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Romanian Mathematical Olympiad

Number theory Difficulty 5.3 AIME, harder Prove it Romania

We will call a positive integer exquisite if it is a multiple of the number of its divisors (for instance, $12$ is exquisite because it has $6$ divisors and $12$ is a multiple of $6$ ).

a) Find the largest exquisite two digit number.

b) Prove that no exquisite number has its last digit $3$ .

Solution

a) We check the numbers decreasingly: $99 = 3^2 \cdot 11$ , $99$ has $6$ divisors and $6 \nmid 99$ ; $98 = 2 \cdot 7^2$ , $98$ has $6$ divisors and $6 \nmid 98$ ; $97 = 97$ , $97$ has $2$ divisors and $2 \nmid 97$ ; $96 = 2^5 \cdot 3$ , $96$ has $12$ divisors and $12 \mid 96$ . So $96$ is the largest exquisite two digit number.

b) Let $X$ be a positive integer with its last digit $3$ .

Then $X$ is odd and $X$ is not a perfect square, and a non-perfect square has an even number of divisors. Since an odd number cannot be a multiple of an even number, $X$ cannot be exquisite.

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