A number theory problem by Kaleem Kħặŋ
Number Theory
Level
3
What is the smallest positive integer which have 7 (positive) divisors including itself?
The answer is 64.
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1 solution
When thinking about a number with exactly 7 positive factors, you start thinking about factors and factor pairs and it should become evident that since there are an odd amount of factors, one of the factor pairs must be composed of the same number twice, else there would be an even amount of factors. Since there are an odd amount, one of the factor pairs would be a number times itself. Hence, we deduce that this is a perfect square. Then by brute force, going through the perfect squares in ascending order, one can find the smallest positive integer with this attribute, namely, 64.
There was only one other perfect square less than this with 7 factors, but it had more than seven factors as it was 36 and it had different combinations of 2 and 3 to make more factors. Hence, I was also intuitively guided to a number whose prime factorization came down to one number to the 6th power, so there would only be 7 combinations of that number (since we're including the 0th power as well), which would be another way of arriving at 64 without testing each number less than it.