How many even integers n, where 100 <= n <= 200, are divisib

Please check the solution as attached.

Answer: Option C

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

GMATinsight's Solution

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

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.

First look for the first multiple of 9 in the given range

Since 99 is a multiple of 9, next multiple will be 108 = 9*12
Last multiple of 9 will be 198 = 9*22
This gives us 22 - 12 + 1 = 11 multiples of 9 of which 6 are even (we start with 108 (even) and end with 198 (even))

Of these 11 multiples, 2 multiples will be multiples of 7 too i.e. 9*14 and 9*21 though only one of them will be even. So there is an overlap of 1 with even multiples of 7.

Multiples of 7 in the range start from 105 = 7*15
Last multiple is 196 = 7*28
This gives us 28 - 15 + 1 = 14 multiples of which 7 will be even.

This gives us 6 + 6 = 12 even multiples of 9 or 7.

The range has 101 total integers of which 51 are even. 12 of them are multiples of 9 or 7 so 51 - 12 = 39 are not.

Answer (C)

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