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No, 120 is not the smallest number that is divisible by 1,2,3,4,5 and 6.

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So, what is then?

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what is it?

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ya!!!

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what is it?

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120

Q: Is 120 the smallest number divisible by 1 2 3 4 5 and 6?

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2/6 or 1/3

AnswerThe probability that a randomly chosen [counting] number is not divisible by 2 is (1-1/2) or 0.5. One out of two numbers is divisible by two, so 1-1/2 are not divisible by two.The probability that a randomly chosen [counting] number is not divisible by 3 is (1-1/3) = 2/3.Similarly, the probability that a randomly chosen [counting] number is not divisible by N is (1-1/N).The probability that a random number is not divisible by any of 2, 3 or 6 can be reduced to whether it is divisible by 2 or 3 (since any number divisible by 6 can definitely be divided by both and so it is irrelevant). This probability depends on the range of numbers available. For example, if the range is all whole numbers from 0 to 10 inclusive, the probability is 3/11, because only the integers 1, 5, and 7 in this range are not divisible by 2, 3, or 6. If the range is shortened, say just from 0 to 1, the probability is 1/2.Usually questions of this sort invite you to contemplate what happens as the sampling range gets bigger and bigger. For a very large range (consisting of all integers between two values), about half the numbers are divisible by two and half are not. Of those that are not, only about one third are divisible by 3; the other two-thirds are not. That leaves 2/3 * 1/2 = 1/3 of them all. As already remarked, a number not divisible by two and not divisible by three cannot be divisible by six, so we're done: the limiting probability equals 1/3. (This argument can be made rigorous by showing that the probability differs from 1/3 by an amount that is bounded by the reciprocal of the length of the range from which you are sampling. As the length grows arbitrarily large, its reciprocal goes to zero.)This is an example of the use of the inclusion-exclusion formula, which relates the probabilities of four events A, B, (AandB), and (AorB). It goes like this:P(AorB) = P(A) + P(B) - P(AandB)In this example, A is the event "divisible by 2", and B is the event "divisible by 3".

The probability of an even must be a real number in the range [0, 1]. 120% is outside that range.

If the die is fair then for a single roll, the probability is 1/2.

Because we are only modeling one event, all six outcomes of the die are equally possible. The probability of rolling a four (or, for that matter, any number) is 1/6, or .166666 repeating. Now, since we are modeling 120 rolls, the theoretical number of outcomes of four (or, again, any number) is 1/6 * 120 = 20 outcomes. The second sentence of the problem is unnecessary.

Related questions

120

60

No (60 is)

No.First off, 120 is not divisible by 9. (The sum of the digits of 120 is "3"; a number is divisible by nine if and only if the sum of its digits is divisible by nine.) For that matter, it's not divisible by 36 either; since 36 is itself divisible by 9, any number which is not divisible by 9 cannot be divisible by 36.The smallest number that's divisible by 1, 2, 3, 6, 9, 12 and 36 is "36" (it's also divisible by 4, which you left out).

Every number is divisible by 1. Including negative numbers, that's an infinite list. There is no smallest.

8

The smallest number that is divisible by the numbers 1 through 10, except 7 and 8, is 180.

12

-7

101

21

The sum of its digits is 1+2+0 = 3 which is divisible by 3. So 120 is divisible by 3. The number 120 ends in a 0 and so it is divisible by 5.