Hi kop
Let's rewrite it this way
2+ 4 + 6 + ..... + k-1 = 2 + 2x2 + 2x3 + .....+ 2x(k-1)/2
= 2( 1 + 2 + 3 + ..... + (k-1)/2 )
Plug (C) in, we have 2x41x20 < 79x80 => K need to be larger
Plug (D) in, we have 2x39x79 < 79x80 => K need to be larger again
=> we are left with (E)
Check (E) if you want to : plug (E) we have 2x( 80x39 + 40) = 79*80
Regard
consecutive numbers
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dominhtri1995
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theCodeToGMAT
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k is ODD
Total Even Numbers = k/2 (considering INT value)
Sum of Even Numbers = (n)(n+1)
Sum of Odd Numbers = n^2
(k/2)(k/2+1) = 79*80
So comparing both ..
k/2 = 79
k = 158
So, k+1 = 159 (since we considered INT value)
[spoiler]{E}[/spoiler]
Total Even Numbers = k/2 (considering INT value)
Sum of Even Numbers = (n)(n+1)
Sum of Odd Numbers = n^2
(k/2)(k/2+1) = 79*80
So comparing both ..
k/2 = 79
k = 158
So, k+1 = 159 (since we considered INT value)
[spoiler]{E}[/spoiler]
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GMATGuruNY
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Sum of evenly spaced integers = (number of integers)(average of biggest and smallest)kop wrote:The sum of the even numbers between 1 and k is 79*80, where k is an odd number, then k=?
(A) 79
(B) 80
(C) 81
(D) 157
(E) 159
Prediction:
79 = the number of even integers and 80 = the average of the biggest and smallest.
Let x = the biggest integer.
Smallest integer = 2.
If the average of the biggest and smallest is 80, we get:
(x+2)/2 = 80.
x+2 = 160
x = 158.
Implication:
79*80 represents the sum of the even integers between 2 and 158, inclusive.
To confirm:
Number of even integers between 2 and 158, inclusive = 79.
Average of biggest and smallest = 80.
Sum = (number)(average) = 79*80.
Since k is the next biggest ODD integer, k=159.
The correct answer is E.
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alt way
n/2 * {2a+(n-1)d = 79*80 = 6320
we know that n = 2 and d = 2
n^2 - n - 6320
n^2 + 80n - 79n - 6320
n = 79 , -80
79 even values means 2*79 total values ,required no is odd hence 159 ans
n/2 * {2a+(n-1)d = 79*80 = 6320
we know that n = 2 and d = 2
n^2 - n - 6320
n^2 + 80n - 79n - 6320
n = 79 , -80
79 even values means 2*79 total values ,required no is odd hence 159 ans
