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.
There are 3 such numbers.
9
There are 12 multiples of 8 in 1 to 100.
There are floor(100/8)=12 multiples of 8 between 1 and 100. 12/100*100=12%
There are 30 multiples of 30 that fall between 100 and 1,000.
20
128!
9
There are 67 multiples of 6 and 50 multiples of 8 in that range. Their total, 117, will include numbers that are both.
100% because all integers are multiples of 1
There are 12 multiples of 8 in 1 to 100.
There are floor(100/8)=12 multiples of 8 between 1 and 100. 12/100*100=12%
10 is two in multiples of 2 to 100
101
8,16,24,32,40,48,56,64,72,80,88,96
8,16,24,32,40,48,56,64,72,80,88,96
Usually all the integers (counting numbers) from 1 to 100.Usually all the integers (counting numbers) from 1 to 100.Usually all the integers (counting numbers) from 1 to 100.Usually all the integers (counting numbers) from 1 to 100.
10 (they are 105, 126, 147, 168, 189, 210, 231, 252, 273, and 294)