Question

What is k if : k!+48=48((k+1)^m)

Answer 1

Answer:

The value of k is greater than or equal to 0, i.e. k≥7.

Step-by-step explanation:

The given equation is

$k!+48=48((k+1)^m)$

The value of k must be a positive integer because k! is defined for k≥0, where k∈Z.

Subtract 48 from both the sides.

$k!=48((k+1)^m)-48$

$k!=48((k+1)^m-1)$

$k!=48(k+1-1)(\frac{(k+1)^m-1}{(k+1)-1})$

Using $[\frac{r^m-1}{r-1}=r^{m-1}+r^{m-2}+...+1]$, we get

$k!=48k((k+1)^{m-1}+(k+1)^{m-2}+...+1)$

Divide both sides by 48k.

$\frac{k!}{48k}=(k+1)^{m-1}+(k+1)^{m-2}+...+1$

$\frac{k(k-1)!}{48k}=(k+1)^{m-1}+(k+1)^{m-2}+...+1$

$\frac{(k-1)!}{48}=(k+1)^{m-1}+(k+1)^{m-2}+...+1$

Note: The value of m can be 0 or 1.

The value of k is positive integer, so the right hand side of the above equation must be a positive integer.

Since RHS of the equation is positive integer, therefore (k-1)! is completely divisible by 48.

$k-1\geq 6$

Add 1 on both sides.

$k\geq 6+1$

$k\geq 7$

Therefore the value of k is greater than or equal to 0.