PUMaC 2021 · 团队赛 · 第 2 题

2. Let k ∈ Z be the smallest positive integer with the property that k is a positive 2

0 lcm( x,y ,z ) ′ integer for all values 1 ≤ x ≤ y ≤ z ≤ 121. If k is the number of divisors of k , find the number ′ of divisors of k . Proposed by: Frank Lu Answer: 174 We consider what this means with respect to a given prime power. Consider a prime power p. Notice then that, if v ( n ) is the power of p in the prime factorization of n, we have that v ( k )+ p p min( v ( x ) , v ( y )) + min( v ( y ) , v ( z )) − max( v ( x ) , 2 v ( y ) , v ( z )) ≥ 0 . Now, if we can force the p p p p p p p powers of v ( y ) to be maximal, and v ( x ) , v ( z ) to be minimal, we’re done. Now, we have 30 p p p primes less than 121 : 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 . Of these prims, notice that all but the first 5 contribute 3 divisors for each to the value of k. As for the last, notice that 2 contributes 2 · 6 + 1 = 13 , 3 contributes 9 , 5 , 7 contribute 5 . But for 11 , notice that either v ( y ) = 1 , meaning that we get a power p that is 0 + 0 − 2 = − 2 , or v ( y ) = 2 , yielding 0 + 2 − 4 = − 2; either way, we have that this p 26 2 28 2 ′ contributes 3 factors. Our answer is hence 3 · 13 · 9 · 5 = 3 · 5 · 13 for k . Thus, we see ′ that the number of divisors in k is equal to 29 · 3 · 2 = 174 . N 9 m