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16 Sep 2014, 00:56
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If in a six-digit integer \(n\), \(f(k)\) represents the value of the \(k_{th}\) digit from the leftmost digit, is \(n\) is divisible by 7? (For example, \(f(4)\) corresponds to the value of the hundreds digit of \(n\).)
(1) \(f(1) = f(4), f(2) = f(5), f(3) = f(6)\)
(2) \(f(1) = f(2) = f(3) = f(4) = f(5) = f(6)\)
(1) \(f(1) = f(4), f(2) = f(5), f(3) = f(6)\)
(2) \(f(1) = f(2) = f(3) = f(4) = f(5) = f(6)\)
16 Sep 2014, 00:56
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Official Solution:
If in a six-digit integer \(n\), \(f(k)\) represents the value of the \(k_{th}\) digit from the leftmost digit, is \(n\) is divisible by 7? (For example, \(f(4)\) corresponds to the value of the hundreds digit of \(n\).)
(1) \(f(1) = f(4), f(2) = f(5), f(3) = f(6)\)
The above condition implies that the last three digits of \(n\) are the same as the first three digits of \(n\). Therefore, \(n = abc,abc = abc*1000 + abc = abc*(1000 + 1) = abc*1001\). Since 1001 is divisible by 7 (1,001 = 143*7), then \(n = abc*1001\) must also be divisible by 7. Sufficient.
(2) \(f(1) = f(2) = f(3) = f(4) = f(5) = f(6)\)
The above condition implies that all digits of \(n\) are the same. Therefore, \(n = aaa,aaa = aaa*1000 + aaa = aaa*(1000 + 1) = aaa*1001\). Since 1001 is divisible by 7 (1,001 = 143*7), then \(n = aaa*1001\) must also be divisible by 7. Sufficient.
Answer: D
If in a six-digit integer \(n\), \(f(k)\) represents the value of the \(k_{th}\) digit from the leftmost digit, is \(n\) is divisible by 7? (For example, \(f(4)\) corresponds to the value of the hundreds digit of \(n\).)
(1) \(f(1) = f(4), f(2) = f(5), f(3) = f(6)\)
The above condition implies that the last three digits of \(n\) are the same as the first three digits of \(n\). Therefore, \(n = abc,abc = abc*1000 + abc = abc*(1000 + 1) = abc*1001\). Since 1001 is divisible by 7 (1,001 = 143*7), then \(n = abc*1001\) must also be divisible by 7. Sufficient.
(2) \(f(1) = f(2) = f(3) = f(4) = f(5) = f(6)\)
The above condition implies that all digits of \(n\) are the same. Therefore, \(n = aaa,aaa = aaa*1000 + aaa = aaa*(1000 + 1) = aaa*1001\). Since 1001 is divisible by 7 (1,001 = 143*7), then \(n = aaa*1001\) must also be divisible by 7. Sufficient.
Answer: D
martinnotceoyet
Joined: 09 Dec 2013
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21 Oct 2014, 07:32
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Quote:
Sorry if I am wrong, but I believe the example should be F(4) is the value of the thousands digit of N .
