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.
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2No, 11 is not. Neither is 143 = 11 * 13. – Will Jagy Jun 30 '13 at 21:14
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2Well, the smallest counterexample is $1$. The next is $121$. – André Nicolas Jun 30 '13 at 21:17
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yes that's why i said OTHER THAN prime numbers. and 121 is combination of 11.11 – user1467270 Jun 30 '13 at 21:18
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Not unless you're an engineer. – TonyK Jun 30 '13 at 21:19
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But $1$ is not prime. And $11$ is not one of $2,3,5,7$. – André Nicolas Jun 30 '13 at 21:19
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I don't see any question here. – Erick Wong Jun 30 '13 at 21:35
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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.
amWhy
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That's what i am saying we achive every number with the combination of prime numbers? right? – user1467270 Jun 30 '13 at 21:19
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Yes, every number is a product of prime numbers or else is a prime: but there are infinitely many prime numbers other than $2, 3, 5, 7$. – amWhy Jun 30 '13 at 21:21
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1@Babak $\quad \overset{\large\vdots;}{\cdots} + \vdots + \underset{\large;\vdots}{\cdots}$ – amWhy Jul 05 '13 at 00:32
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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 $7$.
Lord Soth
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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 $