Tasty Logarithm 6

Option 1:

Let us begin by understanding $\lambda(x)$ , specifically $\lambda(10)$ .

To find all integer divisors of 10, we factor it into primes, $10 = 2^1 5^1$ . Every divisor of 10 will have the form $2^a 5^b, \ 0 \leq a,b \leq 1$ . In general, every divisor of $10^n$ will have the form $2^a 5^b, \ 0 \leq a,b \leq n$ .

Thus we may describe $\lambda(10) = \lambda(2^1 5^1) = (2^0 5^0) (2^0 5^1) (2^1 5^0) (2^1 5^1) = 2^2 5^2 = 10^2.$

More generally, $\lambda(2^a 5^b) = \begin{array}{c c c c c c c c} 2^0 5^0 & \cdot & 2^0 5^1 & \cdot & 2^0 5^2 & \cdots & 2^0 5^b & \cdot\\ 2^1 5^0 & \cdot & 2^1 5^1 & \cdot & 2^1 5^2 & \cdots & 2^1 5^b & \cdot\\ \vdots & & \vdots & & \vdots & \ddots & \vdots \\ 2^a 5^a & \cdot & 2^a 5^1 & \cdot & 2^a 5^2 & ... & 2^a 5^b & \\ \end{array}$

Thus, $\lambda(2^a 5^b) = [(2^0)^{b+1}(2^1)^{b+1}...(2^a)^{b+1}] \cdot [(5^0)^{a+1}(5^1)^{a+1}...(5^b)^{a+1}] \\ = 2^{(0 + 1 + 2 + ... + a)(b+1)} \cdot 5^{(a+1)(0 + 1 + 2 + ... + b)} \\ = 2^{a(a+1)(b+1)/2} \cdot 5^{(a+1)b(b+1)/2}.$

$\lambda(2^n 5^n) = \lambda(10^n) = 2^{n(n+1)^2/2} \cdot 5^{n(n+1)^2/2} = 10^{n(n+1)^2/2}.$

So, $\lambda(\lambda(\lambda(10))) = \lambda(\lambda(10^{1(1+1)^2/2})) = \lambda(\lambda(10^2)) \\ = \lambda(10^{2(3)^2/2}) = \lambda(10^9) \\ = 10^{9(10)^2/2} = 10^{450}. QED.$

Option 2: Write python script. Sip martini.

First off, $\lambda(m)={m^{\frac{n(m)}{2}}}$ , where n(m) is the number of divisors of m. Check out @Calvin Lin 's proof of this equation here .

So $\lambda(10)=10^{\frac 4 2}=100.$

$\Rightarrow \lambda(100)=100^{\frac 9 2}=10^9$ .

Since $10^9=2^9\times5^9$ , $n(10^9)=(9+1)\times(9+1)=100$

$\Rightarrow \lambda(10^9)=(10^9)^{50}=10^{450}$

Hence $n=\boxed{450}.$

In response to challenge master:

This is how I see the proof of the first equation: $\lambda(m)={m^{\frac{n(m)}{2}}}$ , where n(m) is the number of divisors of m.

Let d be a divisor of m. Unless $d^2=m$ , there must be another divisor D such that $d\times D=m$ . So unless m is a square, the divisors of m can be paired up such that each pair's product is m. In this way, we can see that the product of divisors of m $={m^{\frac{n(m)}{2}}}$ . When m is a square, all the divisors other than the square root (let the square root be d) can be paired up and we'll be left with the product looking like this:

$m^{\frac{n(m)-1}{2}}\times d =m^{\frac{n(m)-1}{2}}\times m^{\frac 1 2} = m^{\frac{n(m)}{2}}$

As an example, consider 100 and its factors: 1, 2, 4, 5, 10, 20, 25, 50, 100. The product can be written as:

$(1\times100)(2\times50)(4\times25)(5\times20)\times10.$

So we can see that the divisors can be paired up and the square root (if it is an integer) stands alone. Hence the equation.

Python 2.7:

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def l(m):
    product = m
    if m == 1:
        return 1
    for test in xrange(2, m/2+1):
        if m % test == 0:
            product *= test
    return product
x = l(l(l(10)))
n = 1
while 10**n != x:
    n += 1
print "Answer:", n

I am defining $f(n)$ for that product function.

We can easily figure out that $f(10)=100$ .

We know $100$ has $9$ factors.Now we can for $4$ of $(n,q)$ such that $n*q=100$ with $n$ not equal to $q$ .

Now product of numbers satisfying these conditions is $10^{8}$ .

But we also have $10$ as a factor of $100$ which is not included in those pairs we formed.So final product is $10^{9}$ .i.e $f(100)=10^{9}$ .

Now we have to find $f(10^{9})$ .

We can find number of factors of $10^{9}$ as $100$ .As no. of factors of $10^{9}$ is even i.e $100$ we can form $50$ pairs such that $x,y$ having $x*y=10^{9}$ .

As there are $50$ pairs we have our answer as $10^{450}$ .