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Need help with problem on numbers

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Need help with problem on numbers

by msd_2008 » Wed Jul 16, 2008 4:54 pm
Q. For any positive integer n, the sum of the first n positive integers equals [n(n+1)]/2. What is the sum of all even integers between 99 and 301?

A.10,100 B.20,200 C.22,650 D.40,200 E.45,150

OA is B. Can someone explain how that answer has been arrived at?

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MSD
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by sudhir3127 » Thu Jul 17, 2008 3:30 am
i got the answer as 40,200, D

here it goes .. sum of nterms = n/2( first term + last term)
series is 100+102+..........+300

n= 300-100+1= 201

201/2(100+300)=40,200


can some confirm how the OA is B. 20200
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by parallel_chase » Thu Jul 17, 2008 3:50 am
sudhir3127 wrote:i got the answer as 40,200, D

here it goes .. sum of nterms = n/2( first term + last term)
series is 100+102+..........+300

n= 300-100+1= 201

201/2(100+300)=40,200


can some confirm how the OA is B. 20200

Answer is indeed B.


n is not 201 it is 101. 201 includes odd and even integers. We only have to find out the even integers.

(100+300)/2 = 200

200*101 = 20200

Hope its clear now.
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by sudhir3127 » Thu Jul 17, 2008 3:57 am
can you tell me how to find n?
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by moliver » Thu Jul 17, 2008 3:58 am
how about (300-100)*202/2?

the answer is b = 20.200

or try with this:
the sum of the the even numbers from 0 to 10
0+2+4+6+8+10 = 30
= (10-0)*6/2

0 to 100
= (100-0)*51/2 = 2550
but we include the "100", and we need it so 2550-100 = 2450

0 to 300
= (300-0)*151/2 = 22650

Note from 0 to 300 we have 300 number but we need only the even numbers plus 1 because we start from a even number and finish in a even number : 151 = 300/2 + 1

no we have: 22650 - 2450 = 20200

I am sure that is going to be a easier explanation, but this could be a first option.
I check the answer with the excel and is correct.
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by parallel_chase » Thu Jul 17, 2008 4:03 am
sudhir3127 wrote:can you tell me how to find n?

(300-100)/2 + 1 =101

Similarly if you have to find multiples of 8 between 1- 1000

(1000-8)/8 + 1 = 125

Let me know if you still have any questions.
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by sudhir3127 » Thu Jul 17, 2008 4:07 am
kewl.... got it .. thank u
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by louvre » Thu Jul 17, 2008 5:24 am
parallel_chase
Similarly if you have to find multiples of 8 between 1- 1000

( 1000- 8 ) /8 + 1 = 125
sorry for hijacking the post.
Coming back to the number of multiples between 2 numbers...I made a small simulation.

Find the number of multiples of 6 between 100 and 1000?

I don't know the direct formula, so
(no. of multiples of 6 from 1 to 1000) - (no. of multiples of 6 from 1 to 100)
( ( (1000-6)/6 ) + 1 ) - ( ( (100-6)/6 ) + 1 )
(165.66+1) - (15.66+1) = 182 (ignoring the fractional part).

May I say that?
Thanks in advance
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by louvre » Thu Jul 17, 2008 5:26 am

(165.66+1) - (15.66+1) = 182 (ignoring the fractional part).
(165.66+1) - (15.66+1) = 150

I should avoid such things in the exam.
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by parallel_chase » Thu Jul 17, 2008 5:44 am
louvre wrote:

(165.66+1) - (15.66+1) = 182 (ignoring the fractional part).
(165.66+1) - (15.66+1) = 150

I should avoid such things in the exam.
Well all i can say you can derive such formulas as and when required.

I am not very much into formulation, i dont like to remember formulas except in geometry. Well even in geometry if you know the formulas for rectangle and circle you can practically derive any formula out of that.

Because at the time of GMAT you may forget the formula and fall into the trap answer of GMAT.

Therefore it is necessary to understand the logic behind each and everything you do.
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by anksanks » Thu Jul 17, 2008 7:39 am
This one is a simple arithmetic progression problem.

first time = a = 100.
last term = l = 300
difference = d = 2 (as alternate terms are there, 100, 102, 104)

now, l = a + (n-1) * d
=> 300 = 100 + (n-1) * 2
=> n = 101

now sum = S = (n/2) * [2a + (n-1) * d ]
=> S = (101/2) * [2*100 + (101-1) *2 ]
=> S = (101/2)*[400] = 101 * 200 = 20200
=> S = 20200
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