Problem 6

Determine the least possible value of

f(1998)

, where

f:\mathbb{N}\to\mathbb{N}

is a function such that for all

m,n\in\mathbb{N}

f\left(n^{2}f(m)\right)=m\left(f(n)\right)^{2}.

Step 6 of 6: Minimize the target value

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In plain words

Pair the largest exponent with the smallest available prime.

1998=2\cdot3^3\cdot37,\qquad f(1998)\ge3\cdot2^3\cdot5=120

Detailed analysis

To minimize the weighted product under a prime involution, pair $2$ with $3$ and pair $37$ with the smallest remaining prime $5$ . This gives $f(2)=3$ , $f(3)=2$ , $f(37)=5$ , so $f(1998)=3\cdot2^3\cdot5=120$ . Any other pairing assigns a larger available prime to one of the weights and cannot improve the product. Thus the least possible value is $120$ .