Radical writing

Start with $\sqrt{a+\sqrt{b+c\sqrt{3}}} = 1+\sqrt{3}.$ Square both sides: $a + \sqrt{b+c\sqrt{3}} = 4 + 2\sqrt{3}.$ Subtract $a$ : $\sqrt{b+c\sqrt{3}} = (4-a) + 2\sqrt{3}.$ Square both sides: $b + c\sqrt{3} = (4-a)^2+12 + 4(4-a)\sqrt{3}$

Now, if $c \neq 4(4-a)$ , then $\sqrt{3} = \frac{(4-a)^2+12-b}{c-4(4-a)}$ would be rational. Since it's known to be irrational, we must have both $c=4(4-a),\qquad b= (4-a)^2+12$ Further, since we have assumed $a,c$ are positive integers, we must have $a \in \{1,2,3\}$ . We now consider each case:

• $a=1$ : Then $b= 21$ , $c=12$ , so $abc = 252$
• $a=2$ : Then $b= 16$ , $c=8$ , so $abc = 256$
• $a=3$ : Then $b=13$ , $c=4$ , so $abc = 156$ .

We can see that the only digit appearing in all three in base 10 is $\boxed{5}$ .

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Sidenote: When you square both sides of an equation, you can introduce extraneous solutions only if one side is negative prior to squaring. Therefore, to show each of these is a valid solution to the initial equation, it's enough to note that for $1\leq a\leq 3$ , both sides of the given equation are positive prior to squaring both sides.