Olympiad Maths Prep

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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, 1212 is exquisite because it has 66 divisors and 1212 is a multiple of 66).

a) Find the largest exquisite two digit number.

b) Prove that no exquisite number has its last digit 33.

Solution

a) We check the numbers decreasingly: 99=321199 = 3^2 \cdot 11, 9999 has 66 divisors and 6996 \nmid 99; 98=27298 = 2 \cdot 7^2, 9898 has 66 divisors and 6986 \nmid 98; 97=9797 = 97, 9797 has 22 divisors and 2972 \nmid 97; 96=25396 = 2^5 \cdot 3, 9696 has 1212 divisors and 129612 \mid 96. So 9696 is the largest exquisite two digit number.

b) Let XX be a positive integer with its last digit 33.

Then XX is odd and XX 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, XX cannot be exquisite.

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