2019 AIME II Problems/Problem 9

Problem 9

Call a positive integer $n$ $k$ -pretty if $n$ has exactly $k$ positive divisors and $n$ is divisible by $k$ . For example, $18$ is $6$ -pretty. Let $S$ be the sum of positive integers less than $2019$ that are $20$ -pretty. Find $\tfrac{S}{20}$ .

Solution

Every 20-pretty integer can be written in form $n = 2^a 5^b k$ , where $a \ge 2$ , $b \ge 1$ , $\gcd(k,10) = 1$ , and $d(n) = 20$ , where $d(n)$ is the number of divisors of $n$ . Thus, we have $20 = (a+1)(b+1)d(k)$ , using the fact that the divisor function is multiplicative. As $(a+1)(b+1)$ must be a divisor of 20, there are not many cases to check.

If $a+1 = 4$ , then $b+1 = 5$ . But this leads to no solutions, as $(a,b) = (3,4)$ gives $2^3 5^4 > 2019$ .

If $a+1 = 5$ , then $b+1 = 2$ or $4$ . The first case gives $n = 2^4 \cdot 5^1 \cdot p$ where $p$ is a prime other than 2 or 5. Thus we have $80p < 2019 \implies p = 3, 7, 11, 13, 17, 19, 23$ . The sum of all such $n$ is $80(3+7+11+13+17+19+23) = 5760$ . In the second case $b+1 = 4$ and $d(k) = 1$ , and there is one solution $n = 2^4 \cdot 5^3 = 2000$ .

If $a+1 = 10$ , then $b+1 = 2$ , but this gives $2^9 \cdot 5^1 > 2019$ . No other values for $a+1$ work.

Then we have $\frac{S}{20} = \frac{80(3+7+11+13+17+19+23) + 2000}{20} = 372 + 100 = \boxed{472}$ .

-scrabbler94

2019 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2019 AIME II Problems/Problem 9

Problem 9

Solution

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