find the sum of numbers between 100 and 10000 which are divisible by 3 ?

Question

explain by giving the formula.

Answer 1

By inspection, the first such number is 102 and the last such number is 9999. So the sum is 102 + 105 + 108 + 111 + ... + 9996 + 9999. You can rewrite these numbers as 102 + (102 + 3) + (102 + 2*3) + (102 + 3*3) + ... + (102 + 3298*3) + (102 + 3299*3). To rewrite the last two numbers, I took (9999 - 102) and divided it by 3. So, now we have a series that is easier to work with. You can see that there are 3300 terms, and you can rewrite the sum as 3300*102 + 3*(1 + 2 + 3 + ... + 3298 + 3299). So, how do you sum the numbers from 1 to 3299? It's quite easy, because there is a formula. The sum of numbers from 1 to n is equal to (n + 1)*n/2. So this sum is 3300*3299/2 = 5443350. Now we have 3300*102 + 3*5443350 = 16666650.

You can also approximate this value by using nothing but the (n+1)*n/2 formula. Apply it for all the numbers up to 10000 (with n = 10000), then subtract the sum for all the numbers up to 100 (with n = 100) since they are not included, and then divide the result by three. You get (10001*5000 - 101*50) / 3 = 16666650, which happens to be exactly correct. However, in general, you may be slightly off. For example, suppose you wanted the sum of numbers from 1 to 11 that were divisible by 3. The total sum from 1 to 11 is 12*5.5 = 66, and 66/3 = 22, while the sum requested is 3 + 6 + 9 = 18.

Answer 2

This problem is not as difficult as it first seems. It's nothing more than the sum of an arithmetic progression.

The first number between 100 and 10000 which is divisible by 3 is 102. So that is the first term in our progression. The last number in this range is 9999. We can then use the following formula to find the number of terms in the progression:

A = a + (n-1) d, where A is the last term, a the first term, n the number of terms and d the difference, which in this case is 3.

9999 = 102 + (n-1) 3
9897 = (n-1) 3
(9897)/3 = (n-1)
3299 = (n-1)
3300 = n

The formula for the sum of an arithmetic progression is:

S = n[a + (n-1) d/ 2].

We have all the information we need to find the sum now, so all we have to do is plug them directly into the formula.

S = 3300[102 + (3299) 3/2]
= 336600 + 16330050
= 16,666,650.

A rather large sum indeed.

On the other hand, if one wanted to be facetious, one could say that the sum of the numbers divisible by 3 which lie between 100 and 10000 is zero because, as your question is literally posed, there are no numbers.

Answer 3

Arithmetic Formula

an = a1 + (n - 1)d
an = 102 + 3(n - 1)

9999 = 102 + 3(n - 1)
9897 = 3(n - 1)
n - 1 = 3299
n = 3300

Using that, we can now use the Arithmetic Summation Formula

Sn = (n/2)(a1 + an)
S(3300) = (3300/2)(102 + 9999)
S(333000) = 1650(10101)
S(333000) = 16666650

So the sum of the numbers between 100 and 10000 that are divisible by 3 is 16666650

Answer 4

the 1st term divisible by 3 and above 100 is 102
the last term is 9999

as this is an AP (arithmetic progression) 102,105,108....9999
the common difference is 3
Hence the sum can be calc. by
Sum=(n/2)(2a+(n-1)d) OR sum=(n/2)(a+l)
where
n is the number of terms, a the first term and d is the common difference l=last term

To find the number of terms
Last term=a+(n-1)*d

On solving n=3300

Substituting in the sum eqn.

The answer is
16666650

Answer 5

the sum of number divisible will start from the term 102 which is divisible by 3
and the last term would be 9999.

We can also determine the number of terms as following

9999- 102 / 3 = 3300 terms

Since this is an arithmetic series we could write

SN = number of terms *(first term + last term)/ 2

SN is the sum

so it will be as following

SN = 3300 * (102+9999)/2

SN = 3300 * 5050.5

which is

16,666,650

Answer 6

these numbers are 102, 105, 108...., 9990, 9993, 9996, 9999 or
3*(34, 35, 36...3330, 3331, 3332, 3333)

were this a sum from 1 to 3333 the answer would be
3333*3334/2=5,556,111
if we subtract the sum from to 33 we get 33*34/2=561
subtracting this from the answer above we get
5556111-561=5,555,550

multiplying by 3 (which we factored out earlier)
5555550*3=16,666,650

Answer 7

I used DavidK93's approach and worded it differently:

102 is 3*34; 9999 is 3*3333.
Therefore, the sum you seek is 3 times the sum of all the integers from 34 to 3333.
Average is1683.5; the number of integers is 3300.

Sum of all integers 34 to 3333 = 3300*1683.5 =5555550
Multiply by 3 to get 16,666,650.

Answer 8

These integers are 102,105,...,9999.
All are of the form 3k, k=34,35,...,3333.

Use the formula for the sum of the first n positive integers:
1+2+...+n = n(n+1)/2.

So the sum is 3(34+35+...+3333) =
3[(1+2+...+3333)-(1+2+...+33)]=
3[3333(3333+1)/2-33(33+1)/2] =
3/2*[11112222+1122] = 16670016

Answer 9

100

Answer 10

165,150 (this is for 100 to 1000) oops...

for 100 to 10,000 it's

16,666,650