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

1 answer:

6 0

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

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Answer:

Step-by-step explanation:

Hello!

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d) x tilde

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