, 2016

Since $405=3^4\times5$, then $405$ is divisible by $3^4$ but is not divisible by $3^5$. Thus, $f(405)=4$. First, we find all factors of 3 which exist in the product $1\times2\times3\times4\times5\times6\times7\times8\times9\times10$. The multiples of 3 are the only numbers which contain factors of 3. The multiples of 3 in the given product are 3, 6 and 9. Rewriting the given product, we get $$\begin{align*} & & \hspace{-1cm}1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10\\ &=& 1 \times 2 \times 3 \times 4 \times 5 \times (2\times3) \times 7 \times 8 \times (3\times3) \times 10\\ &=& 3^4\times(1 \times 2 \times 4 \times 5 \times 2 \times 7 \times 8 \times 10).\end{align*}$$ Since the product in parentheses does not include any factors of 3, then the largest power of 3 which divides the given product is $3^4$, and so $f(1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10)=4$. First, we count the number of factors of 3 included in $100!$. Every multiple of 3 includes least 1 factor of 3. The product $100!$ includes 33 multiples of 3 (since $33\times3=99$). Counting one factor of 3 from each of the multiples of 3 (these are $3,6,9,12,15,18,\dots,93,96,99$), we see that $100!$ includes at least 33 factors of 3. However, each multiple of $3^2=9$ includes a second factor of 3 (since $9=3^2, 18=3^2\times2$, etc.) which was not counted in the previous 33 factors. The product $100!$ includes 11 multiples of 9 (since $11\times9=99$), and thus there are at least 11 additional factors of 3 in 100!. Similarly, $100!$ includes 3 multiples of $3^3=27$, each of which contribute an additional factor of 3 (these are $27=3^3, 54=3^3\times2$, and $81=3^4$). Finally, there is one multiple of $3^4=81$ which contributes one more factor of 3. Since $3^5>100$, then $100!$ does not include any multiples of $3^5$ and so we have counted all possible factors of 3. Thus, $100!$ includes exactly $33+11+3+1=48$ factors of 3, and so $100!=3^{48}\times t$ for some positive integer $t$ that is not divisible by 3. Counting in a similar way, the product $50!$ includes 16 multiples of 3, 5 multiples of 9, and 1 multiple of 27, and thus includes $16+5+1=22$ factors of 3. Therefore, $50!=3^{22}\times r$ for some positive integer $r$ that is not divisible by 3. Also, $20!$ includes $6+2=8$ factors of 3, and thus $20!=3^{8}\times s$ for some positive integer $s$ that is not divisible by 3. Therefore, $N=\dfrac{100!}{50!20!}=\dfrac{3^{48}\times t}{(3^{22}\times r)(3^{8}\times s)}=\dfrac{3^{48}\times t}{(3^{30}\times rs)}=\dfrac{3^{18}\times t}{rs}$. Since we are given that $N$ is equal to a positive integer, then $\dfrac{3^{18}\times t}{rs}$ is a positive integer. Since $r$ and $s$ contain no factors of 3 and $3^{18}\times t$ is divisible by $rs$, then it must be the case that $t$ is divisible by $rs$. In other words, we can re-write $N = \dfrac{3^{18}\times t}{rs}$ as $N = 3^{18}\times \dfrac{t}{rs}$ where $\dfrac{t}{rs}$ is an integer. Since each of $r$, $s$ and $t$ does not include any factors of 3, then the integer $\dfrac{t}{rs}$ is not divisible by 3. Therefore, the largest power of 3 which divides $\dfrac{100!}{50!20!}$ is $3^{18}$, and so $f(N)=18$. Since $f(a)=8$, then the exponent of the largest power of 3 that divides $a$ is 8. That is, $a=3^8m$ for some positive integer $m$ and 3 does not divide $m$. Since $f(b)=7$, then the exponent of the largest power of 3 that divides $b$ is 7. That is, $b=3^7n$ for some positive integer $n$ and 3 does not divide $n$. Substituting and simplifying, we get $$a+b=3^8m + 3^7n=3^7(3m+n)$$ Since 3 divides $3m$ but 3 does not divide $n$, then 3 does not divide the sum $3m+n$. That is, $3m+n$ is not a multiple of 3 and so the largest power of 3 that divides $a+b$ is $3^7$. Therefore, $f(a+b)=7$.