Problem 762

4. Since the number 1 is four times more frequent than the numbers that appear the least, the number of occurrences of the number 1 is a multiple of 4.

2 POINTS

If the number 1 appeared 8 times, then the numbers that appear the least would appear 2 times, and the numbers 7 and 3 would each appear 6 times. This would mean we already have 24 numbers, which is not possible.

1 POINT

It is easy to conclude that the number of these numbers would be even greater if the number of occurrences of the number 1 were a larger multiple of 4 (e.g., 12, 16,...).

1 POINT

This means that the number 1 appears 4 times, the two consecutive numbers appear 1 time each, and the numbers 7 and 3 appear 3 times each.

This gives 12 numbers on the flag, so each of the remaining numbers appears 2 times.

The two consecutive numbers with the least number of occurrences can be:

a) 4 and 5,

b) 5 and 6,

c) 8 and 9.

1 POINT

Case a) would give a sum of 93, case b) a sum of 91, and case c) a sum of 85.

1 POINT

Therefore, the number 1 appears 4 times, the numbers 7 and 3 appear 3 times each, the numbers 5 and 6 appear 1 time each, and the numbers 0, 2, 4, 8, and 9 appear 2 times each.

1 POINT 10 POINTS

5.

!

Let $|\angle BAC| = \alpha$.

Since point $S$ lies on the perpendicular bisector of side $\overline{AC}$, $|AS| = |CS|$, meaning triangle $\triangle ASC$ is isosceles.

This means $|\angle SCA| = |\angle CAS|$.

Triangle $\triangle AEC$ is a right triangle, so $|\angle ECA| + |\angle CAE| = 90^\circ$.

Thus, $\frac{\alpha}{2} + \alpha = 90^\circ$ or $\alpha = 60^\circ$.