There are total of \(\frac{200-100}{2}+1=51\) even numbers in the range \(100\leq{n}\leq{200}\) (check this: totally-basic-94862.html?hilit=multiple%20range );

There are total of \(\frac{198-108}{18}+1=6\) even multiple of 9 (so multiple of 18) in the range \(100\leq{n}\leq{200}\);

There are total of \(\frac{196-112}{14}+1=7\) even multiple of 7 (so multiple of 14) in the range \(100\leq{n}\leq{200}\);

As there is 1 even multiple of 7 AND 9 (overlap) in the range \(100\leq{n}\leq{200}\), which is \(2*7*9=126\) then there are total of \(6+7-1=12\) even multiples of 7 OR 9 in the range \(100\leq{n}\leq{200}\);

So there are \(51-12=39\) even numbers which are not divisible neither by 7 nor by 9.

Answer: C.
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Even

From 16 x 7 all the way up to 28 x 7 comes up to 7 such numbers

From 12 x 9 all the way up to 22 x 9 which comes up to 6 such numbers

From 2 x 63 - 1 such number

Total number of even numbers divisible by either 7 or 9: 7 + 6 - 1 = 12


From 100 to 200 there are 51 even numbers as the series start from 100.

Number neither divisible: 51 - 12 = 39

Odd

From 15 x 7 all the way up to 27 x 7 which count up to 7 numbers
From 13 x 9 all the way up to 21 x 9 which count up to 5 numbers
From 63 x 3 - 1 number

Total = 7 + 5 - 1 = 11

From 100 to 200 there are 50 odd numbers

Total number of odd numbers not divisible: 50 - 11 = 39

Pushpinder Gill
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Please check the solution as attached.

Answer: Option C

Please check the video for the step-by-step solution.

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Hi Bunuel,

A very stupid question - instead even integers, how do you get the number of odd integers between 100 and 200?

Thanks!
Uve

Here is the solution for odd numbers



Total Integers from 100 to 200 inclusive = 101 (51 even and 50 odd)
Total Odd Integers from 100 to 200 = 50

Total No. that are multiple of 7 from 1-200 = 200/7 = 28 (i.e. 14 even and 14 odd)
Total No. that are multiple of 7 from 1-100 = 100/7 = 14 (i.e. 7 even and 7 odd)
Total No. that are ODD multiple of 7 from 100-200 = 14-7 = 7

Total No. that are multiple of 9 from 1-200 = 200/9 = 22 (i.e. 11 even and 11 odd)
Total No. that are multiple of 9 from 1-100 = 100/9 = 11 (i.e. 5 even and 6 odd)
Total No. that are ODD multiple of 9 from 100-200 = 11-6 = 5

Total No. that are odd multiple of 7 and 9 will be multiple of 63 and count of such numbers from 100-200 = 1

So Total No. that are odd multiple of 7 and/or 9 from 100-200 = 7+5-1 = 11

So Total No. that are NOT odd multiple of 7 and/or 9 from 100-200 = 50-11 = 39

Answer: Option C
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from 100 to 200, there are 51 even nos.

51/7= 7 R2
51/9= 5 R6

7+5=12
51-12=39

Option C

I took the long route but later I realised that there is an easy way to do this.

Total number of even integers = ((200-100)/2)+1 = 51

To find out even multiples of 9 and 7, find multiples of 18 and 14 b/w 100 and 200.

Number of multiples of 18 = ((198 - 108)/18) + 1 = 5 + 1 = 6
Number of multiples of 14 = ((196-112) / 14) + 1 = 6 + 1 = 7

Remove the multiples of LCM of 18 and 14 i.e. 126 = 1

Therefore, answer is 51 - 6 - 7 + 1 = 39

Solution -
Number even integers n, where \(100 \leq n \leq 200\) = 51 (including 100 and 200)

There are 7 even integers divisible by 7 between 100 to 200 - 112, 126, 140, 154, 168, 182 and 196.
Notice the consecutive difference between the above integers is 14.

There are 6 even integers divisible by 9 between 100 to 200 - 108, 126, 144, 162, 180 and 198.
Notice the consecutive difference between the above integers is 18.

126 has been repeated in the list.

Hence, even integers n, where \(100 \leq n \leq 200\), which are divisible neither by seven nor by nine are -
51 - 7 - 5 = 39.

Answer should be option A.
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