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26 May 2017, 04:48
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If the average (arithmetic mean) of 5 integers is 0.2 smaller than the greatest number, what is the least possible number of even numbers among the integers?
A. One
B. Two
C. Three
D. For
E. Five
A. One
B. Two
C. Three
D. For
E. Five
26 May 2017, 04:49
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Bookmarks
Official Solution:
If the average (arithmetic mean) of 5 integers is 0.2 smaller than the greatest number, what is the least possible number of even numbers among the integers?
A. One
B. Two
C. Three
D. For
E. Five
If you say 5 integers are such that \(a < b < c < d < e\),
you get \(\frac{a+b+c+d+e}{5} = e-0.2\), and if you multiply 5 on both sides, you get \(a+b+c+d+e = 5e-1\), \(a+b+c+d = 4e-1 = odd\). If so, in this case, you get \(a+b+c+d = odd\). Since \(odd+odd+odd+even = odd\), maximum of 3 numbers among these can become an odd numbers and \(4e = 4odd = even\), so e is also an odd number. From \(a, b, c, d,\) and \(e,\) the maximum number of odd numbers is four, and the maximum number of even numbers is at least one. Therefore, the answer is A.
Answer: A
If the average (arithmetic mean) of 5 integers is 0.2 smaller than the greatest number, what is the least possible number of even numbers among the integers?
A. One
B. Two
C. Three
D. For
E. Five
If you say 5 integers are such that \(a < b < c < d < e\),
you get \(\frac{a+b+c+d+e}{5} = e-0.2\), and if you multiply 5 on both sides, you get \(a+b+c+d+e = 5e-1\), \(a+b+c+d = 4e-1 = odd\). If so, in this case, you get \(a+b+c+d = odd\). Since \(odd+odd+odd+even = odd\), maximum of 3 numbers among these can become an odd numbers and \(4e = 4odd = even\), so e is also an odd number. From \(a, b, c, d,\) and \(e,\) the maximum number of odd numbers is four, and the maximum number of even numbers is at least one. Therefore, the answer is A.
Answer: A
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