Authors: $\underline{\text { Muraykin M.V., Berlov S.L., Bogdanov I.i. }}$

There are three commissions of bureaucrats. It is known that for each pair of bureaucrats from different commissions, among the members of the remaining commission, there are exactly 10 bureaucrats who are acquainted with both, and exactly 10 bureaucrats who are not acquainted with both. Find the total number of bureaucrats in the commissions.
Consider a graph where the vertices are bureaucrats, and two bureaucrats from different commissions are connected by a red edge if they are acquainted, and by a blue edge otherwise. Let the number of bureaucrats in three commissions be \(a\), \(b\), and \(c\). Consider any two bureaucrats \(A\) and \(B\) from the first two commissions. If they are acquainted, there are exactly 10 triangles \(A B C\) where all edges are red. Similarly, if \(A\) and \(B\) are not acquainted, there are exactly 10 triangles \(A B C\) where all edges are blue. Therefore, the total number of monochromatic triangles is \(10 a b\). Similarly, it is also \(10 a c\) and \(10 b c\), so \(a = b = c\).

For three bureaucrats \(A\), \(B\), and \(C\) from different commissions, we say that \(B\) and \(C\) are similar for \(A\) if \(A\) is connected to \(B\) and \(C\) by edges of the same color. For each bureaucrat, we find the number of pairs similar to him and calculate the sum \(s\) of all these numbers in two ways.

On the one hand, each pair of bureaucrats \(B\) and \(C\) from different commissions is similar to exactly 20 bureaucrats; there are a total of \(3 a^2\) such pairs, so

\[ s = 60 a^2 \]. On the other hand, in any monochromatic triangle \(A B C\), every two bureaucrats are similar to the third; if the triangle \(A B C\) is multicolored (for example, the edge \(A B\) differs in color from the others), then there is exactly one pair \((A, B)\) similar to the third. Since the number of monochromatic triangles is \(10 a^2\) and the number of multicolored triangles is \(a^3 - 10 a^2\), then \(s = 30 a^2 + (a^3 - 10 a^2)\). Therefore, \(60 a^2 = a^3 + 20 a^2\), from which \(a = 40\).

## Answer

120 bureaucrats.