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How many positive three digit integers are divisible by neither 2 nor 3?
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.
How many natural numbers between 150 and 300 are divisible by 9?
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.
How many numbers between 100 to 500 are divisible by 6?
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.
How many three digits natural numbers are divisible by 4?
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.
What is 29 divisible by nine?
29 is not divisible by 9. 29 divided by 9 = 3 and 2/9
What r the 3 numbers that have 2 4 and 8 as factors?
Calculate the least common multiple of the three numbers. Any multiple of that has all those numbers as factors.
What is the set of all real 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 (-∞, ∞).
What are the numbers the letters r on?
all numbers with three and four in it
What is the set of the real 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 (-∞, ∞).
What is are all the integers?
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, ...}.
How do you find the median of numbers?
Put the numbers in order. Add them all together then divide by the amount of numbers in the line. For example: 1 2 3 4 5 6 = 21 21 divide by 6 is 3 r 3 so your answer is 3.3
Which number is divisible by 5 and 9?
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.
