Calculation:
Concept:
Divisibility law of 2 ⇒ A number divisible by 2 if its last digit is 0, 2, 4, 6, or 8.
Divisibility law of 3 ⇒ A number divisible by 3, if the sum of its digit is divisible by 3
Divisibility law of 5 ⇒ A number divisible by 5 if its last digit is 0 or 5
Given:
Number of digits divisible by 2 which starting from 2 to 1000 = Total 500 = A
Number of digits divisible by 3 which starting from 3 to 999 = Total 333 = B
Number of digits divisible by 5 which starting from 5 to 1000 = Total 200 = C
Number of digits divisible by 2 and 3 together which starting from 6 to 996 = Total 166 = AB
Number of digits divisible by 3 and 5 together which starting from 15 to 990 = Total 66 = BC
Number of digits divisible by 5 and 2 together which starting from 10 to 1000 = Total 100 = CA
Number of digits divisible by 2, 3 and 5 altogether which starting from 30 to 990 = Total 33 = ABC
So the count of numbers which are divisible by either 2, 3 or 5 from 1 to 1000 will be,
Now, the number of integers from 1 to 1000 that are not divisible by any of the numbers 2, 3, 5, is
= 1000 - |A| - |B| - |C| + |AB| + |AC| + |BC| - |ABC|.
= 1000 - (500 + 333 + 200 - 166 - 66 - 100 + 33)
= 1000 - (734)
= 266
Additional Information
Divisibility laws of 4 ⇒ A number is divisible by 4 if its last 2 digits divisible by 4.
Divisibility law of 6 ⇒ A number divisible by 6 if the number divisible by 2 and 3 both.
Divisibility law of 7 ⇒ Take the last digit of the number, double it Then subtract the result from the rest of the number If the resulting number is evenly divisible by 7, so is the original number divisible by 7.
Divisibility law of 8 ⇒ A number divisible by 8 if its last three-digit is divisible by 8
Divisibility law of 9 ⇒ A number is divisible by 9 if the sum of its digit is divisible by 9.
A number is chosen at random from the first 1,000 positive integers. What is the probability that it's divisible by 3,5, or 7?
So I started off by breaking the problem up and having: divisible by 3: p(a) divisible by 5: p(b)
divisible by 7 p(c)
I know I'm going to apply the exclusion inclusion principle, but how do I find out how many numbers are divisible by each without going through all the numbers between 1 and 1000?
Wesley S. How many integers from 1 to 1000, inclusive, are divisible by neither 2 nor 5? Notice the words "neither and nor" in the problem.
2 Answers By Expert Tutors
Jonathan C. answered • 02/09/17
Experienced General Mathematics Tutor
I will be using a trick very similar to what you would see in probability class:
The number of integers divisible by 2 from 1 to 1000 is 1000/2 = 500.
The number of integers divisible by 5 from 1 to 1000 is 1000/5 = 200.
The number of integers divisible by 2 AND 5 from 1 to 1000 is 1000/lcm(2,5) = 1000/10 = 100.
The number of integers divisible by 2 OR 5 from 1 to 1000 is 500 + 200 - 100 = 600. <-- union minus intersection from probability
The number of integers divisible by NEITHER 2 NOR 5 from 1 to 1000 is 1000 - 600 = 400. <-- complement from probability
So your answer is 400 integers.
These integers are the numbers ending in 1, 3, 7, and 9 (just in case you are curious).
Kenneth S. answered • 02/09/17
Expert Help in Algebra/Trig/(Pre)calculus to Guarantee Success in 2018
If not divisible by 2, they're odd. If not divisible by 5, they don't end in 5 (numbers such as 10 and 20 are already eliminated by the elimination of all evens).
Now think about all of the numbers from 1 to 1000 that remain after these deletions.
To get a handle on this problem, you might want to first apply this criterion to just the numbers from 1 to 100 (this is the idea of scaling down a hard problem on the first analysis of it).