Mathematics Olympiad Preparation

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.

In a divisor, the exponent of the first prime may be selected from 0 through a, giving a + 1 choices. The second exponent may be selected from 0 through b, giving b + 1 choices. The choices are independent, so multiply: (a + 1)(b + 1).

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 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.

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}.

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.

First count all shortest routes by choosing the positions of the rightward steps. Then count forbidden routes through the marked point. Their number is the product of the route count to that point and the route count from it to the finish. Subtract the forbidden routes from the total.

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.

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.

First calculate the area of the complete outer rectangle by multiplying its side lengths. Then calculate the area of the cut-out rectangle. Subtract the cut-out area from the complete area.

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.

Adding A, B and C counts students in two laboratories twice and those in three laboratories three times. Subtract twice the three pairwise intersections. Students in all three have then been removed too many times, so add the triple intersection three times. Use A + B + C − 2(AB + AC + BC) + 3ABC.

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.

The numerator is the number of shaded cells and the denominator is the total number of equal cells. Divide both numbers by their greatest common divisor.

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.

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.

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.

When travelling once around any convex polygon, its exterior angles total 360°. A regular polygon has equal exterior angles, so divide 360° by the size of one exterior angle.

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.

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.