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2,000/9 = 2222/9
There are 222 of them.
A simpler way to look at it is multiply 9 times a number and keep going up until the product is over 2000. (9 * 223) is 2007. One less (9 * 222) is 1998, so therefore there are 222 integers between 1 and 2000.
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How many integers between 1 and 150 are divisible by both 4 and 5?
There are seven of them. They are: 20, 40, 60, 80, 100, 120 and 140.
Prove that one of every 3 consecutive positive integers is divisible by 3?
Let three consecutive integers be n, n+1 and n+2. If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3 as these numbers will respectively leave remainders of 1 and 2. If n is not divisible by 3 then it will leave a remainder of 1 or 2. If n leaves a remainder of 1, then n+1 leaves a remainder of 2 and n+2 is therefore divisible by 3. If n leaves a remainder of 2, then n+1 is divisible by 3 and n+2 is not divisible by 3 as it leaves a remainder of 1.
How many integers from 1 to 1000 are divisible by 30 but not by 16?
There are 1000 integers from 1 to 1000. Only the multiples of 30 of these integers are divisible by 30, such as 30, 60, 90, 120, 150, 180, 210, 240, ..., 990. Since the LCM of 16 and 30 is 240, then only 7 integers from 1 to 1,000 are divisible by 30 but not by 16, which are 30, 60, 90, 120, 150, 180, and 210. (Prime factorization: 30 = 2 x 3 x 5 16 = 24 The LCM of 30 and 16 is 24 x 3 x 5 = 240.)
How many integers from 1 through 100 are multiples of 4 and multiples of 7?
Here is the way to solve this problem. Since the number has to be divisible by 7 and 4 this means that the number has to be divisible by 28. The smallest multiple of 28 is 281 and the largest multiple through 1-100 is 283. the since the one is inclusive you do 3-1+1 which equals 3. 3 integers is the answer.
Prove that the two consecutive positive integer is divisible by 2?
There is something missing from your question: what about the two consecutive integers? Is the proof required that one of them is divisible by 2? Or that their product is (which amounts to the same thing)? Showing that one of two consecutive number is divisible by 2: Suppose your two numbers are n and (n+1). If n is divisible by 2, ie n = 2k, the result is shown. Otherwise assume n is not divisible by 2. In this case n = 2m+1. Then: (n+1) = ((2m+1)+1) = 2m + 2 = 2(m+1) which is a multiple of 2 and so divisible by 2. QED. Showing that exactly one of two consecutive integers is divisible by two is shown above with the addition to the first part: "as (n+1) = 2k+1 is not divisible by two and so only n is divisible by 2." To show the product is divisible by 2, show either n is divisible by 2 or (n+1) is as above, then the result follows as one of n and (n+1) is divisible by 2 and so their product is.
How many integers are there between 1 and 1000 are divisible by 3?
333 integers.
How many integers between 1 and 2018 are not divisible by 2 and are not divisible by 5?
Between 1 and 2018, 1008 are not divisible by 2.1613 are not divisible by 5.
Is 46 divisible by 1?
All integers are divisible by 1.
How many integers from 1 to 600 are divisible by 5 or 7?
There are 189.
How many positive integers between 1 and 250 both inclusive that are divisible by any of the integers 2357?
193 of them are divisible by one (or more) of the given numbers.
Is 1 divisible of 6?
All integers are divisible by one.
What is a divisible by 1?
All positive integers are divisible by one.
How many integers in 1.1000 are divisible by 2?
500 numbers in the range [1, 1000] are divisible by 2.
How many integers from 1 -100 inclusive are divisible by 5 and 9?
Divisible by both? 2 numbers... 45 and 90
What numbers over 100 are divisible by 1?
All integers are evenly divisible by 1.
How many perfect squares of integers from 1 to 15 inclusive are divisible by 3?
5 of them.
What numbers are divisible by 1?
All positive integers.
