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Hi,

This question no need to do any calculations, we can just solve this by elimination the options.

If you know the seven tables(multiples), then it is pretty straight forward, in a hundred there are 14 multiples (7 * 14 = 98) and since the question asking for even multiples it has to be less than 14
multiples (In-fact 7 even multiples because half even and half odd). So the only answer choice has to be 7.

Also, first three digit number divisible by 7 is 105(7*15) and the last number (less than 200) is 196(28*7).

If you are looking for a standard approach, then,

To find the number of elements(if they are evenly spaced), use the below expression,

(L – A )/D + 1 = to count the number of elements

L is the last element, A is the first element and D is the common difference.

So, (196 – 105)/ 7 + 1 = 14.

Since only even numbers , so divide the above count by 2. So the answer is E(7).
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Bunuel
How many even integers between 100 and 200, inclusive, are divisible by 7?

A. 20
B. 18
C. 16
D. 14
E. 7


An even integer divisible by 7 is a multiple of 14.
Neither endpoint of the given range -- 100, 200 -- is a multiple of 14.
For this reason, the multiples of 14 between 100 and 200, inclusive, can be counted simply by dividing 14 into the difference between the two endpoints:
\(\frac{(200-100)}{14} = \frac{100}{14} = \frac{50}{7} = 7 \frac{1}{7}\)
Since 14 divides into the difference 7 times, there are 7 multiples of 14 between 100 and 200, inclusive.

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Bunuel
How many even integers between 100 and 200, inclusive, are divisible by 7?

A. 20
B. 18
C. 16
D. 14
E. 7

If an even integer is divisible by 7, then it has to be divisible by 14 also. Therefore, we are looking for integers in the range of 100 to 200 that are divisible by 14. The number of such integers is:

(196 - 112)/14 + 1 = 7

(Note that finding the multiples of 14 rather than multiples of 7 eliminates finding the odd multiples of 7, which would have to be subtracted out, anyway.)

Answer: E
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Bunuel
How many even integers between 100 and 200, inclusive, are divisible by 7?

A. 20
B. 18
C. 16
D. 14
E. 7

Asked: How many even integers between 100 and 200, inclusive, are divisible by 7?

Even integers divisible by 7 are multiples of 14.

14*7 = 98
14*8 = 112
14*14 = 196

14*8,......14*14
Number of even integers divisible by 7 = 14-8 + 1 = 7

IMO E
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As we have to find number of even integers, let's find the number of even integers from 100 to 200, inclusive.

(200-100)/2 + 1 = 51

Out of these 51 numbers, every 7th number will be divisible by 7.

51/7 = 7

Hence, 7 numbers
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We can even use this approach as the numbers should be between 100 to 200 we can use 7 table so there will be 10 numbers which will be divisible by 7 from 1 till 70. Now we need for 80,90 and 100 so from 70 to 80 there is only 1 number which is divisible by 7 then again 80 to 90 we have 1 number divisible by 7, then 90 to 100 2 number so total number will
10+1+1+2 which is 14 and now we need even number which will be 14/2= 7

Please let me know if this is correct approach?
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