First find the sum of all numbers which are divisible by 3, and are less than 61. i.e. there are $\left \lfloor \frac{61}{3} \right \rfloor = 20$ numbers divisible by 3 and are less than 61.

So, their sum

$S_1 = 3 + 6 + 9 + ...... + (20$ terms $) = 3(1 + 2 + 3 + 4 + .... + 20) = 3 \times \frac{20 \times 21}{2} = 630$

Now, the sum of all numbers which are divisible by 5, and are less than 61. i.e. there are $\left \lfloor \frac{61}{5} \right \rfloor = 12$ numbers divisible by 5 and are less than 61.

So, their sum

$S_2 = 5 + 10 + 15 + ...... + (12$ terms $) = 5(1 + 2 + 3 + 4 + .... + 12) = 5 \times \frac{12 \times 13}{2} = 390$

But, there are some numbers which are divisible by both $3$ and $5$ and have been included twice. So, we have to subtract them once. The common multiples can be obtained by taking $LCM(3, 5) = 15$ . Hence, there are $\left \lfloor \frac{61}{15} \right \rfloor = 4$ numbers which are divisible by both 3 and 5 and are less than 61.

Their sum

$S_3 = 15 + 30 + ...... + (4$ terms $) = 15(1 + 2 + 3 + 4) = 150$

So, the final answer is

$S = S_1 + S_2 - S_3 = 630 + 390 - 150 = \boxed{870}$

That's the answer!

The numbers divisible by 3 in the given range are $3,6,9......60$ . The sum of these numbers is

$\displaystyle 3+6+9+.....+60=3(1+2+3+....+20)=3\times \frac{20\times 21}{2}=630$

The numbers divisible by 5 in the given range are $5,10,15....60$ . The sum of these numbers is

$\displaystyle 5+10+15+.......+60=5(1+2+......+12)=5\times \frac{12\times 13}{2}=390$

The sum of integers either divisible by 3 or 5 is not simply 390+630. If we do this, we double-count those numbers divisible by both 3 and 5. In other word, we need to subtract those numbers which are divisible by 15. The numbers divisible by 15 are $15,30,45$ and $60$ . The sum of these numbers is

$\displaystyle 15+30+45+60=150$

Hence, our final answer is $630+390-150=\fbox{870}$ .

There are 20 multiples of 3 in the given interval, so to find that sum we simply add the positive integers from 1 to 20 and multiply the result by 3; this yields $N_3=630$ .

There are 12 multiples of 5 in the given range, so we can find the sum in a similar manner: $N_5=390$ .

From these we can conclude that $N_3+N_5=1020$ . However, we counted multiples of 15 in both $N_3$ and $N_5$ ; hence we will subtract the multiples of 15 ( $N_{15}=150$ )from our previously obtained sum.

Therefore, the required sum is $N=N_3+N_5-N_{15}=870$ .

The number divisible by $3$ for value $N$ are $3, 6, 9, 12 .... 60$ which is an A.P. series.

Then the 1st number of the series $a_1 = 3$ ; the common difference $d_1 = 3$ ; the total number of the sequence $n_1 =\frac{60}{3} = 20$ [60 is the last number, because 61 can not divisible by 3]

So, the sum of all sequence if the A.P. $s_1 = \frac{n_1}{2} [2a_1 + d_1(n_1-1)] \implies \frac{20}{2} [2 \times 3 + 3(20-1)] \implies10 \times (6 + 57) \implies 630$

Using the same method : The number divisible by $5$ for value $N$ are $5, 10, 15, 20 .... 60$ which is also an A.P. series.

Then the 1st number of the series $a_2 = 5$ ; the common difference $d_2 = 5$ ; the total number of the sequence $n_2 =\frac{60}{5} = 12$ [60 is the last number, because 61 is not divisible by 5]

So, the sum of all sequence if the A.P. $s_2 = \frac{n_2}{2} [2a_2 + d_2(n_2 -1)] \implies \frac{12}{2} [2 \times 5 + 5(12-1)] \implies 6 \times (10 + 55) \implies 390$

And the number we counted twice in these two sequence are value of $N$ from $1 to 61$ which are multiple of $3 \times 5 or 15$ . And there is 4 multiples of 15 between 1 and 61 and there sum is $15 + 30 + 45 + 60 = 150$

We get the sum of all integers value of $N = 630 + 390 - 150 \implies \boxed{870}$

We need to find the sum of the sequences: $3, 6, 9 , 12, 15, 18, 21\ldots 57,60$ and $5, 10, 15, 20 \ldots 55, 60$ and subtract out the common terms. The sum of the first sequence is just $\dfrac{(60+3)20}{2} = 630$ , and the sum of the second is $\dfrac{(5+60)12}{2} = 390$ . Summing this gives $1020$ . Now, the common terms are multiples of 15, namely 15,30,45, and 60 within the given range. Subtracting this from 1020 gives the answer $\boxed{870}$ , done.

The sum of numbers who are divisible by $3$ in the given limits is: $3+6+9+12+.......+60$ which is an arithmetic progression with $a_1=3$ , $d=3$ and $n=20$ .Using its formula for sum we obtain $S=630$ .In this sum we have also calculated the numbers which are divisible by $3$ and $5$ at the same time and their sum is $S=15+30+45+60=150$ .The sum of numbers which are divisible by 5 is $S=5+10+15+........+60$ .This is also an arithmetic progression with $a_1=5$ , $d=5$ and $n=12$ .Using its formula for sum we obtain $S=390$ . In this sum we have also calculated those numbers who are divisible by $3$ and $5$ in the same time but we have counted them once and that's why we are going to substract $150$ from $390$ to obtain $240$ .Therefore the desired sum is $630+240=870$ .

integers that are divisible by 3 from 1 to 61 are an Arithmetic Progression starts from 3 and ends at 60 with an increment of 3 then their summation are S = n(a + L)/2 = 20*(3+60)/2=630

integers that are divisible by 5 from 1 to 61 are an Arithmetic Progression starts from 5 and ends at 60 with an increment of 5 then their summation are S = n(a + L)/2 = 12*(5+60)/2=390

integers that are divisible by 5 and 3 from 1 to 61 are an Arithmetic Progression starts from 5x3=15 and ends at 60 with an increment of 15 then their summation are S = n(a + L)/2 = 4*(15+60)/2=150

so the required summation is =630+390-150=870

Integers between 1 and 61 which are divisible by 3 and 5.

Integers divisible by 3 are 3,6,9,...60. (20 terms). Sum = 630

Integers divisible by 5 are 5,10,15,...60 (12 terms). Sum =390. And we can see that multiples 15 are counted twice in multiples of 3 & 5 both.

Therefore, multiples of 15 (Sum = 150) should be removed from the sum of multiples of 3 & 5.

Net sum = 630 + 390 - 150 = 870. (Ans)

Hello all,

as the range for N is [1,61],

we know that N is divisible by 3 or 5,the numbers that are divisible by both 3 & 5 are 15,30,45 and 60,

therefore,for N divisible by 3 are 3,6,9,12,........60,

60/3=20(20 terms up to 60),

S(20 terms)=20/2.[2(3)+(19)(3)]=630(amongst the sum included the sum of 15,30,45,60 together),

as for N that is divisible by 5,we know that the N are = 5,10,15,20,25,30,35,40,45,60(but we already added up for 15,30,45,60),sum of N/5=5+10+20+25+35+40+50+55=240

Therefore, N(sum of integers)=630 + 240 = 870....

thanks...

I did it using Bluej ....

class Divisors

{ public static void main(String args[]) {
int c=0; int i =0; for(i=1;i<=61;i++) { if(i%3 ==0 || i%5 ==0) { c=c+i;

    }
}
System.out.println(c);

} }

The numbers that are divisible by $3$ are multiples of three in the range $1 \leq N \leq 61$ , thus, $3,6,9, \dots 60$ There are $20$ such terms that we need to add.

The numbers that are divisible by $5$ are multiples of five in the range $1 \leq N \leq 61$ , thus, $5,10,15 \dots 60$ . There are $12$ such terms.

But, we overcount numbers that are divisible by both, and these are precisely the numbers divisible by $15$ , like $15,30, \dots 60$ . There are $4$ such terms.

Thus, we need to find the sum of these series. We observe that these are just arithmetic series. The sum of arithmetic series of the form $a_1 + a_2 + \dots + a_n$ is given by $\dfrac{n(a_1 + a_n)}{2}$

Thus, we have to find

$\dfrac{20(3 + 60)}{2} + \dfrac{12(5 + 60)}{2} - \dfrac{4(15 + 60)}{2}$ which is $\boxed{870}$

There are floor(61/3)=20, floor(61/5) = 12 and floor (61/(3 * 5))=4 numbers divisible by 3, 5, 15 respectively.

Now, sum of the numbers divisible by 3 is = (20/2)*(2 * 3 + (20-1) * 3) = 630

sum of the numbers divisible by 5 is = (12/2)*(2 * 5 + (12-1) * 5) = 390

sum of the numbers divisible by 15 is = (4/2)*(2 * 15 + (4-1) * 15) = 150.

So total sum of the numbers divisible by 3 or 5 = 630 + 390 - 150 = 870

Multiples of $3$ : $3, 6, 9, ... ,54, 57, 60$

Multiples of $5$ : $5, 10, 15, ... ,50, 55, 60$

Multiples of $15$ : $15, 30, 45, 60$

Hence, sum of all integers $N$ = $(3 + 6 + 9 + ... + 54 + 57 + 60) + (5 + 10 + 15 + ... + 50 + 55 + 60) - (15 + 30 + 45 + 60)$

$= 3(1 + 2 + 3 + ... + 18 + 19 + 20) + 5(1 + 2 + 3 + ... + 10 + 11 + 12) - 15(1 + 2 + 3 + 4)$

$= 3 \times \frac{20 \times 21}{2} + 5 \times \frac{12 \times 13}{2} - 15 \times \frac{4 \times 5}{2}$

$= 3 \times 210 + 5 \times 78 - 15 \times 10$

$= 630 + 390 - 150$

$= \boxed{870}$

1. Numbers which divisible by 3 are: 3, 6, 9, ... 60. Since $60=3\times20$ then there are 20 numbers incuded to the sequence. The sum of all those numbers are $3+6+9+...+60=3(1+2+3+...+20)=3\times\frac {20.21}{2}=630$
2. Numbers which divisible by 5 are: 5, 10, 15, ... 60. Since $60=5\times12$ then there are 12 numbers incuded to the sequence. Analogous to point (1), we yield $5\times\frac {12.13}{2}=390$
3. Adding the value got from point (1) and (2) gives us the sum of $1020$ . But note that in our calculation, we've doublecounted the sum of numbers which are divisible by 3 and 5, those are, 15, 30, 45, and 60 whose the sum is equal $150$
4. Therefore the sum of all integers $N$ is $1020-150=\boxed {870}$

include<iostream>

include<conio.h>

using namespace std; main(){ int b,hasil=0; int batas=61; for(int i=1;i<b;i++){ if((i%3==0)||(i%5==0)){ hasil +=i; }} cout<<hasil; getch(); }

3 ---> 3, 6, 9, 12, 15, ... , 60 = 3/2 . 20 . 21 = 630 ; 5-----> 5,10,20,25,35,40,50,55 = 240. Thus, 630 + 240 = 870

sum(3+6+12+......+60)+ sum(5+10+15+......60)-sum(15+30+45+60)

Divisible by 3 are: 3, 6, 9, ..., 60.

Divisible by 5: 5, 10, 15, ..., 60.

Divisible by 3 & 5: 15, 30, 45, 60.

By P.I.E., we have that [Sum of divisible by 3 or 5] = [sum of divisible by 3] + [sum of divisible by 5] - [sum of divisible by 3 & 5] = 3(1+2+...+20)+5(1+2+...+12)-15(1+2+3+4)=870

by using the formula 1+2+...+n=n(n+1)/2.

public static calculateSum() {

int sum=0; for(int i=1;i<=61;i++) { if(i%3==0||i%5==0) sum=sum+i; } System.out.println(sum);

}