Find the least positive integer which is divisible by the first five natural numbers.
Answer
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Hint: In this question we have to find the least positive integer which is divisible by the first five natural numbers. We will first see what the first five natural numbers are. We know that natural numbers are numbers which start from $1$ onwards, we will then find the lowest common multiple of the natural numbers. The lowest common multiple of the first five natural numbers will give us the required solution.
Complete step by step answer:
We have to find the positive integer which is divisible by the first five natural numbers.
We know that natural numbers are numbers that start from $1$ onwards and do not consist of fractions, decimals and irrational numbers. They only consist of integers therefore; we have the first five consecutive natural numbers as:
$\Rightarrow 1,2,3,4,5$
Now we have to find a number which is divisible by all the given numbers. We know that all numbers are divisible by $1$, therefore we have the remaining numbers as:
$\Rightarrow 2,3,4,5$
Now since we have the number $2$, all even numbers are divisible by $2$ therefore, we will consider all the even multiples of $3,4,5$, we get:
Even multiples of $3$ are: $3,6,12,18,24,30,36,42,48,54,60,66$
Now $4$ is an even number therefore, we will consider all of its multiples, we get:
multiples of $3$ are: $4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64$
since $5$ is an odd number, we will consider the even multiples of $5$, we get:
Even multiples of $5$ are: $10,20,30,40,50,60,70$
Now out of all the multiples we can see that the number $60$ is the first occurring number. Therefore, the least common multiple of $1,2,3,4,5$ is $60$.
Therefore, the least positive integer which is divisible by first five natural numbers is $60$.
Note: It is to be noted that there is no other positive number which is divisible by $60$. In this question we have been given the numbers as natural numbers. There are also other sets of numbers such as integers, which consists of all positive and negative numbers, real numbers are the superset of all the other sets since they contain all positive, negative, fractions, decimals, rational and irrational numbers.
Complete step by step answer:
We have to find the positive integer which is divisible by the first five natural numbers.
We know that natural numbers are numbers that start from $1$ onwards and do not consist of fractions, decimals and irrational numbers. They only consist of integers therefore; we have the first five consecutive natural numbers as:
$\Rightarrow 1,2,3,4,5$
Now we have to find a number which is divisible by all the given numbers. We know that all numbers are divisible by $1$, therefore we have the remaining numbers as:
$\Rightarrow 2,3,4,5$
Now since we have the number $2$, all even numbers are divisible by $2$ therefore, we will consider all the even multiples of $3,4,5$, we get:
Even multiples of $3$ are: $3,6,12,18,24,30,36,42,48,54,60,66$
Now $4$ is an even number therefore, we will consider all of its multiples, we get:
multiples of $3$ are: $4,8,12,16,20,24,28,32,36,40,44,48,52,56,60,64$
since $5$ is an odd number, we will consider the even multiples of $5$, we get:
Even multiples of $5$ are: $10,20,30,40,50,60,70$
Now out of all the multiples we can see that the number $60$ is the first occurring number. Therefore, the least common multiple of $1,2,3,4,5$ is $60$.
Therefore, the least positive integer which is divisible by first five natural numbers is $60$.
Note: It is to be noted that there is no other positive number which is divisible by $60$. In this question we have been given the numbers as natural numbers. There are also other sets of numbers such as integers, which consists of all positive and negative numbers, real numbers are the superset of all the other sets since they contain all positive, negative, fractions, decimals, rational and irrational numbers.
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