Question

Find the sum of 3 digit numbers which are divisible by 5

Answer 1

According the question we have to take some 3 digit numbers.

Ok,  lets take some  3 digits numbers,

here they are,

1) 555

2)785

3) 677

4) 889

5)997

6)221

Now we have to find the sum of each number, which will be divisible by 5.

555 = 5 +5 +5 = 15 ÷ 5 = 3

785= 7 + 8 + 5= 20 ÷ 5 = 4

677 = 6 + 7+ 7 = 20 ÷5 = 4

889 = 8 + 8 +9 = 25 ÷5 = 5

997 = 9 + 9 + 7 = 25 ÷5 = 5

221 = 2 + 2 +1 = 5 ÷ 5 =1

Hence here are the sum of 3 digit numbers which are divisible by 5.

Ans :- Here ,the sum of 3 digit  numbers,555,785,667,889,997,221, numbers which are divisible by 5 .

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Answer 2

Answer:

The Correct answer would be 98550. The sum of 3 digit numbers which are divisible by 5 is 98550.

Step-by-step explanation:

Given: Numbers are of three digits like 100, 101, 102, etc.

           Divisible by 5, like 100, 105, 110, etc.

To find:The sum of all the three digit numbers divisible by 5.

Now, for finding the sum of all the 3-digit numbers divisible by 5, we have to first know the exact number of three digits that are divisible by 5. For the same, we'll use the Arithmetic Progression.

We know that first 3 digit number and very last 3 digit number that are both divisible by 5 are 100 and 995.

Hence, we know from the formula of nth term of an A.P:

                aⁿ = a + (n-1)d

where, aⁿ is the nth term of A.P, n is the number of terms, a is the first term and d is the common difference.

Taking, aⁿ = 995, a = 100 and d = 5, we get:

995 = 100 + (n-1)5

179 = n-1

n = 180

Therefore, total number of 3-digit terms divisible by 5 is 180.

Now, for the sum, we'll apply the sum formula for an A.P.

    sⁿ = $\frac{n}{2} [2a +(n-1)d]$       or            sⁿ = $\frac{n}{2} (a+l)$ (where l is the last term)

We'll go with the second one as it is more convenient.

Hence, $s^{180}$ = $\frac{180}{2} (100+995)$

                   = 90 × 1095

                  = 98550

 

Hence, the sum would be 98550.