Math
posted by Anon .
Take a four digit number (abcd) and repeat it to make an eight digit number (abcdabcd). This eightdigit number always has at least three different prime factors, unless you start with a certain fourdigit number. Which number is it?

Repeating the 4 digit number is the same as multiplying by 10001, which factors into 73*137 (both prime). The only way 10001n does not have some other prime factor is if n also has only prime factors of of 73 and/or 137. Since 73*137 is 5 digit number, and both 137 and 73 have too few digits, the only product left is 73² = 5329.
53295329 = 5329*10001 = (73²)(73)(137) = (73³)(137)
...is the number with only two prime factors.
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