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If j and k are even integers and j< k
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Updated on: 18 Oct 2019, 10:53
Updated on: 18 Oct 2019, 10:53
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If j and k are even integers and \(j < k\), which of the following equals the number of even integers that are greater than j and less than k ?
a. \(\frac{(k-j-2)}{2}\)
b. \(\frac{(k-j-1)}{2}\)
c. \(\frac{(k-j)}{2}\)
d. \((k-j)\)
e. \((k+j)\)
The explanation says since j and k are even integers, it follows that k = j + 2n . Where did this come from? I don't see how they got k = j+2n; having trouble translating it.
a. \(\frac{(k-j-2)}{2}\)
b. \(\frac{(k-j-1)}{2}\)
c. \(\frac{(k-j)}{2}\)
d. \((k-j)\)
e. \((k+j)\)
The explanation says since j and k are even integers, it follows that k = j + 2n . Where did this come from? I don't see how they got k = j+2n; having trouble translating it.
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Joined: 16 May 2014
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GRE 1: Q165 V161
Re: Translating Algebraic Representation
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05 Aug 2015, 08:48
05 Aug 2015, 08:48
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Expert Reply
Explanation
Even number is defined as those integers which are divisible by 2. Therefore consecutive integers are for example : 2, 4, 6, 8,.... As you can see the one of the basic properties of even numbers is that when an even number is subtracted from another even number the result will be an even number.
So we can write given k,j are even integers and k>j:
k = j + 2n, where n is an integer.
However it must be noted n doesn't represent the number of integers between k and j. There is a counting problem there. The number of even integers between k and j is (n-1).
Foe example if we check the concept to determine the number of even numbers between say 2 and 8. We can easily determine that there are only two namely, 4 and 6.
8 = 2 + 2 *3 (k = j+ 2n)
In this particular case n = 3. So the number of even integers between k and j is n-1 where n = \(\frac{(k-j)}{2}\). Therefore n-1 = \(\frac{(k-j - 2)}{2}\). The correct choice is A.
Re: If j and k are even integers and j< k
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18 Oct 2019, 05:08
18 Oct 2019, 05:08
gonewindy wrote:
If j and k are even integers and \(j < k\), which of the following equals the number of even integers that are greater than j and less than k ?
a. (k-j-2)/2
b. (k-j-1)/2
c.(k-j)/2
d.(k-j)
e. (k+j)
The explanation says since j and k are even integers, it follows that k = j + 2n. Where did this come from? I don't see how they got k = j+2n; having trouble translating it.
a. (k-j-2)/2
b. (k-j-1)/2
c.(k-j)/2
d.(k-j)
e. (k+j)
The explanation says since j and k are even integers, it follows that k = j + 2n. Where did this come from? I don't see how they got k = j+2n; having trouble translating it.
May someone explain this in more detailed way ?
Does n represent the number of even numbers between J and K ?
