Question

find the sum of all 3 digit numbers devisible by 3 and 5

Answer 1

Answer:

Step-by-step explanation:

The greatest 3 digit number divisible by 3 :999

The smallest 3 digit number divisible by 3

:102

Now , first term= 102

Last term =999

d=3 ( as we need to find numbers divisible by3)

Now,

999=102+(n-1)*3

999-102=3*(n-1)

897/3=n-1

299+1=n

300=n

Now, greatest 3 digit number divisible by 5: 995

Smallest 3 digit number divisible by 5:

100

First term = 100

Last term=995

d=5( as we need to find numbers divisible by 5)

995=100+(n-1)*5

895=(n-1)*5

895/5=n-1

179+1=n

180=n

Also factors of 3*5 will be common

Then smallest 3 digit number divisible by 15:105

Largest 3 digit no is 990

990=105+(n-1)*15

885=(n-1)*15

885/15=(n-1)

59+1=n

60=n

Now, 180+300-60 is and

420 is ans