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Find all even natural numbers which can be written as a sum of two odd composite numbers.

Please help in in solving the above problem.

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Let $n\geq 100$ an even number. Consider the quantities $n-91$, $n-93$ and $n-95$, one of these is a multiple of 3, and not exactly 3 cause $100-95>3$, then is a composite odd number. Observing thet 91,93 and 95 are composite, you conclude that every $n\geq 100$ works. Now check directly the remaining numbers, and you have the solution.

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Nice solution (+1). Welcome at Math.SE. – drhab Jan 19 '14 at 16:42
+1 To finish this (without further optimization) requires processing around $\color{#c00}{50}$ or more possible exceptional cases. $\ $ Much faster is to use $\,40\,$ vs. $\,100,\,$ which employs only $\,\color{#c00}7\,$ cases - see my answer. – Bill Dubuque Jan 19 '14 at 19:01

$n\ge 40\,\Rightarrow\, 3$ divides one of $\,n\!-\!a,\ a \in\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \overbrace{\{9,25,35\}}^{\large\qquad\ \equiv\ \{0,\ \ 1,\ \ 2 \}\,\pmod{\! 3}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!$ so $\ n = (n-a)+ a\,$ works, by $\,3\mid n-a > 3.\,$ Else $\,39 \ge n = a + b,\,$ wlog $a \le b\,$ so $\,a< 20\,$ so $\,a = \color{#c00}9$ or $\color{#0a0}{15}\,$ (the only odd composites $< 20).$ Hence the only $\,n\le 39\,$ that work are

$$\begin{eqnarray}\color{#c00}9+\!\!\!\!\!\!\!\!\!\!\overbrace{\{9,15,21,25,27\}}^{\large {\rm odd\ composites}\ b\,\in\, {\bf [}\color{#c00}9,\ 39-\color{#c00}9{\bf ]}}\!\!\!\!\!\!\!\!\!\! &=& \{18,24,30,34,36\}\\ \\ \color{#0a0}{15}+ \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\underbrace{\{15,21\}}_{\large {\rm odd\ composites}\ b\,\in\, {\bf [}\color{#0a0}{15},\ 39-\color{#0a0}{15}{\bf ]}}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &=& \{30,36\}\quad\ \ \, {\bf QED}\end{eqnarray}$$

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What do you mean by "wlog?" I'm guessing "without loss of generality," right? – apnorton Jan 19 '14 at 18:07
@anorton Yes, we may choose notation so that $\,a\,$ denotes the least summand. – Bill Dubuque Jan 19 '14 at 18:25

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