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Oh, dude, to find out how many integers from 1 to 2000 are divisible by 9, you just need to divide 2000 by 9 and see how many times it goes in evenly. So, 2000 divided by 9 is like 222 with a remainder, but we only care about the whole number part, which is 222. So, there are 222 integers from 1 to 2000 that are divisible by 9.
Well honey, to find out how many integers from 1 to 2000 are divisible by 9, you divide 2000 by 9 and round down to the nearest whole number. So, 2000 divided by 9 is 222 with a remainder of 2. That means there are 222 integers from 1 to 2000 that are divisible by 9.
Oh, what a happy little question! To find out how many integers from 1 to 2000 are divisible by 9, we can use a simple formula. The formula is (last number - first number) / divisor + 1. In this case, it would be (2000 - 1) / 9 + 1 = 223 integers. Isn't that just delightful?
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What is the total number of integers between 100 and 300 that are divisible by 3?
To find the total number of integers between 100 and 300 that are divisible by 3, we first determine the smallest and largest integers in this range that are divisible by 3. The smallest integer divisible by 3 is 102, and the largest is 297. To find the total number of integers between 102 and 297 that are divisible by 3, we calculate (297-102)/3 + 1, which equals 66. Therefore, there are 66 integers between 100 and 300 that are divisible by 3.
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.
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.)
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 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.
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.
How many integers in 1.1000 are divisible by 2?
500 numbers in the range [1, 1000] are divisible by 2.
What is a divisible by 1?
All positive integers are divisible by one.
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.
How many positive integers less than 2101 are divisible by atleast one of the primes?
Except for 1, all of them.
