Prove that $2024!$ is divisible by a) $2024^2$; b) $2024^8$. ($n!=1\cdot 2 \cdot 3 \cdot ... \cdot n$) Z.Smysl
Problem
Source: Assara 2024 J5
Tags: number theory
lksb
15.10.2024 00:52
$2024=2^{3}\cdot 11\cdot 23$, as $2024\mid2023!$, $2024^2\mid2024!$
$2024=2^3\cdot11\cdot23\implies 2024^8=2^{24}\cdot 11^8 \cdot 23^8$
By Polignac's formula, we have that:
$v_2(2024!)=1012+506+253+126+63+31+15+7+3+1=2017$
$v_{11}(2024!)=184+16+1=201$
$v_{23}(2024!)=88+3=91$
Therefore, $2024^8\mid2024!$, in fact, $2024^{91}\mid2024!$
megarnie
15.10.2024 01:02
It suffices to prove part b. Note that \[ 2024^8 = (1 \cdot 2024) \cdot (2 \cdot 1012) \cdot (4 \cdot 506) \cdot (8 \cdot 253) \cdot (11 \cdot 184) \cdot (22 \cdot 92) \cdot (23 \cdot 88)\cdot (44 \cdot 46) \mid 2024!\]
AshAuktober
15.10.2024 05:13
Legendres formula works.
k2i_forever
15.10.2024 07:45
b) V_2(2024!)=1012+506+253+126+63+31+15+7+3+1=2017 V_2(2024^8)=24 V_11(2024!)=184+16+1=201 V_11(2024^8)=8 V_23(2024!)=88+3=91 V_23(2024^8)=8 Thus Trivial