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Medium level - sum of integers

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by Brent@GMATPrepNow » Thu Mar 05, 2009 4:42 pm
It's dinner time, so I better post my solution before I go.
The answer is A.

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by pakaskwa » Thu Mar 05, 2009 7:25 pm
For any arithmetic progression, the sum is

S = n(a1+an)/2

Where n is the number of sequence. For the question, there are 50 numbers from 1,3,5,... to 99, half of 100.
a1=1, a50=99,

So S=50(1+99)/2=2500

But the interesting part is that, why the sum is not half of 5050?
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by satish.nagdev » Fri Mar 06, 2009 12:38 am
pakaskwa wrote: But the interesting part is that, why the sum is not half of 5050?
because every even integer falls after odd and its value is +1 of the odd number, so 50 even numbers in set [1 to 100] result in sum of even numbers = sum of odd numbers +50

its like
1 2
3 4
....
.....
99 100

and thanks pakaskwa your formula for finding sum was quick one :)
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by pakaskwa » Fri Mar 06, 2009 3:08 pm
That makes sense! I didn't think enough.

By the way, I think Brent's idea is to let people know that, if they don't remember the formula, how they can get the correct numbers.
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Re: Medium level - sum of integers

by logitech » Sun Mar 08, 2009 3:39 pm
Brent Hanneson wrote:If the sum of the integers from 1 to 100 is 5050, then the sum of the odd integers from 1 to 99 is
(A) 2500
(B) 2525
(C) 2550
(D) 2575
(E) 2600
1+2+...+n = (n)x(n+1)/2

1+2+3+4+5+6+7+8+9+10 = (10)(10+1)/2 = 55

1+3+5+7+9 = 25

2+4+6+8+10 = 2( 1+2+3+4+5) = 2x(5)x(5+1)/2 = 30

I used this simple set to explain you the logic behind the calculation.

So:

2+4+6+..100 = 2( 1+2+3...+50) = 2x(50)(51)/2 = 50x51

So the sum of odd numbers is:

5050 - ( SUM OF EVEN ) = (101)x(50) - (50) x (51 ) = 50x50 = 2500
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by deepoe » Thu Apr 16, 2009 2:59 am
I thought this


1+3+5+7+9 = 25


First digit = 10 numbers
Second digit = 10 numbers

So 10x10x25 = 2500

But is this a good way to calculate it?
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