Mathematics Olympiad Preparation

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.

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.

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.

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.

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).

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.

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.

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.

After the discount, 100% − p% of the original price remains. Express that remaining percentage as a decimal and divide the final price by it. Do not simply add the percentage to the final price, because the discount was based on the original price.

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.

A palindrome is determined completely by its first half, including the middle digit when the length is odd. The first digit has d choices because zero is forbidden. Every other free position has d + 1 choices. The second half is then fixed by reflection.

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.

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).

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.

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.

Imagine every number written with the same length by adding leading zeroes. In each digit position, 0, 1, …, 9 occur equally often. One complete digit cycle sums to 45. Multiply this by the number of repetitions in each position and by the number of positions.

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.

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.

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.

Try the number of €5 coins from zero through the greatest possible value. For each choice, check whether the remaining amount is divisible by 2. Every successful choice gives one different payment method.