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Question

(i) How many two-digit numbers are divisible by 3?
(ii) Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5
(iii) How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?

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Solution

(i) How many two-digit numbers are divisible by 3?

The two digits numbers divisible by 3 are, 12,15,18,21,24,.,99
The above numbers are A.P.
So, first number a=12
Common difference d=1512=3
Then, last number is 99
We know that, Tn (last number) =a+(n1)d
99=12+(n1)399=12+3n399=9+3n999=3n3n=90n=90/3n=30
Therefore, 30 two digits number are divisible by 3

(ii) Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5
The natural numbers which are divisible by both 2 and 5 are 110,120,130,140,,999
The above numbers are A.P.
So, first number a=110
Common difference d=120110=10
Then, the last number is 999
We know that, Tn (last number) =a+(n1)d
999=110+(n1)10999=110+10n10999=100+10n999100=10n10n=888n=888/10n=88
The number of natural numbers which are divisible by both 2 and 5 are 88

(iii) How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?

The numbers which are lie between 10 and 300 , when divisible by 4 leave a remainder 3 are 11,15,19,23,27,.299
The above numbers are A.P.
So, first number a=11 Common difference d=1511=4
Then, the last number is 299
We know that, Tn (last number) =a+(n1)d
299=11+(n1)4299=11+4n4299=7+4n2997=4n4n=292n=292/4n=73
The total which are lie between 10 and 300, when divisible by 4 leaves a remainder 3 are 73

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