Hard problem with fractions I can't solve the following problem.
A person is $x$ years old. Find his age if the following is true. In a group of $x$ people each one started taking pictures of each of the others. At some point we know that more than $\frac{1}{2}$ of the people has taken exactly $\frac{1}{2}$ of all of their pictures (which are $x-1$ for every one of them), at the same time more than $\frac{1}{3}$ of them has taken exactly $\frac{1}{3}$ of all of their pictures, and at the same time more than $\frac{1}{7}$ of them has taken exactly $\frac{1}{7}$ of their pictures.
 A: From the pictures taken we know 2, 3 and 7 are all divisors of $x-1$. The smallest number for which this true is 2*3*7 = 42. Hence the smallest $x$ is 43. The next possible $x$ would be 2 * 42 + 1 = 85.
Now $\frac{1}{2}$ of 42 is 21, $\frac{1}{3}$ of 42 is 14 and $\frac{1}{7}$ of 42 is 6. Since more than half,a third, a seventh of the people have taken their pictures, the minimum number of people would be $(21 + 1) + (14 + 1) + (6 + 1) = 45$ which excludes 43 as a solution.
In case of $x = 85$ we get $(42 + 1) + (28 + 1) + (12 + 1) = 85$, which suits perfectly.
I guess we can safely assume the person isn't 127 years old.
A: We know that for the fractions given to be exact, $x-1$ must be divisible by $2$, $3$, and $7$. This leaves $42$, $84$ for $x-1$. Anything larger or smaller than that couldn't reasonably correspond to the age of a person and the size of a group. Let's test $x=43$. The smallest integers bigger than $\frac{43}{7}$, $\frac{43}{3}$, and $\frac{43}{2}$ are $7$, $15$, and $22$. However, if we add these up, we get $44$; that's bigger than $43$, we can't split a group of size $43$ up and still satisfy the problem.
Now let $x=84$. The smallest numbers bigger than $\frac{85}{7}$, $\frac{85}{3}$, and $\frac{85}{2}$ are $13$, $29$, and $43$. They add to $85$.
Therefore, if we let $13$ people shoot $12$ photos, $29$ people shoot $28$ photos, and $43$ people shoot $42$ photos, the criteria in the problem are satisfied. The man will then be $85$ years old.
