I think the answer is: 66 numbers between 1 and 200 are divisible by 3. :) That's what I think the answer is.
There are 100 even numbers between 1 and 200 (inclusive).
200
99 of them. Each and every one of the numbers between 100 and 200 is a multiple of itself and 1.
7,17,27,37,47,57,67,70,71,72,73,74,75,76,77(two sevens),78,79,87,97=20 times
To determine how many times the digit 7 appears between 1 and 885, we must consider each place value separately. In the units place (1-9), there is one 7. In the tens place (10-99), there are 10 occurrences of 7 (17, 27, 37, 47, 57, 67, 77, 87, 97). In the hundreds place (100-199, 200-299, ..., 800-885), there are 100 occurrences of 7. Therefore, the total number of 7s between 1 and 885 is 1 + 10 + 100 = 111.
6 of them with a remainder of 1
The infinitely many real numbers between 1 and 200 except for 1, 8, 27, 64 and 125.
I think the answer is: 66 numbers between 1 and 200 are divisible by 3. :) That's what I think the answer is.
46 of them.
There are 100 even numbers between 1 and 200 (inclusive).
1
200
66 :)
201
99 of them. Each and every one of the numbers between 100 and 200 is a multiple of itself and 1.
24