no the smallest number that is divisible by four different primes would be 210
6 because it is divisible by 2 and 3.
Sol: 24 = 3 x 8, where 3 and 8 are co-primes. The sum of the digits in the given number is 36, which is divisible by 3. So, the given number is divisible by 3. The number formed by the last 3 digits of the given number is 744, which is divisible by 8. So, the given number is divisible by 8. Thus, the given number is divisible by both 3 and 8, where 3 and 8 are co-primes. So, it is divisible by 3 x 8, i.e., 24.
Except for 1, all of them.
No. The smallest 4 primes are 2, 3, 5 and 7 and their product is 210.
If by "least number" you mean "smallest positive integer", then the answer is the product of the three smallest primes: 2x3x5 = 30
No, it is not a prime number, it is divisible by 2, 625, 5, 250, 10, 125, and probabaly more than that. Primes are only divisible by 1 and itself.
Just 7. Any other number divisible by 7 isn't prime.
No. The 4 smallest primes are 2 3 5 7. Multiplying these gives 210. Note that 1 is not regarded as a prime no. If it were to be then no other number could be prime for all of them are divisible by 1.
Only integers (whole numbers) can be primes. A prime is a number that is only divisible by 1 and itself. 43.2 and all other decimal numbers are not divisible by 1.
There is no such number. If you restrict yourself to integers, there are numbers - such as primes, that are not divisible by an integer other than one. And if you do not restrict yourself to intgers, then there is no smallest number sincce given any number, half that number will be smaller and will be a divisor.
Every even number is evenly divisible by 2.
It is 11*13 = 143
The largest integer that is not the product of two or more different primes would be the largest prime number. Because there are an infinite number of prime numbers, there is no largest integer that is not the product of two or more different primes.
2 * 3 * 5 * 7 = 210 is the smallest one divisible by four unique prime numbers. However, there are an infinite number of them, since there are an infinite number of primes.
There is none. The only even prime number (even meaning divisible by two, and primes being numbers divisible only by the integers 1 and themselves) is 2.
No - 51 is divisible by 3.
In any large range of numbers, more so the higher and larger the range is, there will be more composites than primes. It is far more likely that a given number will be divisible by some number less than the number itself, the higher you go.
100 is divisible by 1, 2, and 5. 1 is not itself a prime number; 101 is a prime; 102/2 =51 102/3 =34 102/17=6
8 has three primes (2x2x2) in its factorization, and 30 is the smallest number having three different primes (2x3x5) in its factorization.
Yes. To prove this, we must first assume the answer to be no. If there are a finite number of primes, there must be a largest prime. We'll call this prime number n. n! is n*(n-1)*(n-2)*...*3*2*1. n!, therefore, is divisible by all numbers smaller than or equal to n. It follows, then that n!+1 is divisible by none of them, except for 1. There are two possibilities: n!+1 is divisible by prime numbers between n and n!, or it is itself prime. Either way, we have proved that there are prime numbers greater than n, contradicting our initial assumption that primes are finite, proving that the number of primes is infinite.