Question

find the number of three digit odd number divisible by 3​

Answer 1

First 3 digit number divisible by 3 = a = 102

Last 3 digit number divisible by 3 = an = 999

Common difference = d =3

n = no. of 3 digit numbers divisible by 3

n/2 = no. of 3 digit odd nos. divisible by 3 (as odd nos.(n/2) + even nos.(n/2) = total nos.(n) )

an = a + (n-1)d

999 = 102 + (n-1)3

897/3 = n-1

n = 299+1 = 300

Thus, n/2 = 300/2 = 150

Answer 2

There are 300 3-digit odd numbers that are divisible by 3.

Given:

3-digit odd numbers that are divisible by 3.

To Find:

The number of 3-digit odd numbers divisible by 3​.

Solution:

To solve this problem, let us look at the divisibility test of 3.

The divisibility test of 3 states that a number is divisible by 3 only if the sum of its digits is divisible by 3.

For example, the number 123 is divisible by 3 as the sum of its digits = 1+2+3 = 6 is divisible by 3.

The first 3-digit number that is divisible by 3 is 102. The last 3-digit number that is divisible by 3 is 999. Since all the numbers occur after an interval of 3.

We know that a sequence is said to be in arithmetic progression if the difference between every two consecutive terms is constant. The sequence 102, 105,....999 is in arithmetic progression.

The first term, a = 102.

The common difference, d = 3.

The $n^{th}$ term of an AP is $a_{n}$ = a + (n-1)d.

We need to find the number of terms 'n' here.

Since 999 is a term of the sequence, we can equate it with $a_{n}$.

⇒ 999 = a + (n-1)d.

⇒ 999 = 102 + (n-1)3 = 99 + 3n

⇒ 900 = 3n.

⇒ n = 300.

This means that there are 300 terms in the given sequence.

∴ There are 300 3-digit odd numbers that are divisible by 3.

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