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The sum of the digits of a two-digit number is multiplied by $$8$$ and the result is found to be $$13$$ more than the number. Then the two-digit number is
Let the tens digit be $$a$$ and the units digit be $$b$$. Then $$8(a+b)=10a+b+13$$, which gives $$7b=2a+13$$. The only valid digit pair is $$a=4$$ and $$b=3$$, so the number is $$43$$, which is prime.
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