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.
Chat with our AI personalities
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?
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.
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.
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.
Well, honey, let me break it down for you. To find the number of integers from 1 to 1000 that are divisible by 30 but not by 16, you first need to figure out how many numbers are divisible by 30. Then, subtract the numbers that are divisible by both 30 and 16. Finally, you'll have your answer. So get those calculators out and start crunching those numbers!
There are seven of them. They are: 20, 40, 60, 80, 100, 120 and 140.
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.
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.