Question

find the sum of all the 3 digit natural number which are divisible by 13.

Answer 1

Heya mate, Here is ur answer

a(first term) = 104

d(common difference)=13 (divisiblity by 13)

an(last term) =988

a+(n-1)d =988

104+ (n-1)13=988

(n-1)13=988-104

(n-1)13=884

n-1=884/13

n-1= 68

n=68+1

n=69.

Sn = n/2 (a+an)

$sn = \frac{69}{2} \times (104 + 998)$


$sn = \frac{69}{2} \times 1102$

$sn = 69 \times 551$



$sn = 38019$



$<b><u>So, sum of 3 digits number divisible by 13 =38019 . </b></u>$

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Answer 2

1000/13 = 76 and change, so there are 76 such numbers less than 1000 
100/13 = 7 and change, so there are 7 such number less than 100 
So there are 69 3 digit number of this form, the first of which is 13*8 = 104. 
They all have form a = 13n where n is a member of the ordinals 8,9,10...76 
If we add up pairs of number, the first and the last, the second smallest and second largest etc, then there are (76-8+1)/2 = 69/2 such pairs, each of which has value (76+8)*13. 
So the total sum is (69/2) * 1092 = 37674