Question

find the number of natural numbers between 101 and 999 which are divisible by 2 and 5

Answer 1

89 natural number

explanation:-
no. divisible by 2&5 must be divisible by 10
Form an AP of natural number present between 101 & 999

AP = 110, 120, 130,............, 990
from the AP we concluded
a = 110
d = 10
nth term of the AP = 990
a+(n-1)d=990
110+(n-1)10=990
(n-1)10=990-110
(n-1)= 880/10
n-1=88
n=88+1
n=89
natural number present between 101 & 109 = 89

Answer 2

the answer is in the attachment please see the attachment

I hope it will help you ❤