IMO training problem

Consider each positive integer less than 1000 to be a three-digit number by prefixing 0s to numbers with fewer than 3 digits. The sum of the products of the digits of all such positive numbers is

$(0 \cdot 0 \cdot 0 + 0 \cdot 0 \cdot 1 + ... + 9 \cdot 9 \cdot 9) - 0 \cdot 0 \cdot 0$

Note that each product appears once, thrice or 6 times depending on the number of repeated digits it has, so we get that the product is $=(0 + 1 + ... + 9)^3 - 0$ .

However, $p(n)$ is the product of non-zero digits of $n$ . The sum of these products can be found by replacing 0 by 1 in the above expression, since ignoring 0's is equivalent to thinking of them as 1's in the products. (Note that the final 0 in the above expression becomes a 1 and compensates for the contribution of 000 after it is changed to 111.) Hence

$S = 46^3 - 1 = (46 - 1)(46^2 + 46 + 1) = 3^3 \cdot 5 \cdot 7 \cdot 103$

and the largest prime factor is 103.

Lets rewrite the first $99$ and then we will multiply that by $1$ , then by $2$ , $\ldots$ , then by $9$ . Lets say that

$S_{1} = [(1 + 2 + \ldots + 9) + 1 + 1(1 + 2 + \ldots + 9) + 2 + 2(1 + 2 + \ldots + 9) + \ldots + 9 + 9(1 + 2 + \ldots + 9)]$

This is equivalent to

$S_{1} = 1(45) + 1(46) + 2(46) + \ldots + 9(46)$

$S_{1} = 45(47)$

$S_{1}$ is the sum from $1$ to $99$ . We'll have extra numbers in our sum ( $100$ , $200$ , \ldots, $900$ ), and hence when we multiply $45\cdot 47$ by the hundreds digit of the numbers, we need to add $1$ . So, it's the same than multiplying $46\cdot 46$ which is equal to $45\cdot 47 + 1$ . So,

$S = S_{1} + (S_{1} + 1) + 2(S_{1} + 1) + \ldots + 9(S_{1} + 1)$

$S = 2115 + 45(2116)$

We can factorize carefully in many ways, but i like this one:

$S = 2115(46) + 45 = 9(235\cdot 46 + 5)$

$S = 3^{3} \cdot 5 \cdot 7 \cdot 103$

And therefore, the largest prime factor is:

$\boxed {103}$ .

When $0$ digits are equal to $0$ , we sum through $\sum_{a=1}^9\sum_{b=1}^9\sum_{c=1}^9abc=45^3$ . When $1$ digit is equal to $0$ , we sum through $\sum_{a=1}^9\sum_{b=1}^9ab=45^2$ , but there are three cases, accounting for $3\cdot45^2$ . When there are $2$ digits, it's $3\cdot45$ . All $3$ cannot be $0$ . Hence, the sum is $45^3+3\cdot45^2+3\cdot45=46^3-1$ which has the largest prime factor of $\boxed{103}$ .

Let $k\in\mathbb{N}$ , and $N_k := \{10^{k-1};\:\ldots;\: 10^k-1\}$ be the set of positive integers with exactly $k$ digits. Define

$\begin{aligned} S_k &:=\sum_{i=1}^{10^k-1}p(i), &&& S_1 &=\sum_{i=1}^9 i= \frac{9\cdot 10}{2}=45&&&&&\left|\:\text{We need to compute }S_3\right. \end{aligned}$

Check for a recursive relation. Notice any integer in $N_{k+1}$ consists of a $\text{\yellow{non-zero first digit}}$ , followed either by an integer in $\green{N_k}$ , or $\red{0}$ : $\begin{aligned} S_{k+1} &= S_k + \sum_{i\in N_{k+1}} p(i) = S_k + \left(\yellow{1 + \ldots + 9}\right) \cdot \left(\green{S_k} + \red{1}\right)=S_k+45(S_k+1)=46S_k + 45 \end{aligned}$

By induction, we get $\begin{aligned} S_k &= 46^k - 1, &&&&& S_3 &=46^3-1=97335=3^3\cdot 5\cdot 7 \cdot\boxed{103} \end{aligned}$

Let x,y,z be nonzero digits.

$\sum_{x=1}^{9} \sum_{y=1}^{9} \sum_{z=1}^{9}p(\overline{xyz})=45^3$

$\sum_{x=1}^{9} \sum_{y=1}^{9}  p(\overline{xy0})= \sum_{x=1}^{9} \sum_{z=1}^{9}  p(\overline{x0z})= \sum_{y=1}^{9} \sum_{z=1}^{9}  p(\overline{yz})=45^2$

$\sum_{x=1}^{9}  p(\overline{x00})= \sum_{y=1}^{9}  p(\overline{y0})= \sum_{z=1}^{9}  p(\overline{z})=45$

$45^3+3×45^2+3×45=45(45^2+3×46)=45×2163=3^3×5×7×\boxed{103}$

This is quite a brute force ansatz with code for matlab. The final sum will be 97335 which can be found to have 103 as it's largest primefactor.

The variables stand for the german words for

h = Hunderter = hundreds

z = Zehner = tens

e = Einser = ones

finalsum = 0;
tempsum = 0;
for h = 0:9
for z = 0:9
    for e = 0:9
        if h == 0
            if z == 0
                tempsum = e;
            elseif z > 0
                if e == 0
                    tempsum = z;
                elseif e > 0 
                    tempsum = z*e;
                end
            end
        elseif h > 0
            if z == 0
                if e == 0
                    tempsum = h;
                elseif e > 0 
                    tempsum = h*e;
                end
            elseif z > 0
                if e == 0
                    tempsum = h*z;
                elseif e > 0 
                    tempsum = h*z*e;
                end
            end
        end
        finalsum = finalsum + tempsum;
    end
end
end
finalsum