A tribute to Ramanujan

Let $1729^{1729^{1729}}=k$ Given series can be written as:-

$\mathsf{172917291729+\sum_{n=1}^{k} \lfloor \frac{172917291729}{5^n} \rfloor}......(1)$

Now observe the term $\sum\frac{blah}{5^n}$ in equation (1) deeply for a minute, it is actualy counting the largest power of $5$ which can divide $172917291729!$ without leaving any remainder .Now we will use the following well known result:-

For an integer $j$ and a prime $p$ largest power of $p$ which can divide $j!$ without leaving any remainder is :- $Ord_p(j!) = \frac{j-\sigma_p(j)}{p-1}$

Note that the notation $Ord_p(j!)$ is used to represent largest power of $p$ which can divide $j!$ and $\sigma_pn$ denotes the sum all digits of $j$ when expressed in base $p$

Now we are in a better situation because now can calculate the value of $\sum\frac{blah}{5^n}$ easily by using this theoram

Hence :-

$\mathsf{\sum_{n=1}^k \lfloor \frac{172917291729}{5^n}\rfloor = \frac{172917291729-\sigma_5(172917291729)}{4}}....(2)$

Now:-

$\mathsf{172917291729 = \overline{(10313113311313404)_5}}$ Hence $\sigma_5(172917291729)=33$ ,putting this value in equation (2) we get:-

$\mathsf{\sum_{n=1}^k \lfloor \frac{172917291729}{5^n}\rfloor = \frac{172917291729-33}{4}}$

And finaly putting this value in equation (1):-

$\sum_{n=0}^k\lfloor\frac{172917291729}{5^n}\rfloor = 216146614653$

Hence $\lambda = 216146614653$

Also $\lambda = 125012501×1729+424$

Hence

$\boxed{\lambda\mod(1729)=424}$

Also note that we don't need to worry about the value of $k$ because when value of $5^n$ will grow larger then $172917291729$ each floor operation will yeild the answer 0 so it does not effect our result

If you know any other mathod for solving this question ,please post it.

Once n is sufficiently large all the terms become zero. Also the fact that we need to find $\lambda \pmod{1729}$ simplifies the work. We can immediately throw out n=0.

Note that $172917291729=100010001 \cdot 1729$ .

So for the first four terms

$\lfloor \dfrac{172917291729}{5^n} \rfloor \equiv \lfloor \dfrac {1729} {5^n} \rfloor \pmod{1729}$

And for terms five to eight

$\lfloor \dfrac{172917291729}{5^n} \rfloor \equiv \lfloor \dfrac {17291729} {5^n} \rfloor \pmod{1729}$

And now we can simply calculate these terms, getting: 345 +69 +13 +2 +5533 +1106 +221 +44 +88533 +17706 +3541 +708 +141 +28 +5 +1=117996 $117996 \equiv 424 \pmod{1729}$