There are 199 multiples of 5 in that range.
There are 128 multiples of 7 in that range.
There are 28 numbers on both lists.
1000 - 299 = 701
There are 17,999 such numbers.
675 of them.675 of them.675 of them.675 of them.
108 -999, if you consider negative numbers.
Of the 729 numbers that satisfy the requirement of positive integers, 104 are divisible by 7.
there are three odd numbers
There are 17,999 such numbers.
Oh, dude, let me break it down for you. So, to find the numbers divisible by both 3 and 4, you need to find the numbers divisible by their least common multiple, which is 12. From 100 to 999, every 12th number is divisible by 12. So, you just need to find how many multiples of 12 are there between 100 and 999. It's like a piece of cake, you got this!
109
675 of them.675 of them.675 of them.675 of them.
Between 100 and 999 there are 448.
3 digits numbers: from 100 to 999 The first number greater than 100 and divisible by 7 is 105 The last number lower than 999 and divisible by 7 is is 987 (987-105)/7=126 intervals of 7-length In fact 127 as numbers are to be counted, not intervals
108 -999, if you consider negative numbers.
100. The highest 3 digit number divisible by 9 is 999 (111 times). The highest 2 digit number divisible by 9 is 99 (11 times). The difference is 100.
831
The first number in this range divisible by 9 is 9 itself...9 = 1 x 9 The last number in the range is 999 = 111 x 9 So there are 111 numbers between 1 and 1000 that are divisible by 9.
There are 300 three digit numbers that are divisible by neither 2 nor 3. There are 999 - 100 + 1 = 900 three digit numbers. 100 ÷ 2 = 50 → first three digit number divisible by 2 is 50 × 2 =100 999 ÷ 2 = 499 r 1 → last three digit number divisible by 2 is 499 × 2 = 998 → there are 499 - 50 + 1 = 450 three digit numbers divisible by 2. 100 ÷ 3 = 33 r 1 → first three digit number divisible by 3 is 34 × 3 = 102 999 ÷ 3 = 333 → last three digit number divisible by 3 is 333 × 3 = 999 → there are 333 - 34 + 1 = 300 three digit numbers divisible by 3. The lowest common multiple of 2 and 3 is 6, so these have been counted in both those divisible by 2 and those divisible by 3. 100 ÷ 6 = 16 r 4 → first three digit number divisible by 6 is 17 × 6 = 102 999 ÷ 6 = 166 r 3 → last three digit number divisible by 6 is 166 × 6 = 996 → there are 166 - 17 + 1 = 150 → there are 450 + 300 - 150 = 600 three digit numbers that are divisible by either 2 or 3 (or both). → there are 900 - 600 = 300 three digit numbers that are divisible by neither 2 nor 3.
Of the 729 numbers that satisfy the requirement of positive integers, 104 are divisible by 7.