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26 May 2017, 04:42
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For positive integers \(n\) and \(m (n>m)\), is the average (arithmetic mean) of \(4(10^{n}), 4(10^{n-1}), ......,\) and \(4(10^{n-m})\) an integer?
(1) \(m < 6\)
(2) \(n = 10\)
(1) \(m < 6\)
(2) \(n = 10\)
26 May 2017, 04:42
Kudos
Bookmarks
Official Solution:
If you take the 1st step of the variable approach and modify the original condition and the question, you get
\(mean = \frac{4(10^{n-m})+4(10^{n-m+1})+\ldots +4(10^{n-1})+4(10^{n})}{n-(n-m)+1} = integer\)?, or
\(mean = \frac{4(10^{n-m})(10^{m+1}-1)}{m+1} = integer?\)
The formula used in the denominator is "the number of consecutive integers=last term-first term+1" and the formula used in the numerator is a geometric sequence. In other words,
From \(a_{1}+a_{1}r+a_{1}r^{2}+\ldots .+a_{1}r^{k-1} = \frac{a_{1}(r^{k}-1)}{r-1}\), you substitute \(a_{1}=4(10^{n-m}), r=10\), and \(k = n - (n-) + 1 = m + 1\).
In the original condition, there are 2 variables \((n, m)\), and in order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer. In fact, C is the answer, but this is an integer question, one of the key questions, so you should apply "CMT 4(A: if you get C too easily, consider A or B)".
In the case of 1),
\(m=1, \frac{4(10^{n-1})(10^{2}-1)}{9(1+1)} = 2(10^{n-1})(11) = integer\)
\(m=2, \frac{4(10^{n-2})(10^{3}-1)}{9(2+1)} = \frac{4(10^{n-2})(999)}{9(3)} = 4(10^{n-2})(37) = integer\)
\(m=3, \frac{4(10^{n-3})(10^{4}-1)}{9(3+1)} = \frac{4(10^{n-3})(9,999)}{9(4)} = 4(10^{n-3})(1,111) = integer\)
\(m=4, \frac{4(10^{n-4})(10^{5}-1)}{9(4+1)} = \frac{4(10)(10^{n-5})(99,999)}{9(5)} = 4(5)(10^{n-5})(11,111) = integer\)
\(m=5, \frac{4(10^{n-5})(10^{6}-1)}{9(5+1)} = \frac{4(10^{n-5})(999,999)}{9(6)} = 2(10^{n-5})(37,037) = integer\), hence yes, it is sufficient.
In the case of con 2), if \(m = 1\) as above, it is yes, but \(m = 6\), \(\frac{4(10^{10-6})(10^{7}-1)}{9(6+1)} \ne integer\), hence no, it is not sufficient. The answer is A.
Answer: A
If you take the 1st step of the variable approach and modify the original condition and the question, you get
\(mean = \frac{4(10^{n-m})+4(10^{n-m+1})+\ldots +4(10^{n-1})+4(10^{n})}{n-(n-m)+1} = integer\)?, or
\(mean = \frac{4(10^{n-m})(10^{m+1}-1)}{m+1} = integer?\)
The formula used in the denominator is "the number of consecutive integers=last term-first term+1" and the formula used in the numerator is a geometric sequence. In other words,
From \(a_{1}+a_{1}r+a_{1}r^{2}+\ldots .+a_{1}r^{k-1} = \frac{a_{1}(r^{k}-1)}{r-1}\), you substitute \(a_{1}=4(10^{n-m}), r=10\), and \(k = n - (n-) + 1 = m + 1\).
In the original condition, there are 2 variables \((n, m)\), and in order to match the number of variables to the number of equations, there must be 2 equations. Therefore, C is most likely to be the answer. In fact, C is the answer, but this is an integer question, one of the key questions, so you should apply "CMT 4(A: if you get C too easily, consider A or B)".
In the case of 1),
\(m=1, \frac{4(10^{n-1})(10^{2}-1)}{9(1+1)} = 2(10^{n-1})(11) = integer\)
\(m=2, \frac{4(10^{n-2})(10^{3}-1)}{9(2+1)} = \frac{4(10^{n-2})(999)}{9(3)} = 4(10^{n-2})(37) = integer\)
\(m=3, \frac{4(10^{n-3})(10^{4}-1)}{9(3+1)} = \frac{4(10^{n-3})(9,999)}{9(4)} = 4(10^{n-3})(1,111) = integer\)
\(m=4, \frac{4(10^{n-4})(10^{5}-1)}{9(4+1)} = \frac{4(10)(10^{n-5})(99,999)}{9(5)} = 4(5)(10^{n-5})(11,111) = integer\)
\(m=5, \frac{4(10^{n-5})(10^{6}-1)}{9(5+1)} = \frac{4(10^{n-5})(999,999)}{9(6)} = 2(10^{n-5})(37,037) = integer\), hence yes, it is sufficient.
In the case of con 2), if \(m = 1\) as above, it is yes, but \(m = 6\), \(\frac{4(10^{10-6})(10^{7}-1)}{9(6+1)} \ne integer\), hence no, it is not sufficient. The answer is A.
Answer: A
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