Mathematics Olympiad Preparation

One complete cycle lasts for the work time plus the charging time. Divide the total available time by the duration of one cycle and keep only the whole-number part, because an unfinished cycle does not count.

From each vertex, segments may be drawn to every other vertex except itself and its two neighbours, leaving n − 3 diagonals. This counts every diagonal from both endpoints, so divide by 2: n(n − 3) / 2.

Add the total distance moved to the starting position. Because the route is circular, divide by the number of stops and use the remainder. A remainder of zero corresponds to the final numbered stop.

A weight on the opposite pan increases the counterweight, while a weight beside the object reduces the counterweight needed. Seek an equality: object + weights with it = weights on the opposite pan. Verify by adding both sides.

Turn each statement into an arrow from the earlier drone to the later drone. Join all arrows into a single order of five drones. The drone in the third position is the answer.

List selected positions in increasing order. At least one unused position must separate every consecutive pair. Remove one required gap after each of the first k − 1 selections. The problem becomes choosing k positions from n − k + 1, giving C(n − k + 1, k).

A two-digit number with tens digit a and units digit b is 10a + b; its reversal is 10b + a. Their difference is 9(a − b). Divide the difference between the two numbers by 9.

Count the 1 × 1 squares and the 2 × 2 squares separately. A strip with two rows cannot contain larger squares. Add the counts for the two possible sizes.

We need n such that n(n + 1) equals the given product. The square root of the product lies between the two integers, so estimate it, take the nearby smaller integer and verify by multiplying it by the next integer.

The old total equals mean · number of results. There is no need to recalculate it fully: find the change in the replaced result and divide that change by the number of results. Add the resulting change in the mean to the old mean.

Count positions divisible by the first or the second step, but not by both. Lights reached by both robots are toggled twice and return to off. Therefore add the two individual counts and subtract twice the number of common multiples.

Every rectangle is determined by choosing two vertical and two horizontal grid lines. If there are v vertical and h horizontal lines, multiply the numbers of choices: v(v − 1) / 2 · h(h − 1) / 2.

In the worst case, one fewer than the required number can be drawn in every colour without completing a group. This gives number of colours · (required amount − 1). One additional bead must complete at least one group.

Count one-digit and two-digit page numbers separately. Pages 1 through 9 use 9 digits. Every page from 10 through the final page uses 2 digits. Add the two amounts.

Subtract those attending both from the first club, and do the same for the second. Add the sizes of the two resulting disjoint groups. In short: A + B − 2 · both.

Each match corresponds to one pair of distinct teams. With n teams, each team can be paired with n − 1 others, but that counts every pair twice. Therefore calculate n(n − 1) / 2.

You may add the sides of all squares and remove internal shared edges: every shared edge is subtracted twice from the initial total. Alternatively, trace only the outside boundary and count its unit segments.

Find the difference between the two amounts. Moving one cell from the larger group to the smaller reduces the difference by two: one side loses one and the other gains one. Therefore divide the original difference by 2.

Small cubes with exactly two painted faces lie on the edges of the large cube but not at its corners. Each of the 12 edges contains n − 2 such cubes. Therefore calculate 12(n − 2).

List possible pairs (first spinner; second spinner). Increase the first number one at a time and choose the second so the sum stays fixed. Count only pairs whose two numbers lie in the stated range. Because the spinners are distinct, reversing a pair gives a different outcome.