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How many natural numbers are in between 120 and 519 which are exactly divisible by both 3 and 7 ?
- a)19
- b)17
- c)15
- d)21
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
How many natural numbers are in between 120 and 519 which are exactly...
Solution:
Step 1: Find the first natural number which is divisible by 3 and 7.
The LCM of 3 and 7 is 21. Therefore, the first natural number which is divisible by both 3 and 7 is 21.
Step 2: Find the last natural number which is divisible by 3 and 7.
The last natural number which is less than or equal to 519 and divisible by both 3 and 7 is 504. This can be obtained by finding the largest multiple of 21 which is less than or equal to 519. 21 × 24 = 504.
Step 3: Find the number of natural numbers between 21 and 504 which are divisible by 3 and 7.
The required numbers are 21, 42, 63, 84, …, 504.
To find the number of terms in this sequence, we use the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
where a1 is the first term, d is the common difference, and an is the nth term.
In this case, a1 = 21, d = 21, and an = 504. We need to find n.
504 = 21 + (n - 1)21
Simplifying this equation, we get:
(n - 1)21 = 483
n - 1 = 23
n = 24
Therefore, there are 24 terms in the sequence.
Step 4: Find the number of terms between 120 and 519.
The first term of the sequence which is greater than or equal to 120 is 126. The last term of the sequence which is less than or equal to 519 is 504. Therefore, we need to find the number of terms between 126 and 504.
To do this, we subtract the number of terms before 126 from the total number of terms in the sequence.
The first term of the sequence is 21, which is less than 126. Therefore, we need to find the number of terms in the sequence from 21 to 126.
To find this, we use the formula for the sum of the first n terms of an arithmetic sequence:
Sn = n/2(2a1 + (n - 1)d)
where Sn is the sum of the first n terms, a1 is the first term, d is the common difference, and n is the number of terms.
In this case, a1 = 21, d = 21, and the last term of the sequence less than or equal to 126 is 105. Therefore, we need to find the sum of the first 5 terms of the sequence:
S5 = 5/2(2 × 21 + (5 - 1)21)
S5 = 5/2(42 + 84)
S5 = 315
Therefore, there are 5 terms in the sequence from 21 to 126.
Now, we can find the number of terms between 126 and 504:
Number of terms = Total number of terms - Number of terms before 126 - Number of terms after 504
Number of terms = 24 - 5 - 0
Number of terms = 19
Therefore, there are 19 natural numbers between 120 and 519 which are
Step 1: Find the first natural number which is divisible by 3 and 7.
The LCM of 3 and 7 is 21. Therefore, the first natural number which is divisible by both 3 and 7 is 21.
Step 2: Find the last natural number which is divisible by 3 and 7.
The last natural number which is less than or equal to 519 and divisible by both 3 and 7 is 504. This can be obtained by finding the largest multiple of 21 which is less than or equal to 519. 21 × 24 = 504.
Step 3: Find the number of natural numbers between 21 and 504 which are divisible by 3 and 7.
The required numbers are 21, 42, 63, 84, …, 504.
To find the number of terms in this sequence, we use the formula for the nth term of an arithmetic sequence:
an = a1 + (n - 1)d
where a1 is the first term, d is the common difference, and an is the nth term.
In this case, a1 = 21, d = 21, and an = 504. We need to find n.
504 = 21 + (n - 1)21
Simplifying this equation, we get:
(n - 1)21 = 483
n - 1 = 23
n = 24
Therefore, there are 24 terms in the sequence.
Step 4: Find the number of terms between 120 and 519.
The first term of the sequence which is greater than or equal to 120 is 126. The last term of the sequence which is less than or equal to 519 is 504. Therefore, we need to find the number of terms between 126 and 504.
To do this, we subtract the number of terms before 126 from the total number of terms in the sequence.
The first term of the sequence is 21, which is less than 126. Therefore, we need to find the number of terms in the sequence from 21 to 126.
To find this, we use the formula for the sum of the first n terms of an arithmetic sequence:
Sn = n/2(2a1 + (n - 1)d)
where Sn is the sum of the first n terms, a1 is the first term, d is the common difference, and n is the number of terms.
In this case, a1 = 21, d = 21, and the last term of the sequence less than or equal to 126 is 105. Therefore, we need to find the sum of the first 5 terms of the sequence:
S5 = 5/2(2 × 21 + (5 - 1)21)
S5 = 5/2(42 + 84)
S5 = 315
Therefore, there are 5 terms in the sequence from 21 to 126.
Now, we can find the number of terms between 126 and 504:
Number of terms = Total number of terms - Number of terms before 126 - Number of terms after 504
Number of terms = 24 - 5 - 0
Number of terms = 19
Therefore, there are 19 natural numbers between 120 and 519 which are
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Community Answer
How many natural numbers are in between 120 and 519 which are exactly...
Given
The numbers are between 120 and 519
they are exactly divisible by both 3 and 7
Concept used:
The number divisible by both 3 and 7 is divisible by 21
(LCM of 3 and 7)
Calculation:
The total numbers divisible by 21 in between 1 and 120 is
⇒ 120 / 21 = 5
The total numbers divisible by 21 in between 1 and 519 is
⇒ 519 / 21 = 24
So total number that is divisible by 21 between 120 and 519 is
(24 - 5) = 19
So, the total number is 19
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