If $(2146!)$base $10$=$(x)$base$26$ then what will be the number of consequtive zeroes at the end of $'x'$?
I didn't have any idea about how to go ahead with this question, so this is what I saw in the solution
In base 2, 10 means 2
In base 3, 10 means 3
Similarly in base $26$, $26$ is obtained by multiplying $2$ and $13$
To see the number of zeroes in base 26 we need to see the number of 2s and 13s in 2146!. Why?
On
Because if $2$ appears in $2\,146!$ with the exponent, say, $132$, and $13$ appears in $2\,146!$ with the exponent, say $12$, then the answer will $12$. Just like if$$N=2^{13}\times5^6\times\text{(a number co-prime with $2$ and $5$)},$$then the number of $0$'s in $N$ will be $6$.
On
If we were asked to find the number of zeroes at the end of, say, $36!$ in decimal notation, that is the same as asking for the highest value of $k$ such that $10^k$ divides $36!$
Similarly for your question we need to find the highest value of $k$ such that $26^k$ divides $2146!$. Since $26=2\cdot 13$, we can find the answer for $2$ and $13$ separately and then take the lower value.
Clearly there will a far higher power of $2$ that divides $2146!$ than the maximum power of $13$, so we can focus on finding the multiplicity of $13$ in $2146!$.
As an illustration we can look at the number of zeroes at the end of $456!$ in base-$26$.
$\lfloor 456/13\rfloor = 35$ of the contributing numbers are divisible by $13$, and $\lfloor 456/13^2\rfloor = 2$ of those are divisible by $13^2=169$. None are divisible by $13^3=2197$.
So $13^{37} \mid 456!$ and $13^{38} \nmid 2146!$ and there are $37$ base-$26$ zeroes at the end of $456!$
Let $[x]$ denote the largest integer not exceeding $x.$ For $n\in \mathbb N$ and for prime $p$ the largest integer $k$ such that $p^k$ divides $n!$ is $$k=\sum_{j=1}^{\infty}[np^{-j}].$$ Note that only finitely many terms in the summation are non-zero, as $[np^{-j}]=0$ when $p^j>n.$
Consider the $n$ terms in the product $1\times ... \times n.$ To find $k:\;$ Rather than counting "$1$" for each $m\leq n$ that's divisible by $p$ but not by $p^2,$ and counting "$2$" for each $m\leq n$ that's divisible by $p^2$ but not by $p^3,$ et cetera, $instead$ we can count "$1$" for $every$ $m\leq n$ that's divisible by $p$ (giving $[np^{-1}]$ of them), and count $1$ more for each $m\leq n$ that's divisible by $p^2$ (giving $[np^{-2}]$ of them), et cetera.
The largest power of $13$ that divides $2146!$ is $13^{177}$ because $[2146/13]+[2146/13^2]=165 +[(165+1/13)/13]=165+12,$ and $j\geq 3\implies [2146/13^j]\leq [2146/13^3]=[(165+1/13)/13^2]=0.$
The largest power of $2$ that divides $2146!$ is greater than $2^{177}$. So the largest power of $26$ that divides $2146!$ is $26^{177}.$ So in base $26$ the number $2146!$ ends in exactly $177$ zeroes.