140 Followers Problem - 2

Probability Level 5

$\Large \lfloor \log (x) \rfloor + \lfloor \log (y) \rfloor + \lfloor \log (z) \rfloor = 140$

If the number of positive integral solutions to above equation can be expressed in the form of $a^{b}\times c^{d} \times e^{f} \times g^{h} \times i^{j}$ , where $a$ , $c$ , $e$ , $g$ and $i$ are distinct positive primes and $b$ , $d$ , $f$ , $h$ and $j$ are positive integers. Then find the value of

$a+b+c+d +e+f+g+h+i+j$

Details and Assumptions

$\lfloor k \rfloor$ represents the greatest integer less than or equal to $k$ .
$\log k$ is the logarithm of $k$ to the base 10.

Try its sister problem 140 Followers Problem .

This is my original problem.

This problem is from the set 140 Followers Problems .

The answer is 417.

2 solutions

Surya Prakash
Aug 28, 2015

The number of solutions to this equation is equivalent to finding the coefficient of $t^{140}$ in the expansion,

$(t^{\lfloor \log (1)\rfloor} + t^{\lfloor \log (2)\rfloor} + t^{\lfloor \log (3)\rfloor} + \ldots)^{3}$

$=(9+90t + 900t^{2} + 9000t^{3}+ \ldots)^{3}$

$=729 (1+10t + (10t)^2 + (10t)^3 + \ldots )^{3}$

$=729 \left( \dfrac{1}{1-10t} \right)^{3} \text{(Infinite G.P.)}$

$=729 (1-10t)^{-3}$

$=729 \left( 1+ \binom{3}{1}(10t) + \binom{4}{2}(10t)^{2} + \binom{5}{3}(10t)^{3} + \ldots \right)$

So the coefficient of $t^{140}$ in this expansion is

$=729 \times (10)^{140} \times \binom{142}{140}$

$=2^{140} \times 3^{7} \times 5^{140} \times 71^{1} \times 47^{1}$

Therefore, $a+b+c+d+e+f+g+h+i+j= \boxed{417}$ .

Moderator note:

I'm loving how you're converting these into generating functions, which makes the total much easier to calculate!

Calvin Lin Staff - 5 years, 9 months ago

thank you sir.

Surya Prakash - 5 years, 9 months ago

I've learnt a lot from the first problem he gave. Thanks :)

Figel Ilham - 5 years, 8 months ago

Srijit Chakrabarty
Aug 29, 2015

Now the initial problem is to partition 140 in 3 numbers where any perticular number can be 0. Now this is done in 142C2=141 71=3 47*71.
Let a solution triplet be {a,b,c} , then x is in the range $(10^{a}, 10^{a+1}-1)$ , similarly y is in the range $(10^{b}, 10^{b+1}-1)$ , and z is in the range $(10^{c}, 10^{c+1}-1)$
* Also (a+b+c=140) *
So the total number of Integral solutions are given by $142C2 * (10^{a}*(10-1)*10^{b}*(10-1)*10^{c}*(10-1))$
$= 47*71*3*10^{140}*9^{3}$ $=2^{140}*5^{140}*3^{7}*47^{1}*73^{1}$ .
The solution follows from this.