Bunuel wrote:
For positive integers n and m, is m! < 3^n ?
(1) n = m
(2) n > 3
Is \(m!<3^n=>\frac{m!}{3^n}<1\) -------(1)
now \(m!=m(m-1)(m-2)(m-3)....3*2*1\) and \(3^n=3*3*3\)..... to n terms
so \(\frac{m!}{3^n}=\frac{m(m-1)(m-2)(m-3)....3*2*1}{3*3*3*3.....n terms}\)
Notice that in the above fraction only numerators 2 & 1 are less than 3 and rest all numerators will be either greater than 3 or equal to 3. So if in the numerator we have a number, such as 9, that can take care of the two 3's in the denominator then equation (1) will be definitely greater than 1
Statement 1: \(n=m=4\), then equation
then \(\frac{m!}{3^n}=\frac{4*3*2*1}{3*3*3*3}\), clearly \(\frac{4}{3}\) is marginally greater than \(1\), \(\frac{3}{3}=1\) but \(\frac{2}{3}\) & \(\frac{1}{3}\) are less than \(1\). Hence this fraction will be definitely less than \(1\). So we have a
Yes for equation (1). but if \(n=m=10\)
then \(\frac{m!}{3^n}=\frac{10*9*8*7*6*5*4*3*2*1}{3*3*3*3*3*3*3*3*3*3}\), clearly \(10\) & \(9\) will take care of 3's corresponding to numerators \(2\) & \(1\). Hence this number will definitely be greater than \(1\). so we have a
No for equation (1).
InsufficientStatement 2: nothing mentioned about \(m\).
InsufficientCombining (1) & (2): As we can see from statement 1, we will have both the scenarios, hence
InsufficientOption
E