Mathematics Olympiad Preparation

If all tokens were distinct, n tokens could be arranged in n! ways. Swapping tokens of the same colour does not create a new sequence. Divide the total factorial by the factorial for each repeated colour: n! / (a!b!c!).

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.

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 number is divisible by 9 when the sum of its digits is divisible by 9. Add the visible digits, then find the digit from 0 to 9 that makes the total the next multiple of 9.

You can compare the fractions without converting them to decimals. For ^a⁄_b and ^c⁄_d, compare a · d with c · b. The larger cross-product identifies the larger fraction.

One move changes the number of “+” cards by −2, 0 or +2, so its parity never changes. Compare the initial parity with the two possible final states: all “−” means 0 plus cards, while all “+” means as many plus cards as there are cards. A matching parity means that final state can be reached.

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.

Begin with the single polygon region. Every side or diagonal corresponds to a pair of vertices, and each interior intersection of two diagonals corresponds to four selected vertices. The number of regions is C(n, 4) + C(n, 2) − n + 1.

Any permitted digit may occupy the first position. Each next position has one fewer choice because digits cannot repeat. Multiply the numbers of choices for all positions.

Start with numbers having the first remainder: ra, ra + a, ra + 2a, and so on. Test them in order with the second divisor. The first number that also has the second stated remainder is the least solution.

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.

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.

Check every number in each set by dividing it by the given divisor or using a divisibility rule. A set is correct only if none of its numbers leaves a remainder.

The first cut adds 1 new piece. Each later cut is split by all previous cuts into one more segment than the number of earlier cuts, and every segment creates a new piece. Add 1 + 2 + … + n = n(n + 1) / 2 to the original single piece.

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.

Consider the left edge. It can be covered by one vertical domino, leaving a 2 × (n − 1) board, or by two horizontal dominoes, leaving a 2 × (n − 2) board. Thus F(n) = F(n − 1) + F(n − 2), with F(1) = 1 and F(2) = 2.

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 shortest route has a fixed number of right and upward steps. Arrange all steps in one sequence and choose the positions occupied by one direction. The number of choices is n! / (k!(n − k)!).

From the equation, y = total − x. The upper bound on y creates another lower bound on x. Compare it with the stated bound for x, then count the permitted integer values of x up to the total. Each value of x determines exactly one y.

The first panel may use any colour. Every later panel may use any colour except the one immediately before it. With k colours and n positions, calculate k · (k − 1)^{n − 1}.