made easy

Question

In a set of 500 first odd natural numbers. The total number divisible by 3 or 5 are

Answer 1

let A = set..div by 3

B : set..div by 5

A^B : div by both 3 and 5

A = 3, 9, 15.....495, sum1 = n/2(first+last)(WE CAN'T TAKE 6,12...THESE ARE EVEN)

B= 5, 15, 25.....495 calculate sum2

A^B = 15,45,..... 495calculate sum3

use AUB (div by 3 or 5 )= sum1 + sum2 - sum3

Comments

how you will calculate sum1?

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AP formula
for n
Tn = a + (n-1)d...a and Tn(last term) u know so calculate n then use below formula
then sum =n/2(a+Tn)...n

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@cse23 What is the " sum " to do with the problem stated ?

sum1 => Gives total sum of numbers  ( 1+3+5+7+9)

We need total numbers between 1 and 300  ( 1 , 3, 5, 7 , 9   )= 5

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Not between 1 and 300 but 500. Okay.