The smallest number divisible by first four prime numbers is 2x3x5x7 = 210.
It is 2520.
The smallest three-digit number divisible by the first three prime numbers (2, 3, and 5) and the first three composite numbers (4, 6, and 8) is 120.
The smallest number evenly divisible by the numbers 1-12 is 60. 60 is also be by 0, but not by 13.
the answer is 120 because the first three prime numbers are 2,3,5 and the first three composite numbers are 4,6,8.
The numbers that are divisible by two are infinite. The first four are: 2,4,6,8 . . .
The list of numbers that are divisible by 11 is infinite. The first four are: 11,22,33,44 . . .
Any multiple of 420.
The first prime numbers are 2, 3, 5, 7, 11. When multiplied together the product is 2310. It is the smallest number that has the first five numbers as its factors.
Just multiply the first five prime numbers! 2x3x5x7x11 = 6x35x11 = 210x11 = 2310. Any multiple of this number "has the first five prime numbers as factors", but 2310 is the smallest such number.
There are no top numbers. Suppose there was a top number that was divisible by two. Then that number plus two would be divisible by two and it would be bigger than the top number. So the first number could not have been a top number.
The list of numbers that are divisible by 351 is infinite. The first four are: 351, 702, 1053, 1404 . . .
the first 10 whole numbers are numbers 1 to 10 and in those numbers only 3 numbers are divisible by 3 in which 3, 6 and 9 therefore the probability of from those figures that the numbers won't be divisible by 3 is 7/10 or 70%.
First find the prime factors of all those numbers. Multiply each prime together, using the highest power of the primes if they have exponents.The primes of the given numbers to their highest power are 22, 32, and 5, so the smallest number divisible by all of them is 180 (22 x 32 x 5).
30, which is the smallest positive integer divisible by the first three primes: 2, 3 and 5.
Here's one. It is the first number divisible by the first three prime numbers.
The least common denominator of 25, 15, and 3 is 75. To get this we need to find the smallest number that is divisible by 25, 15, and 3 without any remainders. Since 25 has to go into the number it is easiest to first look at which numbers are divisible by 25 without any remainder. This would be numbers like 25, 50, 75, 100, etc. Then look at which of these numbers are divisible by 15 and 3 without any remainder. The first number that is is the least common denominator. In this case, it is 75.