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Re: If j and k are even integers and j< k [#permalink]
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here you have basically two strategies: or picking numbers or conceptually is a little bit nasty because the latter involves some manipulation and the knowledge of number properties

Picking 14 and 6.

\(14-6=8\)

\(\frac{8-2}{2}=\frac{6}{2}=3\)

In fact, the even numbers between 14 and 6 are 3 : 8,10, and 12.

Or

Notice that even numbers are equally spaced: 2,4,6,........Only answer choice A has an equal space and that guarantees you is the right answer

\(\frac{(k-j-2)}{2}\)

The minus 2 in the numerator

Hope this helps
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If j and k are even integers and j< k [#permalink]
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Sometimes substitution of arbitrary numbers helps to solve these problems. Given the condition, j and k are even (thus, equally placed) with j<k

ex: 2,4,6,8...so on

assume j =2 and k =8. it satisfies the two conditions mentioned above ( both are even and j<k)

You can see all the options are expressed in linear form. We can try to express j and k in a linear form using our assumptions

k =j +2n .

this linear form satisfies the above arrangement. K= 2+2(3) = 2+6 =8 ( n is 0,1,2,3... and n is in third position)

so now between 2 and 8, there are three numbers. - Condition 1

2 and 8 are even. - Condition 2

2<8 -Condition 3

only option a satisfy the above three conditions when we plug in j=2, k= 8

(8-2-2)/ 2 =3
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Re: If j and k are even integers and j< k [#permalink]
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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. \(\frac{(k-j-2)}{2}\)

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.


I think \(\frac{(k-j)}{2}\) and \((k-j)\) is also satisfy the given condition. take take 16 and 4, diff is 12 and diff/2 is 8, which is even no lies in between my considered values. Similar case is observed in D as well
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Re: If j and k are even integers and j< k [#permalink]
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The number of integers between the two numbers INCLUDING the last is given by the formula:

(last-first)/2

The number of integers between the two numbers INCLUDING both the first and the last is given by the formula:

(last-first)/2 + 1

The number of integers between the two numbers EXCLUDING BOTH the first and the last is given by the formula:

(last-first)/2 -1

This is the one we need because it says between the two numbers, and it translates to (last-first-2)/2
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Re: If j and k are even integers and j< k [#permalink]
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