Question

How many integers between 10 and 60, inclusive, can be evenly divided by neither 2 nor 3?

Answer 1

Find total number of integers.

$a_1=10,a_n=60,d=1 \\a_n=a_1+(n-1)d \\60=10+(n-1)1 \\n-1=50 \\n=51$

Find how many integers is divisible by 2.

$a_1=10,a_n=60,d=2 \\a_n=a_1+(n-1)d \\60=10+(n-1)2 \\2(n-1)=50 \\n-1=25 \\n=26$

Eliminate even numbers.

11, 13, 15,..., 57, 59

This array contains 51 - 26 = 25 numbers.

Eliminate numbers before the first number divisible by 3 and after the last number divisible by 3.

15, 17, 19,..., 55, 57

This array contains 25 - 3 = 22 numbers.

Now we should eliminate numbers divisible by 3: 15, 21, 27...

$a_1=15,a_n=57,d=6,n=? \\a_n=a_1+(n-1)d \\57=15+(n-1)6 \\6(n-1)=42 \\n-1=7 \\n=8$

There are 8 such numbers.

Therefore, there are 25 - 8 = 17 numbers that can be evenly divided by neither 2 nor 3