odd) + even = odd\), hence, in the case of (1) and case of (2), it is always yes and sufficient. Therefore, the answer is D.
Answer: D
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