Difficult Math Problem #118 - Number Theory

There are 51 numbers in all.

Total num of numbers divisible by 3 = 51/3 = 17
Total num of numbers divisible by 5 = 51/5 = 11 (we know 100 and 150
fit the bill here, hence 11)
Total num of numbers divisible by 15 = 4 (105, 120, 135, 150)

Using A + B -AnB
total num of numbers divisible by 3 or 5 = 17 + 11 - 4 = 24

Hence total num of numbers not divisible by 3 or 5 = 51-24 = 27

This seems rather round-about. I can't think of a simpler way...

How many integers between 100 and 150, inclusive can be evenly divided by neither 3 nor 5?

The first integer divisible by 3 between 100 and 150 (both inclusive) is 102 (3*34) and the last one is 150 (3*50). This means there are 17 factors of 3 between 100 and 150.

The first integer divisible by 5 between 100 and 150 (both inclusive) is 100 (5*20) and the last one is 150(5*30). Thus there are 11 factors of 5 between 100 and 150.

The first integer divisible by 15 between 100 and 150 (both inclusive) is 105(15*7) and the last one is 150(15*10). Thus there are 4 factors of 15 (The LCM of 3&5) between 100 and 150

51 - [(17+11)-4] = 51 - 24 = 27 factors not divisible by either

OA:

Number of integers that divide 3:

the range is 100-150
Relevant to this case, we take 102 - 150 (since 102 is the first to div 3)
102 = 34*3
150= 50*3, so we have 50-34+1 = 17 multiples of 3

For multiples of 5,
100=5*20
150=5*30
30-20+1 =11

Now we have a total of 27 integers, but we double counted the ones that divide BOTH 3 AND 5, ie 15.

105 is the first to divide 15.
105=15*7
150=15*10
10-7+1 = 4 integers

So our total is 17+11-4 = 24 integers that can be divided by either 3 or 5 or both.

51 integers - 24 integers = 27 that cannot be evenly divided.