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Q: What r all the numbers divisible by 3?

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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.

Every number is divisible by any non-zero number.Any element of the set of numbers of the form 29*k + r, where k is an integer and 0<r<29, is not evenly divisible.

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.

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.

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 (-∞, ∞).

345 is not divisible by 9. A quick rule for checking to see if a number is divisible by 9 is to add all of the number's digits. If their sum is divisible by 9, so is the number.2+3+4 = 9, which is divisible by 9, so 234 is divisible by 9. (234/9 = 16)3+4+5 = 123, which is not divisible by 9, so 345 is not divisible by 9 (345/9 = 38 r 3)5+6+7 = 18, which is divisible by 9, so 567 is divisible by 9 (567/9 = 63)

Calculate the least common multiple of the three numbers. Any multiple of that has all those numbers as factors.

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, ...}.

They are: 2, 3 and 7

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

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.

r3

All those numbers than can b represented as one integer over another integer r rational.

It is y. Every third letter is r, immediately followed by y, and since 117 is divisible by 3, it is r. Therefore, 118 is y.

all you do as a fration it is 2/3 then all you do is divide 3 into 2 which will be 1 r.1

only 1. 0 is divisible by 0 but nothing but 0 is ever because no number multiplied by 0 can be anything other than 0. so only 0 can be divided by 0. with variables Y= all real numbers. R= all numbers other than 0 0xY=0 soo 0/0=0 but because the product is always zero Rx0=0 R/0=undefined so your answer is only 1 number.

No, rational numbers cannot r.

30 and 60

Let the three numbers in GP: a/r, a, ar---------(A)Where '^' is power of . . .Sum of these numbers are:a/r +a +ar = 38a(1\r+1+r) = 38 ---- (1)Product of these numbers are:a^3 = 1728= (12)^3a = 12Putting the value of a in (1) you will get:12(1\r+1+r) = 38And factorising, we getr = 2/3 or r = 3/2Sub. the r and a value in (A), we get8,12,18 or 18,12,8. When a = 12And smallest no. is 8.

If r is an integer then r = 1, 2 or 4.

No. To be divisible by 3, if the sum of the digits is divisible by 3 so is the original number, otherwise the remainder when the sum is divided by 3 is the remainder when the original number is divided by 3. 5 + 0 + 0 = 5 5 ÷ 3 = 1 r 2, so 500 ÷ 3 has remainder 2.

By inverting the numbers. For example, if:2 x 3 = 6 then: 6 / 3 = 2

200 ÷ 3 = 66 r 2 → first multiple of 3 greater than or equal to 200 is 67 (x 3 = 201) 250 ÷ 3 = 83 r 3 → last multiple of 3 less than or equal to 250 is 83 (x 3 = 249) All multiples of 2 are even, so want all multiples of 3 that are not even, namely the odd multiples of 3; so want 67 x 3, 69 x 3, 71 x 3, ... 83 x 3 which are: 201, 207, 213, 219, 225, 231, 237, 243, 249.