Freshman’s Product

If $a,b$ are two nonnegative integers with decimal representations $a=(\dots a_2a_1a_0)$ and $b=(\dots b_2b_1b_0)$ respectively, then the freshman’s product of $a$ and $b$, denoted $a\boxtimes b$, is the integer $c$ with decimal representation $c=(\dots c_2c_1c_0)$ such that $c_i$ is the last digit of $a_i\cdot b_i$.
For example, $234 \boxtimes 765 = 480$.

Let $F(R,M)$ be the sum of $x_1 \boxtimes \dots \boxtimes x_R$ for all sequences of integers $(x_1,\dots,x_R)$ with $0\leq x_i \leq M$.
For example, $F(2, 7) = 204$, and $F(23, 76) \equiv 5870548 \pmod{ 1\,000\,000\,009}$.

Find $F(234567,765432)$, give your answer modulo $1\,000\,000\,009$.

To solve this problem, we need to understand a few important things about the question and the provided examples.

At first, it appears daunting due to the large numbers, but understanding the structure can bring us towards the solution.

Understanding Freshman’s Product:
In the Freshman’s product, we are only concerned about the last digit of the product of corresponding digits in two numbers. In other words, c_i = a_i * b_i (mod 10), where a_i and b_i denote the ith digits of numbers ‘a’ and ‘b’.

Understanding the Function F(R, M):
F(R, M) sums the Freshman’s product of all sequences of integers from 0 to M, for R digits. In other words, we form all possible numbers with R digits from 0 to M and compute their Freshman’s product, and we sum all products at the end.

Solving F(R, M):
Consider we have two R-digit numbers, xx..x (with a total of R ‘x’ digits) and yy..y respectively, and we calculate the Freshman’s product (say P) of these two.
– Each x could take any value from 0 to M.
– Each y could also take any value from 0 to M.
– However, we are interested in the last digit of x*y only.

Note that since multiplication table is periodic modulo 10, we count the occurrence of each digit (from 0 to 9) as the last digit when two numbers are multiplied. Since ‘x’ ranges from 0 to M, then the last digit could be any of 0 to M when multiplied by any digit. However, the digits 0 to M could cycle multiple times if M >= 10, hence we must take M min 9 when calculating. This is because all digits from 0 to M mod 10 will appear equally often, so we can simply take the average product for each digit.

Sum(d=0 to M mod 10) [d*average(0 to 9)] * [M/10 + 1] (if d <= M mod 10) Sum(d=0 to M mod 10) [d*average(0 to 9)] * [M/10] (if d > M mod 10)

For ‘P’, such computation is done independently for each digit. Since we have R such digits, and there are 10^(R-1) such R-digit numbers, we have:
P = R * 10^(R-1) * Sum_{d=0..M mod 10} weight[d] * d

Finally, we find sum of ‘P’ over all such unique pairs of R-digit numbers. There are (1 + M) ^ R such numbers, hence we choose two, so total sum = 0.5 * ((1 + M)^R) * ((1 + M)^R – 1) * P.

Keeping the result modulo 1,000,000,009, you can calculate F(234567, 765432). The exact calculation is computationally intensive and is best left to a computer program.

This method uses combinatorial counting principles and number theory (modular arithmetic) to arrive at a formula for F(234567, 765432) mod (1,000,000,009). The final calculation is best performed by a computer, as it requires handling large numbers.

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