A closed form for the sum $$ s = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $$ is $1 - \frac{a^{n
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For the product of the digits to be divisible by 10, it must have 10 as a factor, of course. Since 10 can't act as a digit, this means two of our digits must be 2 & 5.
No digit can be 0 because this would give a total product of 0.
Neither 2 nor 5 can be our largest because the lowest you can get is 1+2+3=6.
The lowest possible largest digit would have to be 1+2+5=8.
As for 9, this is not an option, and here's why: We need that 2 and 5, but to get 9 our other digit would have to be 2, which is a repeat.
So, our digits are 1, 2, 5, and 8.
It doesn't matter which order they are in. (take a look at the rules)
Therefore, we should put the highest digits in the highest value places.
The answer is
No digit can be 0 because this would give a total product of 0.
Neither 2 nor 5 can be our largest because the lowest you can get is 1+2+3=6.
The lowest possible largest digit would have to be 1+2+5=8.
As for 9, this is not an option, and here's why: We need that 2 and 5, but to get 9 our other digit would have to be 2, which is a repeat.
So, our digits are 1, 2, 5, and 8.
It doesn't matter which order they are in. (take a look at the rules)
Therefore, we should put the highest digits in the highest value places.
The answer is