How many even integers n, where \[100\le n\le 200\], are divisible neither by seven nor by nine?
(A) 40
(B) 37
(C) 39
(D) 38
Answer
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Hint: Even numbers are divisible by 2. We know the formula, the count of numbers that are divisible by \[n\] between the given numbers, \[\dfrac{Highest\,number\,divisible\,by\,n-Lowest\,number\,divisible\,by\,n}{n}+1\] . Use this formula and calculate the count of numbers that are divisible by 2 between 100 and 200. The even numbers that are divisible by 7 must be a multiple of 14. Use this formula and calculate the count of even numbers that are divisible by 7. Similarly, calculate the number of even numbers that are divisible by 9. 126 is the even number that is divisible by both 7 and 9. Now, solve it further and calculate the number of even numbers between 100 and 200 that are neither divisible by 7 nor 9.
Complete step-by-step solution
According to the question, we are given even numbers between 100 and 200. We are asked to find the number of even integers that are neither divisible by 7 nor by 9.
First of all, let us find the number of even numbers between 100 and 200.
We know that even numbers are those numbers that are divisible by 2. So, to find the number of even numbers between 100 and 200, we have to figure out the numbers that are divisible by 2.
We know the formula to get the count of numbers that are divisible by \[n\] between the given numbers, \[\dfrac{Highest\,number\,divisible\,by\,n-Lowest\,number\,divisible\,by\,n}{n}+1\] ………………………………….(1)
Between 100 and 200, we have
The highest number divisible by 2= 200 …………………………………(2)
The lowest number divisible by 2= 100 ………………………………………..(3)
Now, from equation (1), equation (2), and equation (3), we get
The total count of numbers that are divisible by 2 = \[\dfrac{200-100}{2}+1=50+1\] = 51 …………………………………..(4)
The numbers that are divisible by 7 are 7, 14, 21, 28, ………………..
Here we can observe that the even numbers divisible by 7 are 14, 28, ………………..
So, we can say that the even numbers that are divisible by 7 must be a multiple of 14.
So, the count of even numbers between 100 and 200 that are divisible by 7 must be equal to the count of numbers that are divisible by 14 …………………………………..(5)
Between 100 and 200, we have
The highest number divisible by 14 = 196 …………………………………(6)
The lowest number divisible by 14 = 112 ………………………………………..(7)
Now, from equation (1), equation (6), and equation (7), we get
The total count of numbers that are divisible by 14 = \[\dfrac{196-112}{14}+1=6+1\] = 7.
So, the count of even numbers that are divisible by 7 between 100 and 200 is 7 …………………………………..(8)
The numbers that are divisible by 9 are 9, 18, 27, 36, ………………..
Here we can observe that the even numbers divisible by 9 are 18, 36, ………………..
So, we can say that the even numbers that are divisible by 9 must be a multiple of 18.
So, the count of even numbers between 100 and 200 that are divisible by 9 must be equal to the count of numbers that are divisible by 18 …………………………………..(9)
Between 100 and 200, we have
The highest number divisible by 18 = 198 …………………………………(10)
The lowest number divisible by 14 = 108 ………………………………………..(11)
Now, from equation (1), equation (10), and equation (11), we get
The total count of numbers that are divisible by 14 = \[\dfrac{198-108}{18}+1=5+1\] = 6.
So, the count of even numbers that are divisible by 9 between 100 and 200 is 6 …………………………………..(12)
From equation (4), equation (8), and equation (12), we get
The count of even numbers that are neither divisible by 7 or 9 = \[51-7-6=38\] …………………………………(13)
Also, we have one number that is 126 as our even number between 100 and 200 which is divisible by both 7 and 9.
But, in equation (13), the count of 126 has been subtracted twice. So, we have to add one count in equation (13).
The count of even numbers that are neither divisible by 7 or 9 = \[38+1=39\].
Therefore, the count of even numbers that are neither divisible by 7 or 9 is 39.
Hence, the correct option is (C).
Note: For this question, we have to keep one formula into our consideration i.e., the count of numbers that are divisible by \[n\] between the given numbers, \[\dfrac{Highest\,number\,divisible\,by\,n-Lowest\,number\,divisible\,by\,n}{n}+1\] . This formula will make the solution easy to solve and also reduce calculation errors.
Complete step-by-step solution
According to the question, we are given even numbers between 100 and 200. We are asked to find the number of even integers that are neither divisible by 7 nor by 9.
First of all, let us find the number of even numbers between 100 and 200.
We know that even numbers are those numbers that are divisible by 2. So, to find the number of even numbers between 100 and 200, we have to figure out the numbers that are divisible by 2.
We know the formula to get the count of numbers that are divisible by \[n\] between the given numbers, \[\dfrac{Highest\,number\,divisible\,by\,n-Lowest\,number\,divisible\,by\,n}{n}+1\] ………………………………….(1)
Between 100 and 200, we have
The highest number divisible by 2= 200 …………………………………(2)
The lowest number divisible by 2= 100 ………………………………………..(3)
Now, from equation (1), equation (2), and equation (3), we get
The total count of numbers that are divisible by 2 = \[\dfrac{200-100}{2}+1=50+1\] = 51 …………………………………..(4)
The numbers that are divisible by 7 are 7, 14, 21, 28, ………………..
Here we can observe that the even numbers divisible by 7 are 14, 28, ………………..
So, we can say that the even numbers that are divisible by 7 must be a multiple of 14.
So, the count of even numbers between 100 and 200 that are divisible by 7 must be equal to the count of numbers that are divisible by 14 …………………………………..(5)
Between 100 and 200, we have
The highest number divisible by 14 = 196 …………………………………(6)
The lowest number divisible by 14 = 112 ………………………………………..(7)
Now, from equation (1), equation (6), and equation (7), we get
The total count of numbers that are divisible by 14 = \[\dfrac{196-112}{14}+1=6+1\] = 7.
So, the count of even numbers that are divisible by 7 between 100 and 200 is 7 …………………………………..(8)
The numbers that are divisible by 9 are 9, 18, 27, 36, ………………..
Here we can observe that the even numbers divisible by 9 are 18, 36, ………………..
So, we can say that the even numbers that are divisible by 9 must be a multiple of 18.
So, the count of even numbers between 100 and 200 that are divisible by 9 must be equal to the count of numbers that are divisible by 18 …………………………………..(9)
Between 100 and 200, we have
The highest number divisible by 18 = 198 …………………………………(10)
The lowest number divisible by 14 = 108 ………………………………………..(11)
Now, from equation (1), equation (10), and equation (11), we get
The total count of numbers that are divisible by 14 = \[\dfrac{198-108}{18}+1=5+1\] = 6.
So, the count of even numbers that are divisible by 9 between 100 and 200 is 6 …………………………………..(12)
From equation (4), equation (8), and equation (12), we get
The count of even numbers that are neither divisible by 7 or 9 = \[51-7-6=38\] …………………………………(13)
Also, we have one number that is 126 as our even number between 100 and 200 which is divisible by both 7 and 9.
But, in equation (13), the count of 126 has been subtracted twice. So, we have to add one count in equation (13).
The count of even numbers that are neither divisible by 7 or 9 = \[38+1=39\].
Therefore, the count of even numbers that are neither divisible by 7 or 9 is 39.
Hence, the correct option is (C).
Note: For this question, we have to keep one formula into our consideration i.e., the count of numbers that are divisible by \[n\] between the given numbers, \[\dfrac{Highest\,number\,divisible\,by\,n-Lowest\,number\,divisible\,by\,n}{n}+1\] . This formula will make the solution easy to solve and also reduce calculation errors.
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