Obtain the sum of all positive integers up to 1000, which

Question

Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

A. 10,050
B. 5,050
C. 5,000
D. 50,000
E. 55,000

FN · Accepted Answer

last term thats not divisble by 10

995=5+(n-1)10

995=5+10n-10
1000=10n
n=100

sum= 100/2(10+(100-1)10)

this give us 50,000.

D it is

greenoak · Answer

Hi!

For those who asked to explain the formula:

For arithmetic progression,

Sn = n*(a1+an)/2

Our series is 5, 15, 25, … , 995. (a1=5, an=995, d=10).

Only, we need to calculate n. In order to find it, we can use the formula an = a1+(n-1)*d – just put in numbers and solve for n.

maratikus · Answer

the answer is sum of 5*k - sum of 10*n

5 + ... + 1000 = (5+1000)*200/2 = 1,005*100 = 100,500

10 + .. + 1000 = (10+1000)*100/2 = 1,010*50 = 50,500

the answer is 100,500-50,500 = 50,000 -> D

jallenmorris · Answer

D (which is clearly established by now;) )

The way I figured it was this way.

Multipe of 5 is 5, 10, 15, 20, etc. Those not divisible by 2 are 5, 15, 25, 35, etc.

So, there will be 1 number to include for every 10, so 1000 numbers / 10 = 100 numbers to include in the sum.

If you look at 5 + 995 = 1000, 15 + 985 = 1000, 25 + 975=1000, 35 + 965 = 1000...495+505=1000, You do this, then you have 50 pairs that add up to be 1000, so 50 * 1000 = 50,000 (i.e. D)

I tend to see things in patterns rather than formulas.

bajaj · Answer

There are 200 no's divisible by 5 upto 1000 and out of this only 100 nos are not divisible by 2. So the sum we need is (5+15+....+995) or 5(1+3+5+....199). The sum of consequtive n integers is n^2. Therefe for sum of 100 odd nos is 100^2.

Ans = 5 (100^2)=50000 .D

Ans = 5 (100^2)=50000 .D

Oski · Answer

In your example, 1,3,5,..,199 are not consecutive integers

(and I don't see where this results "n^2" come from). What you can write is that the number we look for is \sum_{k=1 and k is odd}^{200} 5*k = \sum_{i=0}^{99} 5(2i+1) = 5 \left( \sum_{i=0}^{99} 2i + \sum_{i=0}^{99} 1 \right) = 5 \left( 2 \sum_{i=0}^{99} i + 100 \right) = 5 \left( 2 \frac{99*100}{2} + 100 \right) = 5 \left( 9,900 + 100 \right) = 50,000

kapilnegi · Answer

5,15,25,.....,995
total terms = 100

sum = n/2

n = 100
a = 5
d = 10

Sum = 50000

D is the answer.

craky · Answer

5+15+25+...+995=
1*5+3*5+5*5+...+199*5=
5*(1+3+5+...+199)

Sum of first n odd numbers:  n^2

#of odd numbers:  n=\frac{(199-1)}{2}+1 = \frac{199-1+2}{2}=100

5*100*100=50000

gmatpapa · Answer

My approach:

Step 1: Find out the sum of all integers divisible by 5. Number of such terms=  \frac{(1000-0)}{5} + 1= 201 . Sum of all such integers=  201*500  (using the sum formula of terms in Arithmetic Progression).. (i)
Step 2: Find out the sum of all integers divisible by 10: Number of such terms=  \frac{(1000-0)}{10}= 101 . Sum of all such integers=  101*500= 101*500  .. (ii)

Now, the multiples of 10 are the ones which are divisible by both 5 and 2. Therefore, we need to subtract (ii) by (i) to arrive at the terms divisible only by 5 and not by 2

Step 3:  201*500 - 101*500 
 =  500*(201-101) 
 =  500*100 
 =  50,000 . Answer.

Thus, the answer is D.

gmatpapa · Answer

This is the quickest approach IMO. you really dont need anything else! 
+1

fluke · Answer

Sequence is: 5,15,25,....995

Common difference d = 10
Number of elements: ((995-5)/10)+1=100
Average: (first+last)/2 = (5+995)/2 = 500

Sum= Number of elements * Average = 100 * 500 = 50000

Ans: "D"

re4Bschool11 · Answer

100 possibilities, evenly spaced by 10. Since its an evenly spaced set, the median (500) is also the mean. Find the median: will be the average between the 50th observation (495) and the 51st observation (505), which is 500. Average * number of observations == total. 500*100 is 50,0000.

nishanthadithya · Answer

Can u explain in brief why do you multiply by 500. Thank you

Posted from my mobile device

Bunuel · Answer

For every evenly spaced set (a.k.a. arithmetic progression), the sum equals to (mean)*(# of terms). 500 is the mean there: (mean)=(first+last)/2=(5+995)/2=500.

Hope it's clear.

GMATinsight · Answer

Total Multiples of 5 from 1 through 1000 = 1000/5 = 200

Half of these 200 multiples of 5 will be even (i.e. divisible by 2) and remaining half will be odd multiples of 5

i.e. Question : 5+15+25+......+955=?

Since it's an Arithmetic Progression (In which difference between any two consecutive terms remain constant) in which  sum of first and last term = Sum of second and Second last term = and so on...

100 such numbers i.e.e 50 such pairs and sum of each pair = 5+955 = 1000

i.e. Sum of all pairs = (100/2)*(5+955) = 50*1000 = 50,000

Answer: Option D

stonecold · Answer

The question is basically asking us the sum of the AP series 
5+15+......995
Number of terms can be calculated from A(n)= A+(n-1)D 
So N=100
Sum =100/2 *   = 50,000
hence D

rahulbanerjee1992 · Answer

Let's first try to figure out what the problem's actually asking us.
Numbers divisible by 5 mean the set : {5,10,15,....,1000}
Numbers divisible by 5 and 2 mean the set : {10,20,30,...,1000}
 OUR CRITERION :  divisible by 5 and not divisible by 2 
So, the set we need to work with is : {5,15,25,35,...995}
This is a generic arithmetic progression problem, where the common difference is 10 and number of numbers is 100*, so we can easily apply the formula,

s=n/2*{2a+(n-1)d} where s=sum of numbers, n=number of numbers a=first term and d=common difference
So, s=100/2*{2*5+(100-1)*10}=50,000 ...OPTION D

*number of numbers is 100 because between 1-10, we have only one number, i.e. 5, now multiply it to our total 1-1000, i.e. 100.

GmatPoint · Answer

The sum of positive integers which are less than 1000 and divisible by 5 but not by 2.
﻿The numbers divisible by 5 are written in their general form: 5*k.
﻿The numbers are also divisible by 2 if k is an even number.
﻿Hence k only takes odd numbers.
﻿The highest odd number k can take so that the number 5*k is less than 1000 is 199.
﻿Hence the sum of the numbers :
5*( 1+3+5+............199)

= Sum of first n odd numbers starting from 1 is   n^2

5*(100*100)﻿

﻿= 50000

luisdicampo · Answer

Deconstructing the Question

We must find the sum of all positive integers up to   1000   that are divisible by  5  and not divisible by  2 .

Numbers divisible by  5  but not by  2  must be odd multiples of  5 .

Thus the sequence is

5, 15, 25, 35, ...

This is an arithmetic sequence with first term

5

and common difference

10

Step-by-step

Find the largest valid number less than or equal to  1000 .

995 = 5 	imes 199

So the sequence is

5, 15, 25, ..., 995

Now compute the number of terms

n = \frac{995 - 5}{10} + 1

n = \frac{990}{10} + 1

n = 99 + 1

n = 100

Now compute the sum of the arithmetic sequence

S = \frac{n}{2}(first + last)

S = \frac{100}{2}(5 + 995)

S = 50 	imes 1000

S = 50\,000

Answer D: 50,000

Obtain the sum of all positive integers up to 1000, which

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