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

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

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

Hi Bunuel,

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

Thanks!
Uve

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)

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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