Problem of the Day


If x,y, and z are chosen at random with replacement from the set \{31,32,33,...,n\}, what is the probability that xyz+xz+xy+x is divisible by 3?

(1) n=81m-39 for some positive integer m

(2) n=771

Reveal Answer

Answer

D. Each statement alone is sufficient.

See the Solution

Solution

[latexpage]

First, factor an $x$ out of $xyz+xz+xy+x$ to get $x(yz+z+y+1)$. Then note the expression inside the parentheses can be factored as $(y+1)(z+1)$, so the entire expression is now $x(y+1)(z+1)$. This will be divisible by 3 if and only if at least one of $x$, $y+1$, and $z+1$ is a multiple of 3.

Statement (1): SUFFICIENT. $81m-39=3(27m-13)$ so $n$ must be a multiple of 3. Because the series starts with a number that is 1 more than a multiple of 3, this ensures that there is an equal number of integers in the set that are divisible by three, one more than a multiple of 3, and two more than a multiple of three. So, the probability of a randomly selected number being in any given one of these sets is exactly $\frac{1}{3}$. The expression will be divisible by three if at least one of the following is true: $x$ is a multiple of 3, $y$ is 2 more than a multiple of 3,$z$ is 2 more than a multiple of 3. This will be true with probability $\frac{1}{3}+\frac{2}{3}\frac{1}{3}+\frac{2}{3}\frac{2}{3}\frac{1}{3}=\frac{19}{27}$

Statement (2) SUFFICIENT. The analysis for statement 1 applies here because 771 is a multiple of 3. We can tell this because its digits add to 15, which is a multiple of 3.


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