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26 May 2017, 04:42
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If n is a 6-digit positive integer, \(abcd96\), where \(a\) is the hundred-thousands digit of \(n\), \(b\) is the ten-thousands digit of \(n\), \(c\) is the thousands digit of \(n\), and \(d\) is the hundreds digit of \(n\), is \(n\) divisible by 36?
(1) The remainder is 3 when \(a + b + c + d\) is divided by 9.
(2) \(a + b + c + d\) is divisible by 3.
(1) The remainder is 3 when \(a + b + c + d\) is divided by 9.
(2) \(a + b + c + d\) is divisible by 3.
26 May 2017, 04:42
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Official Solution:
You need preliminary knowledge to solve this question.
(1) Frist, the remainder of a certain positive integer \(n\) divided either by 3 or 9 is the same as the remainder of the sum of all digits of \(n\) divided by 3 or 9.
(2) Second, the remainder of a certain positive integer \(n\) divided by 4 is the same as the remainder of both the tens digit of \(n\) and the units digit of \(n\) divided by 4. This is because \(100 = 4*25\), and it can be divided to the hundreds digit by 4.
If you take the 1st step of the variable approach and modify the original condition and the question, you get \(n=abc,d96=36t\)? (\(t\)=any positive integer). Since the tens digit and the units digit of \(n=abc,d96\) is 96, according to the second preliminary knowledge (2) above, 96 is divisible by 4. So "is \(n = abc,d96\) divisible by 36?" becomes "is \(n = abc,d96\) divisible by 9?". Also, according to the first preliminary knowledge (1), "is \(n = abc,d96\) divisible by 9?" becomes "is \(a + b + c + d + 9 + 6\) divisible by 9?", or "is \(a + b + c + d + 15\) divisible by 9?".
In the case of con 1), \(a + b + c + d = 9P + 3\) (\(P\)=any positive integer), so using substitution, \(a + b + c + d + 15 = 9P + 3 + 15 = 9(P+2)\) is always divisible, hence yes, it is sufficient.
In the case of con 2), if \(a + b + c + d = 15\), the question is yes, but if \(a + b + c + d = 9\), it is no, hence it is not sufficient. The answer is A.
Answer: A
You need preliminary knowledge to solve this question.
(1) Frist, the remainder of a certain positive integer \(n\) divided either by 3 or 9 is the same as the remainder of the sum of all digits of \(n\) divided by 3 or 9.
(2) Second, the remainder of a certain positive integer \(n\) divided by 4 is the same as the remainder of both the tens digit of \(n\) and the units digit of \(n\) divided by 4. This is because \(100 = 4*25\), and it can be divided to the hundreds digit by 4.
If you take the 1st step of the variable approach and modify the original condition and the question, you get \(n=abc,d96=36t\)? (\(t\)=any positive integer). Since the tens digit and the units digit of \(n=abc,d96\) is 96, according to the second preliminary knowledge (2) above, 96 is divisible by 4. So "is \(n = abc,d96\) divisible by 36?" becomes "is \(n = abc,d96\) divisible by 9?". Also, according to the first preliminary knowledge (1), "is \(n = abc,d96\) divisible by 9?" becomes "is \(a + b + c + d + 9 + 6\) divisible by 9?", or "is \(a + b + c + d + 15\) divisible by 9?".
In the case of con 1), \(a + b + c + d = 9P + 3\) (\(P\)=any positive integer), so using substitution, \(a + b + c + d + 15 = 9P + 3 + 15 = 9(P+2)\) is always divisible, hence yes, it is sufficient.
In the case of con 2), if \(a + b + c + d = 15\), the question is yes, but if \(a + b + c + d = 9\), it is no, hence it is not sufficient. The answer is A.
Answer: A
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