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Question

Question: How many numbers of two digits are divisible by \(3\) ?...

How many numbers of two digits are divisible by 33 ?

Explanation

Solution

Hint: Select the lowest and highest two digit terms divisible by 33. To see whether the series is an A.P or not. Then if it is in A.P solve it by taking the nth{{n}^{th}} term of the A.P. Find the value of nn. You will get the answer.

We use ten digits in the way we count. Sometimes we use one digit to represent a number and sometimes we use more. In this lesson, you'll learn how many digits are in any given numeral, and recognize a digit when you see it. You will also begin learning about place value.
This is the numeral: 5151.These are the digits in that numeral: 55 and 11. The numeral has two digits because two symbols, or digits, make up the numeral 5151. What about this numeral ? 452452. This numeral has three digits: 4,54,5 and 22.
That 77 is a special kind of symbol. It's called a digit. We have ten digits we use to make up all numerals. A numeral is a number written down. These digits are 0,1,2,3,4,5,6,7,80,1,2,3,4,5,6,7,8 and 99. That's it! You can make any numeral you want out of those ten digits.
Just like a red light means 'stop', a 77 means 'seven'. It's a single symbol that represents a numeral. Yes, just like this is a symbol to stop.
We know, first two digit number divisible by 33 is 1212 and the last two digit number divisible by 33 is 9999. Thus, we get 12,15,......,9912,15,......,99.
So the lowest two digit number divisible by 33 is 1212.
Highest two digit number divisible by 33 is 9999.
So we can see the difference between the numbers that are divisible by 3, is 33.
So the above series is in A.P.
We have to find it in terms of nn.
And herea=12,d=3,an=99a=12,d=3,{{a}_{n}}=99
Thus, the nth{{n}^{th}}term of A.P is :
an=a+(n1)d{{a}_{n}}=a+(n-1)d
Where,
a=a=First-term
d=d= Common difference
n=n= number of terms
an=nth{{a}_{n}}={{n}^{th}}term
So now applying the formula for 99, we get,
99=12+(n1)399=12+(n-1)3
Simplifying further we get,
9912=(n1)3 87=3n3 90=3n n=30 \begin{aligned} & 99-12=(n-1)3 \\\ & 87=3n-3 \\\ & 90=3n \\\ & n=30 \\\ \end{aligned}
So we get n=30n=30.
Therefore, the number of two digits divisible by 33 are 3030.

Note: Read the question properly. Also, we should know the lowest and highest two digit terms divisible by 33. So the concepts related to A.P should be clear. Here we have used the concept of A.P that is we have used nth{{n}^{th}} term of A.P which is an=a+(n1)d{{a}_{n}}=a+(n-1)d.