Product of digits

We can generalise this problem. Let $S(n)$ be the sum over n-digit numbers.

For 1-digit numbers, the solution is: $\sum_{i=0}^{9}i=\frac{9(9+1)}{2}=45$ .

For n-digit numbers we have: $S(n)=(\sum_{i=1}^{9}i) S(n-1)=45 S(n-1)$ because all n-digit numbers are equivalent to the 2-digit numbers with the digits 1 to 9 on the front.

So we therefore have that $S(n)=45^n$ which we can show by induction.

Back to the question: $A=S(2)=45^2,B=S(3)=45^3 \Rightarrow \frac{B}{A}=\frac{45^3}{45^2}=45$

Let us first find out individual value of $A$ and $B$

$A=\sum_{n=10^1}^{10^2}f(n)=(1 \times 1)+(1 \times 2)...+(2 \times 1)+(2 \times 2)...$

Taking the leading digits common from $1$ to $9$

$A=1(\sum_{i=1}^{9}i)+2(\sum_{i=1}^{9}i)+3(\sum_{i=1}^{9}i)+4(\sum_{i=1}^{9}i)+5(\sum_{i=1}^{9}i)+6(\sum_{i=1}^{9}i)+7(\sum_{i=1}^{9}i)+8(\sum_{i=1}^{9}i)+9(\sum_{i=1}^{9}i)$

$A=(\sum_{i=1}^{9}i)(1+2+3+4+5+6+7+8+9)=45\times45=45^2$

Now finding $B$ , we will remove number from 100 to 111 as all numbers in between will have a zero at $10_{th}$ place. Similarly we will remove numbers between 200 and 201, 300 and 301, 400 and 401 and so on

$B=\sum_{n=10^2}^{10^3}f(n)=(1\times f(11))+(1\times f(12))...+(2\times f(11))+(2\times f(12))...$

$B=1(f(11)+f(12)+f(13)...f(99))+2(f(11)+f(12)+f(13)...f(99))+3(f(11)+f(12)+f(13)...f(99))...$

$B=(f(11)+f(12)+f(13)...f(99))(1+2+3+4+5+6+7+8+9)=(45^2)(45)=45^3$

On dividing A and B

$\frac{B}{A}=\frac{45^3}{45^2}=\boxed{45}$