How many three digit natural numbers are divisible by 3?
Last updated date: 24th Mar 2023
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Answer
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Hint: To find the number of three digit numbers divisible by 3, we need to form an arithmetic progression with common difference as 3. Then we will find the ${{n}^{th}}$ term of this series which will be the highest three digit number that could be a multiple of 3. Then, we use the formula-
Complete step-by-step answer:
${{a}_{n}}$= a + (n-1) d
${{a}_{n}}$= ${{n}^{th}}$ term of this series
a = first term of the series
n = number of terms in the series
d= common difference (3 in this case)
Thus, to proceed with the above problem (to find the appropriate arithmetic series), we first find the ${{n}^{th}}$ term of this series. The ${{n}^{th}}$ term of the series would be the highest three digit number which is a multiple of 3. To find this number, we divide by 1000 by 3. Doing so, we get 333 as the quotient and 1 as the remainder. Now, to get this number we subtract the remainder (that is 1) from 1000 to get the ${{n}^{th}}$ term of the series. We get, 1000 - 1 = 999.
The next step would be the first term of the series. This term would clearly be 102. This is because it is the smallest three digit number that is divisible by 3.
We have, the first term, a = 102 and ${{n}^{th}}$ term, ${{a}_{n}}$=999. For arithmetic progression, we have the formula-
${{a}_{n}}$= a + (n-1) d
Where, n is the required number of terms in the series and d is the common difference (which is 7 in this case)
999 = 102 + 3(n-1)
n-1=$\dfrac{999-102}{3}$
n-1=299
n=300
Hence, the required number of terms is 300.
Note: Another technique to arrive at the answer is to divide 1000 by 3. We get 333 as the quotient. This implies that the terms are 3, 6, 9, …, 999 (which are 333 terms). However, we have to exclude the terms that are single digit and double digit from this range of numbers. Thus, we now divide 100 by 3 (to remove lower digit numbers). We get 33 as the quotient. This implies that the terms are 3, 6, 9, …, 99 (which are 33 terms). We have to remove these terms since they are not three digit numbers. Thus, we do, 333-33=300. Thus, we get the number of terms as 300.
Complete step-by-step answer:
${{a}_{n}}$= a + (n-1) d
${{a}_{n}}$= ${{n}^{th}}$ term of this series
a = first term of the series
n = number of terms in the series
d= common difference (3 in this case)
Thus, to proceed with the above problem (to find the appropriate arithmetic series), we first find the ${{n}^{th}}$ term of this series. The ${{n}^{th}}$ term of the series would be the highest three digit number which is a multiple of 3. To find this number, we divide by 1000 by 3. Doing so, we get 333 as the quotient and 1 as the remainder. Now, to get this number we subtract the remainder (that is 1) from 1000 to get the ${{n}^{th}}$ term of the series. We get, 1000 - 1 = 999.
The next step would be the first term of the series. This term would clearly be 102. This is because it is the smallest three digit number that is divisible by 3.
We have, the first term, a = 102 and ${{n}^{th}}$ term, ${{a}_{n}}$=999. For arithmetic progression, we have the formula-
${{a}_{n}}$= a + (n-1) d
Where, n is the required number of terms in the series and d is the common difference (which is 7 in this case)
999 = 102 + 3(n-1)
n-1=$\dfrac{999-102}{3}$
n-1=299
n=300
Hence, the required number of terms is 300.
Note: Another technique to arrive at the answer is to divide 1000 by 3. We get 333 as the quotient. This implies that the terms are 3, 6, 9, …, 999 (which are 333 terms). However, we have to exclude the terms that are single digit and double digit from this range of numbers. Thus, we now divide 100 by 3 (to remove lower digit numbers). We get 33 as the quotient. This implies that the terms are 3, 6, 9, …, 99 (which are 33 terms). We have to remove these terms since they are not three digit numbers. Thus, we do, 333-33=300. Thus, we get the number of terms as 300.
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