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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?


Solution

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

The two digits numbers divisible by 3 are, $$12,15,18,21,24, \ldots ., 99$$ 
The above numbers are A.P.
So, first number $$a=12$$ 
Common difference $$d=15-12=3$$
Then, last number is 99 
We know that, $$T_{n}$$ (last number) $$=a+(n-1) d$$
$$\begin{array}{l}99=12+(n-1) 3 \\99=12+3 n-3 \\99=9+3 n \\99-9=3 n \\3 n=90 \\n=90 / 3 \\n=30\end{array}\\$$
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 $$110,120,130,140, \ldots, 999$$ 
The above numbers are A.P.
So, first number $$a=110$$ 
Common difference $$d=120-110=10$$
Then, the last number is 999 
We know that, $$T_{n}$$ (last number) $$=a+(n-1) d$$
$$\begin{array}{l}999=110+(n-1) 10 \\999=110+10 n-10 \\999=100+10 n \\999-100=10 n \\10 n=888 \\n=888 / 10 \\n=88\end{array}\\$$
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 $$11,15,19,23,27, \ldots .299$$
The above numbers are A.P.
So, first number $$a=11$$ Common difference $$d=15-11=4$$ 
Then, the last number is 299 
We know that, $$T_{n}$$ (last number) $$=a+(n-1) d$$
$$\begin{array}{l}299=11+(n-1) 4 \\299=11+4 n-4 \\299=7+4 n \\299-7=4 n \\4 n=292 \\n=292 / 4 \\n=73\end{array}\\$$
The total which are lie between 10 and 300, when divisible by 4 leaves a remainder 3 are 73

Mathematics

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