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$P(n)$ is the product of two digits in the integer $n$. For example,

$P(18)=1\cdot8=8$

$P(50)=5\cdot0=0$

$P(99)=9\cdot9=81$.

Is there any nice way to find the value of $P(10)+P(11)+P(12)+...+P(99)$ instead of counting one by one?

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Yes, your sum is $$ \sum_{i=1}^9\sum_{j=0}^9 P(\overline{ij}) = \sum_{i=1}^9\sum_{j=0}^9i\cdot j=\sum_{i=1}^9i\sum_{j=0}^9j=\sum_{i=1}^9 45i=45\cdot 45 = 2025. $$

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  • $\begingroup$ I don't understand this. Can you explain more for your answer? Thanks $\endgroup$ – MathLOL Jun 15 '15 at 17:18
  • $\begingroup$ The first sum should just go from $1$ to $9$, as the original question starts at $P(10)$. True, it adds $0$ to the sum, but it would be a bit easier to see. $\endgroup$ – Ross Millikan Jun 15 '15 at 17:28
  • $\begingroup$ @RossMilikan You are right, I'll edit it. $\endgroup$ – Pjotr5 Jun 15 '15 at 18:18

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