Question

Sum of integers from 1 to 100 which are divisible by 2 or 5 are

Answer 1

Step-by-step explanation:

5 - twenty numbers are divisible those are 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100

2- fifty numbers are divisible those are 2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,90,92,94,96,98,100

Answer 2

Answer:

Sum of integers from 1 to 100 which are divisible by 2 or 5 = 3050

Step-by-step explanation:

The formula for finding =

$S = \frac{n}{2} ( a + l )$

Where

• n = number of terms
• a = first term
• l = last term

The sum of the integers divisible by 2 is -

$S_{2}$ = 2 + 4 + 6 + ........... 100

Number of terms ( n ) here is 50  because as we know from the numbers

1-100 ,  there are 50 odd nos and 50 even nos. and only even nos are divisible by 2 .

Here first term = 2

Last term = 100

∴ $S_{2}$ = $\frac{50}{2} ( 2 + 100 )$ = 2550

⇒ The sum of the integers divisible by 2 is = 2550

The sum of the integers divisible by 5 is -

$S_{5}$ = 5 + 10 +15 +......... 100

Number of terms here is 20 .. ( counted )

Here first term = 5

Last term = 100

$S_{5} = \frac{20}{2} ( 5 + 100 )$ = 1050

⇒The sum of the integers divisible by 5 = 1050

Now ,

• The sum of  $S_{2}$ +  $S_{5}$ , 2550 + 1050 = 3600 is not the answer
• Because they have common terms in them also.
• now we will have to look for the smallest number that is divisible both by 2 as well as 5
• The common terms in the terms 2 and 5 both will surely be divisible by 10
• Because 10 is divisible by 2 and 5 both

So, we have to find $S_{10}$ and subtract it from the sum of the sum of  $S_{2}$ +  $S_{5}$ to get our answer.

$S_{10}$ = 10 + 20 + 30 +.......100

Number of terms = 10 ( counted )

First term = 10

Last term = 100

$S_{10} = \frac{10}{2} ( 10 + 100 )$ = 550

∴ The answer is

$S_{2}$ +  $S_{5}$ - $S_{10}$ = ( 2550 + 1050 ) - 550

                     =  3600 - 550

                      = 3050

hope it helps ..