Itself and any of its multiples
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.
150 ÷ 9 = 16 r 6 → the first natural number between 150 and 300 divisible by 9 is 9 × 17 (=154)300 ÷ 9 = 33 r 3 → the last natural number between 150 and 300 divisible by 9 is 9 × 33 (=297)→ there are 33 - 17 + 1 = 17 natural numbers between 150 and 300 divisible by 9.
There are 67 numbers between 100 and 500 divisible by 6. The first number greater than 100 divisible by 6: 100 ÷ 6 = 16 r 4 → first number divisible by 6 is 6 × 17 = 102 Last number less than 500 divisible by 6: 500 ÷ 6 = 83 r 2 → last number divisible by 6 is 6 × 83 = 498 → all multiples of 6 between 17 × 6 and 83 × 6 inclusive are the numbers between 100 and 500 that are divisible by 6. → there are 83 - 17 + 1 = 67 such numbers.
The first 3 digit natural number is 100: 100 ÷ 4 = 25 → first 3 digit natural number divisible by 4 is 4 × 25 The last 3 digit natural number is 999: 999 ÷ 4 = 249 r 3 → last 3 digit natural number divisible by 4 is 4 × 249 → number of 3 digit natural numbers divisible by 4 is 249 - 25 + 1 = 225.
29 is not divisible by 9. 29 divided by 9 = 3 and 2/9
Calculate the least common multiple of the three numbers. Any multiple of that has all those numbers as factors.
To find the number of combinations of 3 numbers from a set of 42 numbers, you can use the combination formula ( C(n, r) = \frac{n!}{r!(n-r)!} ). Here, ( n = 42 ) and ( r = 3 ). So, ( C(42, 3) = \frac{42!}{3!(42-3)!} = \frac{42 \times 41 \times 40}{3 \times 2 \times 1} = 11480 ). Therefore, there are 11,480 combinations of 3 numbers from 42 numbers.
The set of all real numbers (R) is the set of all rational and irrational numbers. The set R has no restrictions in its domain and so includes (-∞, ∞).
all numbers with three and four in it
The set of all real numbers (R) is the set of all rational and Irrational Numbers. The set R has no restrictions in its domain and so includes (-∞, ∞).
All integers are the elements of the set of integers, I, which is one of the components of the set of all real numbers, R. I = {..., - 3, -2, -1, 0, 1, 2, 3, ...}.
When you are trying to find a common number that is divisible by two different numbers, you multiply the two together. So, for this example, you multiply 9 and 5, which equals 45. Therefore, 45 would be divisible by both 9 and 5.