Question

42. Find the number of natural numbers between 101 and 999 which are divisible by both 2.
and 5.​

Answer 1

Answer:

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

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Answer 2

$\huge\star\frak{{Answer:-}}$

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