MathLabs

Problem 6

An international society has its members from six different countries. The list of members contains 19781978 names, numbered 1,2,…,19781,2,\ldots,1978. Prove that there is at least one member whose number is the sum of the numbers of two members from his own country, or twice as large as the number of one member from his own country.
Step 5 of 6: Finish with two differences
Previous stepNext step
In plain words

At the last stage, addition or doubling is unavoidable.

d1,d2 are in the sixth country,d2−d1 is in one of the six countriesd_1,d_2\text{ are in the sixth country},\quad d_2-d_1\text{ is in one of the six countries}
Detailed analysis

The final three numbers produce two positive differences d1,d2d_1,d_2 both in the last country. Their difference d2−d1d_2-d_1 is in one of the six countries. If it is in the last country, then d2=d1+(d2−d1)d_2=d_1+(d_2-d_1) gives the desired sum; if it is in an earlier country, unwinding the nested representations gives the same conclusion there. The equality d2=2d1d_2=2d_1 is exactly the allowed doubling case when d2−d1=d1d_2-d_1=d_1.