There are 142 of them and I have neither the time, inclination nor the desire to list them all.
I suggest you start with 7, then add 7, then add another seven and keep going until you exceed 1000.
1000 ÷ 7 = 142 remainder 6 So, there are 142 numbers between 1 and 1000 that are evenly divisible by 7. 1 x 7 = 7 2 x 7 = 14 3 x 7 = 21 ... 142 x 7 = 994
343!
343 is divisible by 1, 7, 49, 343.
Just 7. Any other number divisible by 7 isn't prime.
All 2 digit numbers are divisible by 7: it is simply that some of the quotients are not whole numbers. 10/7 = 1 3/7 99/7 = 14 1/7 So the number of 2-digit numbers which are evenly divisible by 7 are 14-1 = 13.
142 of them.
There are 95 numbers in the range of 1 to 1,000 that are divisible by seven but not by three.
1000 ÷ 7 = 142 remainder 6 So, there are 142 numbers between 1 and 1000 that are evenly divisible by 7. 1 x 7 = 7 2 x 7 = 14 3 x 7 = 21 ... 142 x 7 = 994
There are no numbers that are divisible by 21 but not by 7 so the "or 21" part of the question can be ignored.There are 166 numbers between 1 and 1000 that are divisible by 6. However, 23 of those are also divisible by 7.So 166 - 23 = 143 numbers.
142
7
343!
142 of them.
There are 14 numbers between 1 & 100 that are divisible by 7.
8
One of every three numbers is divisible by three. Therefore, 1000/3 (rounded down) tells you how many numbers in that range are divisible by 3. Similar for the number 7. Now, if you add the two, you'll be counting all the multiples of 21 (which is 3 x 7) double. Therefore, you must calculate 1000 / 21 (rounded down), and subtract that from the previous result.
858