Find the sum of all natural numbers less than 1000 which are not divisible by 5 or 2.?

Answer 1

Sum of numbers up to 1000 = (1/2)*1000*1001 = 500500

Sum of numbers divisible by 2 up to 1000 = 2*(1+2+...500) = 2*(1/2)*500*501 = 250500

Sum of numbers divisible by 5 up to 1000 = 5*(1+2+...200) = 5*(1/2)*200*201 = 100500

Sum of numbers divisible by both 2 and 5 up to 1000
= 10*(1+2+3+...100) = 10*(1/2)*100*101 = 50500

Required sum = 500500 - 250500 - 100500 + 50500 = 200000

Answer 2

First, the sum of ALL the natural numbers from 1 to 1000 is S = n(a + l)/2 or if you prefer, S = n[2a + (n - 1 )d]/2 - in the first version I have written, l is the last term, a the first, n the number of terms, S the sum of the terms; this gives S = (1000/2) x 1001 = 500 x 1001 = 500500. The numbers which ARE divisible by 2 are 2, 4, 6, 8, .......1000 = so the sum of these is 2(1 + 2 + 3 + 4 + ..... + 500) and using the above formula gives S = 250500. The numbers that ARE divisible by 5 are 5, 10, 15, 20,..... 1000 so their sum is 5(1 + 2 + 3 +.... 200), so that S, again by the above formula is 5 x (200/2)(201) = 500 x 201 = 100500. That means that the sum of the numbers divisible by 2 or 5 is 250 500 + 100 500 = 350 500; the total of all the numbers from 1 to 1000 is 500 500, so the total you need is 500 500 - 350 500 which is 150 000.

Answer 3

First sum all the numbers from 1 to 1000:

Sum(1 to n) = (n + 1)*n/2

Sum(1 to 1000) = 1001 * 1000/2 = 500500

Now subtract all the numbers divisible by 2:

2 + 4 + 6 + ..... 2n = 2(n + 1) * n /2 = (n + 1) n

Here n= 500 since the even sum goes up to 1000 = 2*500

Sum(2,4,6,8...1000) = 501 * 500 = 250500

Now it gets a little trickier. To subtract all the numbers left divisible by 5 we need to extract only the odd multiples of 5 since the even ones are divisible by two and have been removed. This leaves:

5 + 15 + 25 + 35 ...... + 995

Factoring out a 5:

5(1 + 3 + 5 + 7...... + 199)

There are exactly 100 terms in the sum.

Rewriting the terms:

5((2-1) + (4-1) + (6-1) ..... + (200-1))

Since there are 100 terms, all the -1 add up to 100 leaving:

5(2 + 4 + 6 + 8 .... + 100 - 100)

Factoring out a 2 and cancelling the 100s::

10(1 + 2 + 3 + 4 ..... + 49) = 10 Sum(1 to 49)

= 10 * 50 * 49 / 2 = 12250

Put it all together and your sum is:

500500 - 250500 - 12250 = 237750

Answer 4

There have been several decent answers, so I decided to propose one that is probably just as good, but perhaps more novel:

Consider adding up just 1-10, with those restrictions
1+3+7+9

For the next ten, 11-20, with those restrictions, we have:
11+13+17+19 = 1 + 3 + 7 + 9 + 40

and then:

21 + 23 + 27 + 29 = 1 + 3 + 7 + 9 + 80

...and so on.


Call the interval 1-10 interval 1
Call the interval 11-20 interval 2 ...etc

For interval n, we have 1+3+7+9+40(n-1)
We have 100 intervals, so we simply have:

100(1+3+7+9) + 40 (0+1+2+3+...+99)

Now a clear pair-off argument helps us with 0+1+2+3+...+99.
1 + 99 = 100
2 + 98 = 100
...49 such pairs ...
48 + 52 = 100
49 + 51 = 100
and the 50 is left over, giving 4950.

You could also use the standard formula given by the link below. Regardless, we're then left with:

100(1+3+7+9) + 40 (4950)

which we can then determine is 200,000

Answer 5

Here's the way to do this.
(1) Add all the numbers less than 1000, i.e.
1+2+....+999
(2) Subtract all numbers divisible by 5, i.e.
5+10+15+.... +995
(3) Subtract all numbers divisible by 2, i.e.
2+4+6+....+998
(4) Add back in all numbers divisible by 10, since you've taken them out twice, i.e.
10+20+....+990

These are all simple arithmetic sequences. If you don't know know how to sum an arithmetic sequence, add up the terms on either end. How many such terms do you have?
eg if you have 3+6+9+12+15+18
the terms on the end sum to 21. So do the next two, so do the next two. So you have 3 lots of 21, total 63.

Hope this helps.

Answer 6

There are two sums of arithmetic series you need to find.

S(1) = 1 + 3 + 5 + ... + 999
This has 500 terms
so S(1) = 500/2 (1 + 999) = 500^2

S(2) = 5 + 15 + 25 + ... + 995
This has 100 terms
S(2) = 100/2 * (5 + 995) = 100*500

Your required sum = S(1) - S(2) = 400*500
= 200,000

Answer 7

the largest multiple of 5 that is less than 1000 = 995

5+(n-1)*5 =995
n = 199

there are 199 multiples of 5 less than 1000.

sum of all multiples of 5 less than 1000 = 5+10+15+...+995 = 199 (5+995) / 2 = 99500



the largest multiple of 2 that is less than 1000 = 998

2+(n-1)*2 = 998
n = 499

there are 499 multiples of 2 less than 1000.

sum of all multiples of 2 less than 1000 = 2+4+6+...+998 = 499 (2+998) / 2 = 249500



multiple of both 2 and 5 = multiple of 10

the largest multiple of 10 that is less than 1000 = 990

10+(n-1)*10 = 990
n = 99

there are 99 multiples of 10 less than 1000.

sum of all multiples of 10 less than 1000 = 10+20+30+...+990 = 99(10+990) / 2 = 49500



sum of all natural numbers less than 1000 = 999(1+999) / 2 = 499500

sum of all natural numbers less than 1000 which are not divisible by 5 or 2 = 499500 - 99500 - 249500 + 49500 = 200000


Beautiful number

Answer 8

Very simple.
Answer is 200000.

1. Add all number from 1 to 1000 = 500500

2. Then subtract sum of numbers which divided by 2 = -50500
3. Then subtract sum of numbers which divided by 5 = -100500

4. As you have removed numbers which divided by 2 & 5 twice so add sum of numbers which devided by 10 = 50500

5. Result is => = 500500 -250500-100500+ 50500 = 200000

Formulas

sum of 1 to n = (n*(n+1))/2
sum of 1 to n which are devided by x = x*(n/x*((n/x)+1))/2

Answer 9

number which are not divisible by 2 means odd number. so the sum of odd number between 1 to 999 is (500)^2=2,50,000.( there are 500 odd no. frm 1 to 999).
but there are odd no. which is divisible by 5 i.e 5,15,25,....,995.
so, sum of this series are 5(1+3+5+.....199) = 5(10000) = 50,000
hence answer is 2,50,000 - 50,000 = 2,00,000

Answer 10

very interesting question, 1) first found sum of odd numbers(div/2), 2) sum of numbers divisible by 5, 3) sum of commen to div/2 & div/5, 4) subtract from total of step 1 & 2
5) sum of total numbers; 6) subtract from total on step 4 figure. You got answer. All steps I have formula except for step no.3