# Prime Numbers and Multiples?

Other than prime numbers are all numbers multiple of 2,3,5 and 7 (Other Prime numbers as well). Suppose like if we need 8 it's the combination of 2.2.2, and 15 as 5.3 etc.

• No, 11 is not. Neither is 143 = 11 * 13. – Will Jagy Jun 30 '13 at 21:14
• Well, the smallest counterexample is $1$. The next is $121$. – André Nicolas Jun 30 '13 at 21:17
• yes that's why i said OTHER THAN prime numbers. and 121 is combination of 11.11 – user1467270 Jun 30 '13 at 21:18
• Not unless you're an engineer. – TonyK Jun 30 '13 at 21:19
• But $1$ is not prime. And $11$ is not one of $2,3,5,7$. – André Nicolas Jun 30 '13 at 21:19

No: $$11\cdot 13 = 143$$

$$13 \cdot 17 = 221$$

$$\vdots$$

The most we can say is what the Fundamental Theorem of Arithmetic tells us: Every integer greater than $1$ is a prime number or a product of prime numbers.

Edit: (Revised question) Yes, except $1$ is neither prime nor is it a product of prime numbers.

• That's what i am saying we achive every number with the combination of prime numbers? right? – user1467270 Jun 30 '13 at 21:19
• Yes, every number is a product of prime numbers or else is a prime: but there are infinitely many prime numbers other than $2, 3, 5, 7$. – Namaste Jun 30 '13 at 21:21
• @amWhy: Links and all +1 – Amzoti Jul 1 '13 at 0:36
• @amWhy: I love that $\vdots$ above + – mrs Jul 5 '13 at 0:27
• @Babak $\quad \overset{\large\vdots\;}{\cdots} + \vdots + \underset{\large\;\vdots}{\cdots}$ – Namaste Jul 5 '13 at 0:32

No, $1763 = 41 \times 43$. For example, if you multiply any two primes other than $2,3,5$ and $7$ (note there are infinitely many primes), you get a number that is "not a combination of" $2,3,5$ or $7$.

No . Fundamental theorem of arithmetic states that every integer greater than 1 is either prime itself or is the product of prime numbers examples:
$22 = 11 \cdot 2$
$1200 = 2^4 \cdot 3 \cdot 5^2$

It's not like that.

Every positive integer(non-prime) is always a multiple of other primes smaller than that.

But it doesn't mean that those will surely have 2,3,5,&7.

for example you can take:

$$121=11\times11$$