2023 SSMO Accuracy Round Problems/Problem 2

Problem

Suppose that the average of all $n$ -digit palindromes is denoted by $P_{n}$ and the average of all $n$ -digit numbers is denoted by $N_{n}.$ Find $\left\lfloor\sum_{n=1}^{100}(P_{n}-N_{n})\right\rfloor.$

Solution

The outermost digits of an $n$ -digit palindrome can range from $1$ to $9$ , each with equal probability (notice that they must be equal due to being a palindrome, so the ones digit cannot be $0$ ), so the average is $5$ . The inner digits can range from $0$ to $9$ , again with equal probability, so their average is $4.5$ . Thus $P_n=5(10^{n-1}+1)+4.5(10+10^2+\ldots+10^{n-2})$ .

However, the $n$ -digit numbers range in the exact same way except that the ones digit can range from $0$ to $9$ . Thus $N_n=5(10^{n-1})+4.5(1+10+10^2+\ldots+10^{n-2})$ .

Then, \begin{align*} P_n-N_n&=5(10^{n-1}+1)+4.5(10+10^2+\ldots+10^{n-2})-5(10^{n-1})-4.5(1+10+10^2+\ldots+10^{n-2})\\ &=5(1)-4.5(1)\\ &=0.5 \end{align*}

However, we must consider one special case: $n=1$ . Here, $0$ is an $n$ -digit number, so the difference between $P_1$ and $N_1$ is $0$ (they are the same set). For all $n>1$ the difference is $0.5$ ; therefore, $\left\lfloor\sum_{n=1}^{100}(P_{n}-N_{n})\right\rfloor=\lfloor0+99\cdot0.5\rfloor=\lfloor49.5\rfloor=\boxed{49}$

~ eevee9406

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2023 SSMO Accuracy Round Problems/Problem 2

Problem

Solution