Problem
NT-B2-M05-P007 Power of \(12\)
#7
★★★☆☆ Level 3 of 5
Find the largest \(k\) such that \(12^k\mid100!\).
Factor \(12=2^2\cdot3\).
We need \(2k\le v_2(100!)=97\) and \(k\le v_3(100!)=48\). Hence \(k\le48\) in both cases, and the answer is \(48\).
Typical example of taking the minimum over prime divisors.